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For this Question: Given a number, you task is to reverse its digits. If the reversed number contains 0s in the beginning, you must remove them as well.First line consists of a single integer denoting n.
Constraint:-
1<=size of n<=100Output is a single line containing reversed number n.Sample Input
123445
Sample Output
544321
Sample Input
16724368
Sample Output
86342761, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str=br.readLine();
int i=str.length()-1;
if(i==0){
int number=Integer.parseInt(str);
System.out.println(number);
}else{
while(str.charAt(i)=='0'){
i--;
}
for(int j=i;j>=0;j--){
System.out.print(str.charAt(j));
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number, you task is to reverse its digits. If the reversed number contains 0s in the beginning, you must remove them as well.First line consists of a single integer denoting n.
Constraint:-
1<=size of n<=100Output is a single line containing reversed number n.Sample Input
123445
Sample Output
544321
Sample Input
16724368
Sample Output
86342761, I have written this Solution Code: n=int(input())
def reverse(n):
return int(str(n)[::-1])
print(reverse(n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number, you task is to reverse its digits. If the reversed number contains 0s in the beginning, you must remove them as well.First line consists of a single integer denoting n.
Constraint:-
1<=size of n<=100Output is a single line containing reversed number n.Sample Input
123445
Sample Output
544321
Sample Input
16724368
Sample Output
86342761, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main()
{
string s;
cin>>s;
reverse(s.begin(),s.end());
int I;
for( I=0;I<s.length();I++){
if(s[I]!='0'){break;}
}
if(I==s.length()){cout<<0;return 0;}
for(int j=I;j<s.length();j++){
cout<<s[j];}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, find the first non-repeating character in the string and return its index. If it does not exist, return -1.First line of the input contains the string s.
<b>Constraints</b>
1<= s. length <= 100000Print the index of the first non- repeating character in a stringInput
s = "newtonschool"
Output
1
Explanation
"e" is the first non- repeating character in a string, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine();
System.out.print(nonRepeatChar(s));
}
static int nonRepeatChar(String s){
char count[] = new char[256];
for(int i=0; i< s.length(); i++){
count[s.charAt(i)]++;
}
for (int i=0; i<s.length(); i++) {
if (count[s.charAt(i)]==1){
return i;
}
}
return -1;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, find the first non-repeating character in the string and return its index. If it does not exist, return -1.First line of the input contains the string s.
<b>Constraints</b>
1<= s. length <= 100000Print the index of the first non- repeating character in a stringInput
s = "newtonschool"
Output
1
Explanation
"e" is the first non- repeating character in a string, I have written this Solution Code: from collections import defaultdict
s=input()
d=defaultdict(int)
for i in s:
d[i]+=1
ans=-1
for i in range(len(s)):
if(d[s[i]]==1):
ans=i
break
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, find the first non-repeating character in the string and return its index. If it does not exist, return -1.First line of the input contains the string s.
<b>Constraints</b>
1<= s. length <= 100000Print the index of the first non- repeating character in a stringInput
s = "newtonschool"
Output
1
Explanation
"e" is the first non- repeating character in a string, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-01-10 12:51:16
**/
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
int firstUniqChar(string s) {
map<char, int> charCount;
int len = s.length();
for (int i = 0; i < len; i++) {
charCount[s[i]]++;
}
for (int i = 0; i < len; i++) {
if (charCount[s[i]] == 1)
return i;
}
return -1;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
string str;
cin>>str;
cout<<firstUniqChar(str)<<"\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a <b>BST</b> and some keys, the task is to insert the keys in the given BST. Duplicates are not inserted. (If a test case contains duplicate keys, you need to consider the first occurrence and ignore duplicates).<b>User Task:</b>
Since this will be a functional problem. You don't have to take input. You just have to complete the function <b>insertInBST()</b> that takes "root" node and value to be inserted as parameter. The printing is done by the driver code.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^3
1 <= node values <= 10^4
<b>Sum of "N" over all testcases does not exceed 10^5</b>Return the node of BST after insertion.Input:
2
3
2 1 3
4
8
2 1 3 N N N 6 4
1
Output:
1 2 3 4
1 2 3 4 6
Explanation:
Testcase 1: After inserting the node 4 the tree will be
2
/ \
1 3
\
4
Inorder traversal will be 1 2 3 4.
Testcase 2: After inserting the node 1 the tree will be
2
/ \
1 3
/ \ / \
N N N 6
/
4
Inorder traversal of the above tree will be 1 2 3 4 6., I have written this Solution Code:
static Node insertInBST(Node root,int key)
{
if(root == null) return new Node(key);
if(key < root.data)
root.left = insertInBST(root.left,key);
else if(key > root.data)
root.right = insertInBST(root.right,key);
return root;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have a box of paint. The paint can be used to cover area of 1 square unit. You have N magic spells with magic value from 1 to N. You can use each spell at max once.
When you use a spell on the box of paint, the quantity of paint in the box gets multiplied by the magic value of that spell. For example, if currently the paint in the box can cover A square units and a spell of B magic value is cast on the box, then after the spell the paint in the box can be used to cover A*B square units of area.
You want to paint a square (completely filled) of maximum area with integer side length with the paint such that after painting the square, the paint in the box gets empty. You can apply any magic spells you want in any order before the painting starts and not during or after the painting.
Find the area of maximum side square you can paint. As the answer can be huge find the area modulo 1000000007.The first and the only line of input contains a single integer N.
Constraints:
1 <= N <= 1000000Print the area of maximum side square you can paint modulo 1000000007.Sample Input 1
3
Sample Output 1
1
Explanation: We apply no magic spell.
Sample Input 2
4
Sample Output 2
4
Explanation: We apply magic spell number 4., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
int si[1000001];
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int ans=1;
int mo=1000000007;
for(int i=2;i<=n;++i){
if(si[i]==0){
int x=i;
int c=0;
while(x<=n){
c+=n/x;
x*=i;
}
if(c%2==0){
ans=(ans*i)%mo;
}
int j=i;
while(j<=n){
si[j]=1;
j+=i;
}
}
else{
ans=(ans*i)%mo;
}
}
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n houses evenly lined up on the street, and each house is beautifully painted.
Given an array of house colors, find the maximum distance between two houses with different colors. The distance between two houses is the absolute difference between their indices.
For example, if the ith house and the jth house have different colors, the distance between them is abs(i- j).
Constraints:
1)Lenght of the list should be greater than 2 and less than 100
2)List elements should be in the range of 0 to 100
3)List should not have all the same elements (eg: 1 1 1)
If the above constraints are not met, then Print("Invalid input")A List of Integers separated by space, which represents the houses with different colors.The output should be an integer representing the maximum distance between two houses with different colors.Sample1:
Input:
1 1 1 6 1 1 1
Output:
3
Explanation:
In the above sample, consider color 1 as a blue house, and color 6 as a red house.
The furthest two houses with different colors are
House[0] has color 1(blue house), and House[3] has color 6(red house).
The distance between them is abs(0 - 3) = 3.
Sample2:
Input:
1 1 1
Output:
Invalid input
Sample3:
Input:
101 102 103
Output:
Invalid input
Sample4:
Input:
1
Output:
Invalid input , I have written this Solution Code: def maxDistance(A):
if len(A) < 2 or len(A) > 100: # edge case: list length outside range
print("Invalid input")
return
elif any(x < 0 or x > 100 for x in A): # edge case: elements outside range
print("Invalid input")
return
elif all(x == A[0] for x in A): # edge case: list with same elements
print("Invalid input")
return
res = 0
for i, x in enumerate(A):
if x != A[0]:
res = max(res, i)
if x != A[-1]:
res = max(res, len(A) - 1 - i)
print(res)
A = [int(item) for item in input("").split()]
maxDistance(A), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b your task is to print the summation of these two numbersInput contains a single line containing two space separated integers a and b.
Constraints:-
1 <= a, b <= 10<sup>20000</sup>Print the sum of a and b.Sample Input:-
3 8
Sample Output:-
11
Sample Input:-
15 1
Sample Output:-
16, I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
import java.math.BigInteger;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
BigInteger sum;
String ip1 = sc.next();
String ip2 = sc.next();
BigInteger a = new BigInteger(ip1);
BigInteger b = new BigInteger(ip2);
sum = a.add(b);
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b your task is to print the summation of these two numbersInput contains a single line containing two space separated integers a and b.
Constraints:-
1 <= a, b <= 10<sup>20000</sup>Print the sum of a and b.Sample Input:-
3 8
Sample Output:-
11
Sample Input:-
15 1
Sample Output:-
16, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-02-03 02:46:30
**/
#include <bits/stdc++.h>
#define NX 105
#define MX 3350
using namespace std;
const int mod = 998244353;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
typedef long long INT;
const int pb = 10;
const int base_digits = 9;
const int base = 1000000000;
const int DIV = 100000;
struct bigint {
vector<int> a;
int sign;
bigint() : sign(1) {}
bigint(INT v) { *this = v; }
bigint(const string &s) { read(s); }
void operator=(const bigint &v) { sign = v.sign, a = v.a; }
void operator=(INT v) {
sign = 1;
if (v < 0) sign = -1, v = -v;
for (; v > 0; v = v / base) a.push_back(v % base);
}
bigint operator+(const bigint &v) const {
if (sign == v.sign) {
bigint res = v;
for (int i = 0, carry = 0; i < (int)max(a.size(), v.a.size()) ||
carry;
i++) {
if (i == (int)res.a.size()) res.a.push_back(0);
res.a[i] += carry + (i < (int)a.size() ? a[i] : 0);
carry = res.a[i] >= base;
if (carry) res.a[i] -= base;
}
return res;
}
return *this - (-v);
}
bigint operator-(const bigint &v) const {
if (sign == v.sign) {
if (abs() >= v.abs()) {
bigint res = *this;
for (int i = 0, carry = 0; i < (int)v.a.size() || carry; i++) {
res.a[i] -= carry + (i < (int)v.a.size() ? v.a[i] : 0);
carry = res.a[i] < 0;
if (carry) res.a[i] += base;
}
res.trim();
return res;
}
return -(v - *this);
}
return *this + (-v);
}
void operator*=(int v) {
if (v < 0) sign = -sign, v = -v;
for (int i = 0, carry = 0; i < (int)a.size() || carry; i++) {
if (i == (int)a.size()) a.push_back(0);
INT cur = a[i] * (INT)v + carry;
carry = (int)(cur / base);
a[i] = (int)(cur % base);
}
trim();
}
bigint operator*(int v) const {
bigint res = *this;
res *= v;
return res;
}
friend pair<bigint, bigint> DIVmod(const bigint &a1, const bigint &b1) {
int norm = base / (b1.a.back() + 1);
bigint a = a1.abs() * norm;
bigint b = b1.abs() * norm;
bigint q, r;
q.a.resize(a.a.size());
for (int i = a.a.size() - 1; i >= 0; i--) {
r *= base;
r += a.a[i];
int s1 = r.a.size() <= b.a.size() ? 0 : r.a[b.a.size()];
int s2 = r.a.size() <= b.a.size() - 1 ? 0 : r.a[b.a.size() - 1];
int d = ((INT)base * s1 + s2) / b.a.back();
r -= b * d;
while (r < 0) r += b, --d;
q.a[i] = d;
}
q.sign = a1.sign * b1.sign;
r.sign = a1.sign;
q.trim();
r.trim();
return make_pair(q, r / norm);
}
bigint operator/(const bigint &v) const { return DIVmod(*this, v).first; }
bigint operator%(const bigint &v) const { return DIVmod(*this, v).second; }
void operator/=(int v) {
if (v < 0) sign = -sign, v = -v;
for (int i = (int)a.size() - 1, rem = 0; i >= 0; i--) {
INT cur = a[i] + rem * (INT)base;
a[i] = (int)(cur / v);
rem = (int)(cur % v);
}
trim();
}
bigint operator/(int v) const {
bigint res = *this;
res /= v;
return res;
}
int operator%(int v) const {
if (v < 0) v = -v;
int m = 0;
for (int i = a.size() - 1; i >= 0; --i)
m = (a[i] + m * (INT)base) % v;
return m * sign;
}
void operator+=(const bigint &v) { *this = *this + v; }
void operator-=(const bigint &v) { *this = *this - v; }
void operator*=(const bigint &v) { *this = *this * v; }
void operator/=(const bigint &v) { *this = *this / v; }
bool operator<(const bigint &v) const {
if (sign != v.sign) return sign < v.sign;
if (a.size() != v.a.size()) return a.size() * sign < v.a.size() * v.sign;
for (int i = a.size() - 1; i >= 0; i--)
if (a[i] != v.a[i]) return a[i] * sign < v.a[i] * sign;
return false;
}
bool operator>(const bigint &v) const { return v < *this; }
bool operator<=(const bigint &v) const { return !(v < *this); }
bool operator>=(const bigint &v) const { return !(*this < v); }
bool operator==(const bigint &v) const { return !(*this < v) && !(v < *this); }
bool operator!=(const bigint &v) const { return *this < v || v < *this; }
void trim() {
while (!a.empty() && !a.back()) a.pop_back();
if (a.empty()) sign = 1;
}
bool isZero() const { return a.empty() || (a.size() == 1 && !a[0]); }
bigint operator-() const {
bigint res = *this;
res.sign = -sign;
return res;
}
bigint abs() const {
bigint res = *this;
res.sign *= res.sign;
return res;
}
INT longValue() const {
INT res = 0;
for (int i = a.size() - 1; i >= 0; i--) res = res * base + a[i];
return res * sign;
}
friend bigint gcd(const bigint &a, const bigint &b) { return b.isZero() ? a : gcd(b, a % b); }
friend bigint lcm(const bigint &a, const bigint &b) { return a / gcd(a, b) * b; }
void read(const string &s) {
sign = 1;
a.clear();
int pos = 0;
while (pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
if (s[pos] == '-') sign = -sign;
pos++;
}
for (int i = s.size() - 1; i >= pos; i -= base_digits) {
int x = 0;
for (int j = max(pos, i - base_digits + 1); j <= i; j++) x = x *
pb +
s[j] - '0';
a.push_back(x);
}
trim();
}
friend istream &operator>>(istream &stream, bigint &v) {
string s;
stream >> s;
v.read(s);
return stream;
}
friend ostream &operator<<(ostream &stream, const bigint &v) {
if (v.sign == -1) stream << '-';
stream << (v.a.empty() ? 0 : v.a.back());
for (int i = (int)v.a.size() - 2; i >= 0; --i) stream << setw(base_digits) << setfill('0') << v.a[i];
return stream;
}
static vector<int> convert_base(const vector<int> &a, int old_digits, int new_digits) {
vector<INT> p(max(old_digits, new_digits) + 1);
p[0] = 1;
for (int i = 1; i < (int)p.size(); i++) p[i] = p[i - 1] * pb;
vector<int> res;
INT cur = 0;
int cur_digits = 0;
for (int i = 0; i < (int)a.size(); i++) {
cur += a[i] * p[cur_digits];
cur_digits += old_digits;
while (cur_digits >= new_digits) {
res.push_back(int(cur % p[new_digits]));
cur /= p[new_digits];
cur_digits -= new_digits;
}
}
res.push_back((int)cur);
while (!res.empty() && !res.back()) res.pop_back();
return res;
}
typedef vector<INT> vll;
static vll karatsubaMultiply(const vll &a, const vll &b) {
int n = a.size();
vll res(n + n);
if (n <= 32) {
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
res[i + j] += a[i] * b[j];
return res;
}
int k = n >> 1;
vll a1(a.begin(), a.begin() + k);
vll a2(a.begin() + k, a.end());
vll b1(b.begin(), b.begin() + k);
vll b2(b.begin() + k, b.end());
vll a1b1 = karatsubaMultiply(a1, b1);
vll a2b2 = karatsubaMultiply(a2, b2);
for (int i = 0; i < k; i++) a2[i] += a1[i];
for (int i = 0; i < k; i++) b2[i] += b1[i];
vll r = karatsubaMultiply(a2, b2);
for (int i = 0; i < (int)a1b1.size(); i++) r[i] -= a1b1[i];
for (int i = 0; i < (int)a2b2.size(); i++) r[i] -= a2b2[i];
for (int i = 0; i < (int)r.size(); i++) res[i + k] += r[i];
for (int i = 0; i < (int)a1b1.size(); i++) res[i] += a1b1[i];
for (int i = 0; i < (int)a2b2.size(); i++) res[i + n] += a2b2[i];
return res;
}
bigint operator*(const bigint &v) const {
vector<int> a5 = convert_base(this->a, base_digits, 5);
vector<int> b5 = convert_base(v.a, base_digits, 5);
vll a(a5.begin(), a5.end());
vll b(b5.begin(), b5.end());
while (a.size() < b.size()) a.push_back(0);
while (b.size() < a.size()) b.push_back(0);
while (a.size() & (a.size() - 1)) a.push_back(0), b.push_back(0);
vll c = karatsubaMultiply(a, b);
bigint res;
res.sign = sign * v.sign;
for (int i = 0, carry = 0; i < (int)c.size(); i++) {
INT cur = c[i] + carry;
res.a.push_back((int)(cur % DIV));
carry = (int)(cur / DIV);
}
res.a = convert_base(res.a, 5, base_digits);
res.trim();
return res;
}
inline bool isOdd() { return a[0] & 1; }
};
int main() {
bigint n, m;
cin >> n >> m;
cout << n + m << "\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b your task is to print the summation of these two numbersInput contains a single line containing two space separated integers a and b.
Constraints:-
1 <= a, b <= 10<sup>20000</sup>Print the sum of a and b.Sample Input:-
3 8
Sample Output:-
11
Sample Input:-
15 1
Sample Output:-
16, I have written this Solution Code: n,m = map(int,input().split())
print(n+m) , In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
InputStreamReader is = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(is);
int n=Integer.parseInt(br.readLine().trim());
if(n%3==1)
System.out.println("Dead");
else
System.out.println("Alive");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: n = int(input().strip())
if n%3 == 1:
print("Dead")
else:
print("Alive"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
if(n%3==1)
cout<<"Dead";
else
cout<<"Alive";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array of N integers a[], and Q queries. For each query, you will be given one element K your task is to print the maximum element from the array which is less than or equal to the given element(Floor), and the minimum element from the array which is greater than or equal to the given element(Ceil).<b>In case of Java only</b>
<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>floorAndCeil()</b> that takes the array a[], integer N and integer k as arguments.
<b>Custom Input</b>
The first line of input contains a single integer N. The Second line of input contains N space-separated integers depicting the values of the array. The third line of input contains a single integer Q. The next Q line of input contains a single integer the value of K.
Constraints:-
1 <= N <= 100000
1 <= K, Arr[i] <= 1000000000000
1 <= Q <= 10000In a new line Print two space-separated integers depicting the values of Floor and Ceil of the given number. If the floor or ceil of the element does not exist print -1.Sample Input:-
5
2 5 6 11 15
5
2
4
8
1
16
Sample Output:-
2 2
2 5
6 11
-1 2
15 -1, I have written this Solution Code: n = int(input())
arr = list(map(int,input().split()))
t = int(input())
def findFloor(arr, l, h, x, res):
if(l<=h):
m = l+(h-l)//2
if(arr[m] == x):
return m
if(arr[m] > x):
return findFloor(arr, l, m-1, x, res)
if(arr[m] < x):
res = m
return findFloor(arr, m+1, h, x, res)
else:
return res
def findCeil(arr, l, h, x, res):
if(l<=h):
m = l+(h-l)//2
if(arr[m] == x):
return m
if(arr[m] < x):
return findCeil(arr, m+1, h, x, res)
res = m
return findCeil(arr, l, m-1, x, res)
else:
return res
for _ in range(t):
x = int(input())
f = findFloor(arr, 0, n-1, x, -1)
c = findCeil(arr, 0, n-1, x, -1)
floor = -1
ceil = -1
if(f!=-1):
floor = arr[f]
if(c!=-1):
ceil = arr[c]
print(floor,ceil), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array of N integers a[], and Q queries. For each query, you will be given one element K your task is to print the maximum element from the array which is less than or equal to the given element(Floor), and the minimum element from the array which is greater than or equal to the given element(Ceil).<b>In case of Java only</b>
<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>floorAndCeil()</b> that takes the array a[], integer N and integer k as arguments.
<b>Custom Input</b>
The first line of input contains a single integer N. The Second line of input contains N space-separated integers depicting the values of the array. The third line of input contains a single integer Q. The next Q line of input contains a single integer the value of K.
Constraints:-
1 <= N <= 100000
1 <= K, Arr[i] <= 1000000000000
1 <= Q <= 10000In a new line Print two space-separated integers depicting the values of Floor and Ceil of the given number. If the floor or ceil of the element does not exist print -1.Sample Input:-
5
2 5 6 11 15
5
2
4
8
1
16
Sample Output:-
2 2
2 5
6 11
-1 2
15 -1, I have written this Solution Code: static void floorAndCeil(int a[], int n, int x){
int it = lower(a,n,x);
if(it==0){
if(a[it]==x){
System.out.println(x+" "+x);
}
else{
System.out.println("-1 "+a[it]);
}
}
else if (it==n){
it--;
System.out.println(a[it]+" -1");
}
else{
if(a[it]==x){
System.out.println(x+" "+x);
}
else{
it--;
System.out.println(a[it]+" "+a[it+1]);
}
}
}
static int lower(int a[], int n,int k){
int l=0;
int h=n-1;
int m;
int ans=n;
while(l<=h){
m=l+h;
m/=2;
if(a[m]<k){
l=m+1;
}
else{
h=m-1;
ans=m;
}
}
return ans;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array of N integers a[], and Q queries. For each query, you will be given one element K your task is to print the maximum element from the array which is less than or equal to the given element(Floor), and the minimum element from the array which is greater than or equal to the given element(Ceil).<b>In case of Java only</b>
<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>floorAndCeil()</b> that takes the array a[], integer N and integer k as arguments.
<b>Custom Input</b>
The first line of input contains a single integer N. The Second line of input contains N space-separated integers depicting the values of the array. The third line of input contains a single integer Q. The next Q line of input contains a single integer the value of K.
Constraints:-
1 <= N <= 100000
1 <= K, Arr[i] <= 1000000000000
1 <= Q <= 10000In a new line Print two space-separated integers depicting the values of Floor and Ceil of the given number. If the floor or ceil of the element does not exist print -1.Sample Input:-
5
2 5 6 11 15
5
2
4
8
1
16
Sample Output:-
2 2
2 5
6 11
-1 2
15 -1, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
signed main(){
int n;
cin>>n;
vector<int> v;
int x;
FOR(i,n){
cin>>x;
v.EB(x);}
int q;
cin>>q;
while(q--){
cin>>x;
auto it = lower_bound(v.begin(),v.end(),x);
if(it==v.begin()){
if(*it==x){
cout<<x<<" "<<x;
}
else{
cout<<-1<<" "<<*it;
}
}
else if (it==v.end()){
it--;
cout<<*it<<" -1";
}
else{
if(*it==x){
cout<<x<<" "<<x;
}
else{
it--;
cout<<*it<<" ";
it++;
cout<<*it;}
}
END;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in line numbered 1 to N from left to right. Each person wants to know the height of the person to left of him having height less than him. If there are multiple such people he wants to know the height of the person closest to him.
If there is no such person report -1.The first line of input contains N, the size of the array.
The second line of input contains N space-separated integers.
Constraints
2 ≤ N ≤ 100000
0 ≤ Arr[i] ≤ 1000000000 (Height can be zero wierd people :p )The output should contain N space separated integers, the ith integer should be the height reported to ith person (-1 if no person to the left is found whose height is less).Sample Input 1
5
1 2 3 4 5
Sample Output 1
-1 1 2 3 4
Sample Input 2
2
1 1
Sample Output 2
-1 -1, I have written this Solution Code: import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Stack;
import java.util.StringTokenizer;
public class Main {
final static long MOD = 1000000007;
public static void main(String args[]) {
FastScanner fs = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int n = fs.nextInt();
int[] a = fs.nextIntArray(n);
Stack<Integer> stck = new Stack<>();
for (int i = 0; i < n; i++) {
while (!stck.isEmpty() && stck.peek() >= a[i]) {
stck.pop();
}
if (stck.isEmpty()) {
out.print("-1 ");
stck.push(a[i]);
} else {
out.print(stck.peek() + " ");
stck.push(a[i]);
}
}
out.flush();
out.close();
}
static class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner() {
try {
br = new BufferedReader(new InputStreamReader(System.in));
st = new StringTokenizer("");
} catch (Exception e) {
e.printStackTrace();
}
}
public String next() {
if (st.hasMoreTokens())
return st.nextToken();
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
e.printStackTrace();
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public String nextLine() {
String line = "";
try {
line = br.readLine();
} catch (Exception e) {
e.printStackTrace();
}
return line;
}
public char nextChar() {
return next().charAt(0);
}
public Integer[] nextIntegerArray(int n) {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public int[] nextIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public long[] nextLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public char[] nextCharArray() {
return nextLine().toCharArray();
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in line numbered 1 to N from left to right. Each person wants to know the height of the person to left of him having height less than him. If there are multiple such people he wants to know the height of the person closest to him.
If there is no such person report -1.The first line of input contains N, the size of the array.
The second line of input contains N space-separated integers.
Constraints
2 ≤ N ≤ 100000
0 ≤ Arr[i] ≤ 1000000000 (Height can be zero wierd people :p )The output should contain N space separated integers, the ith integer should be the height reported to ith person (-1 if no person to the left is found whose height is less).Sample Input 1
5
1 2 3 4 5
Sample Output 1
-1 1 2 3 4
Sample Input 2
2
1 1
Sample Output 2
-1 -1, I have written this Solution Code: import sys
n=int(input())
myList = [int(x) for x in sys.stdin.readline().rstrip().split(' ')]
outputList = []
outputList.append(-1);
for i in range(1,len(myList),1):
flag=False
for j in range(i-1,-1,-1):
if myList[j]<myList[i]:
outputList.append(myList[j])
flag=True
break
if flag==False:
outputList.append(-1)
print(*outputList), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in line numbered 1 to N from left to right. Each person wants to know the height of the person to left of him having height less than him. If there are multiple such people he wants to know the height of the person closest to him.
If there is no such person report -1.The first line of input contains N, the size of the array.
The second line of input contains N space-separated integers.
Constraints
2 ≤ N ≤ 100000
0 ≤ Arr[i] ≤ 1000000000 (Height can be zero wierd people :p )The output should contain N space separated integers, the ith integer should be the height reported to ith person (-1 if no person to the left is found whose height is less).Sample Input 1
5
1 2 3 4 5
Sample Output 1
-1 1 2 3 4
Sample Input 2
2
1 1
Sample Output 2
-1 -1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
stack <long long > s;
long long a;
for(int i=0;i<n;i++){
cin>>a;
while(!(s.empty())){
if(s.top()<a){break;}
s.pop();
}
if(s.empty()){cout<<-1<<" ";}
else{cout<<s.top()<<" ";}
s.push(a);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new magic trick on you, The magic trick goes in 6 steps:-
1. Think of a number X(don't tell Sara)
2. Add A(Given by Sara) to it.
3. Double the sum in your mind.
4. Add an even number B(Given by Sara) to it.
5. Half the amount
6. Subtract the initial number which you had taken from the sum
After this Sara will tell the resulting amount without knowing the initial number, can you print the result for her?
<b>Note: B is always even. </b>You don't have to worry about the input, you just have to complete the function <b>magicTrick()</b>
<b>Constraints</b>:-
1 <= A, B <= 1000Print the resulting amountSample Input:-
3 4
Sample Output:-
5
Sample Input:-
8 4
Sample Output:-
10, I have written this Solution Code: void magicTrick(int a, int b){
cout<<a+b/2;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new magic trick on you, The magic trick goes in 6 steps:-
1. Think of a number X(don't tell Sara)
2. Add A(Given by Sara) to it.
3. Double the sum in your mind.
4. Add an even number B(Given by Sara) to it.
5. Half the amount
6. Subtract the initial number which you had taken from the sum
After this Sara will tell the resulting amount without knowing the initial number, can you print the result for her?
<b>Note: B is always even. </b>You don't have to worry about the input, you just have to complete the function <b>magicTrick()</b>
<b>Constraints</b>:-
1 <= A, B <= 1000Print the resulting amountSample Input:-
3 4
Sample Output:-
5
Sample Input:-
8 4
Sample Output:-
10, I have written this Solution Code: static void magicTrick(int a, int b){
System.out.println(a + b/2);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new magic trick on you, The magic trick goes in 6 steps:-
1. Think of a number X(don't tell Sara)
2. Add A(Given by Sara) to it.
3. Double the sum in your mind.
4. Add an even number B(Given by Sara) to it.
5. Half the amount
6. Subtract the initial number which you had taken from the sum
After this Sara will tell the resulting amount without knowing the initial number, can you print the result for her?
<b>Note: B is always even. </b>You don't have to worry about the input, you just have to complete the function <b>magicTrick()</b>
<b>Constraints</b>:-
1 <= A, B <= 1000Print the resulting amountSample Input:-
3 4
Sample Output:-
5
Sample Input:-
8 4
Sample Output:-
10, I have written this Solution Code: A,B = map(int,input().split(' '))
C = A+B//2
print(C)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length, breadth, and height of a cuboid. Your task is to calculate its Perimeter.
Note:- Formula for the perimeter of the cuboid is 4(Length + Breadth + height)<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Perimeter()</b> that takes integers L, B, and H as parameters.
Constraints:-
1 <= L, B, H <= 100Return the length of the Cuboid.Sample Input:-
L = 3, B = 5, H = 1
Sample Output:-
36
Sample Input:-
L = 1, B = 1, H = 1
Sample Output:-
12, I have written this Solution Code:
L,B,H=input().split()
a=4*(int(L)+int(B)+int(H))
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length, breadth, and height of a cuboid. Your task is to calculate its Perimeter.
Note:- Formula for the perimeter of the cuboid is 4(Length + Breadth + height)<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Perimeter()</b> that takes integers L, B, and H as parameters.
Constraints:-
1 <= L, B, H <= 100Return the length of the Cuboid.Sample Input:-
L = 3, B = 5, H = 1
Sample Output:-
36
Sample Input:-
L = 1, B = 1, H = 1
Sample Output:-
12, I have written this Solution Code: static int Perimeter(int L, int B, int H){
return 4*(L+B+H);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton wants to play Holi with his friends. So he wants to buy three colors for which he needs some money from his father. For buying all three colors he needs at least 75 rupees. For buying 2 colors he needs at least 50 rupees. For buying at least 1 color he needs 25 rupees.
You have to check how many colors did Newton buy. If he buys all three colors print "Newton is very happy", if he buys only two colors print "Newton is happy", if he buys only one color print "Newton is sad" otherwise print "Newton won't play Holi".The first line contains the integer n denoting the amount of money Newton receives from his father.
<b>Constraints</b>
0 ≤ n ≤ 1000Print the output based on the suitable condition.Sample input:
60
Sample output:
Newton is happy
<b>Explanation</b>
Newton will get only two colors as he gets only 60 ruppes from his father., I have written this Solution Code: import java.util.Scanner;
class Solution {
public void num(int money){
if (money < 25) {
System.out.println("Newton won't play Holi");
}
else if (money < 50) {
System.out.println("Newton is sad");
}
else if (money < 75) {
System.out.println("Newton is happy");
}
else {
System.out.println("Newton is very happy");
}
}
}
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int money= scanner.nextInt();
Solution obj = new Solution();
obj.num(money);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is fond of ice cream initially she had N ice creams with her. If Sara eats exactly half of the ice cream she has in a day and the remaining icecreams get tripled each night. How many ice creams does Sara have at the end of D-days?
<b>Note:- </b>
Take the floor value while dividing by 2.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Icecreams()</b> that takes integers N and D as parameters.
<b>Constraints:-</b>
1 <= N <= 100
1 <= D <= 20Print a single integer denoting the number of ice creams at the end of D days.Sample Input 1:-
5 1
Sample Output 1:-
9
</b>Explanation:-</b>
Sara will eat 2 ice creams and the remaining 3 will be tripled i. e 9.
Sample Input 2:-
5 3
Sample Output 2:-
24
<b>Explanation:-</b>
Day 1:- N=5, eaten = 2, rem = 3 => remaining = 3*3 = 9
Day 2:- N=9, eaten = 4, rem = 5 => remaining = 5*3 = 15
Day 3:- N=15, eaten = 7, rem = 8 => remaining = 8*3 = 24, I have written this Solution Code: static void Icecreams (int N, int D){
int x=N;
while(D-->0){
x-=x/2;
x*=3;
}
System.out.println(x);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is fond of ice cream initially she had N ice creams with her. If Sara eats exactly half of the ice cream she has in a day and the remaining icecreams get tripled each night. How many ice creams does Sara have at the end of D-days?
<b>Note:- </b>
Take the floor value while dividing by 2.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Icecreams()</b> that takes integers N and D as parameters.
<b>Constraints:-</b>
1 <= N <= 100
1 <= D <= 20Print a single integer denoting the number of ice creams at the end of D days.Sample Input 1:-
5 1
Sample Output 1:-
9
</b>Explanation:-</b>
Sara will eat 2 ice creams and the remaining 3 will be tripled i. e 9.
Sample Input 2:-
5 3
Sample Output 2:-
24
<b>Explanation:-</b>
Day 1:- N=5, eaten = 2, rem = 3 => remaining = 3*3 = 9
Day 2:- N=9, eaten = 4, rem = 5 => remaining = 5*3 = 15
Day 3:- N=15, eaten = 7, rem = 8 => remaining = 8*3 = 24, I have written this Solution Code: void Icecreams (int N, int D){
int x=N;
while(D--){
x-=x/2;
x*=3;
}
cout << x;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is fond of ice cream initially she had N ice creams with her. If Sara eats exactly half of the ice cream she has in a day and the remaining icecreams get tripled each night. How many ice creams does Sara have at the end of D-days?
<b>Note:- </b>
Take the floor value while dividing by 2.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Icecreams()</b> that takes integers N and D as parameters.
<b>Constraints:-</b>
1 <= N <= 100
1 <= D <= 20Print a single integer denoting the number of ice creams at the end of D days.Sample Input 1:-
5 1
Sample Output 1:-
9
</b>Explanation:-</b>
Sara will eat 2 ice creams and the remaining 3 will be tripled i. e 9.
Sample Input 2:-
5 3
Sample Output 2:-
24
<b>Explanation:-</b>
Day 1:- N=5, eaten = 2, rem = 3 => remaining = 3*3 = 9
Day 2:- N=9, eaten = 4, rem = 5 => remaining = 5*3 = 15
Day 3:- N=15, eaten = 7, rem = 8 => remaining = 8*3 = 24, I have written this Solution Code: void Icecreams (int N, int D){
int x=N;
while(D--){
x-=x/2;
x*=3;
}
printf("%d", x);
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is fond of ice cream initially she had N ice creams with her. If Sara eats exactly half of the ice cream she has in a day and the remaining icecreams get tripled each night. How many ice creams does Sara have at the end of D-days?
<b>Note:- </b>
Take the floor value while dividing by 2.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Icecreams()</b> that takes integers N and D as parameters.
<b>Constraints:-</b>
1 <= N <= 100
1 <= D <= 20Print a single integer denoting the number of ice creams at the end of D days.Sample Input 1:-
5 1
Sample Output 1:-
9
</b>Explanation:-</b>
Sara will eat 2 ice creams and the remaining 3 will be tripled i. e 9.
Sample Input 2:-
5 3
Sample Output 2:-
24
<b>Explanation:-</b>
Day 1:- N=5, eaten = 2, rem = 3 => remaining = 3*3 = 9
Day 2:- N=9, eaten = 4, rem = 5 => remaining = 5*3 = 15
Day 3:- N=15, eaten = 7, rem = 8 => remaining = 8*3 = 24, I have written this Solution Code: def Icecreams(N,D):
ans = N
while D > 0:
ans = ans - ans//2
ans = ans*3
D = D-1
return ans
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two arrays - value and frequency both containing N elements.
There is also a third array C which is currently empty. Then you perform N insertion operation in the array. For ith operation you insert value[i] to the end of the array frequency[i] number of times.
Finally you have to tell the kth smallest element in the array C.First line of input contains N.
Second line contains N integers denoting array - value
Third line contains N integers denoting array - frequency
Fourth line contains single integer K.
Constraints
1 <= N, value[i], frequency[i] <= 100000
1 <= k <= frequency[1] + frequency[2] +frequency[3] +........ + frequency[N]
Output a single integer which is the kth smallest element of the array C.Sample input 1
5
1 2 3 4 5
1 1 1 2 2
3
Sample output 1
3
Explanation 1:
Array C constructed is 1 2 3 4 4 5 5
Third smallest element is 3
Sample input 2
3
2 1 3
3 3 2
2
sample output 2
1
Explanation 2:
Array C constructed is 2 2 2 1 1 1 3 3
Second smallest element is 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
StringTokenizer st = new StringTokenizer(br.readLine());
int[] val = new int[n];
for(int i=0; i<n; i++){
val[i] = Integer.parseInt(st.nextToken());
}
st = new StringTokenizer(br.readLine());
int[] freq = new int[n];
for(int i=0; i<n; i++){
freq[i] = Integer.parseInt(st.nextToken());
}
int k = Integer.parseInt(br.readLine());
for (int i=0; i<n; i++) {
for (int j=i+1; j<n; j++) {
if (val[j] < val[i]) {
int temp = val[i];
val[i] = val[j];
val[j] = temp;
int temp1 = freq[i];
freq[i] = freq[j];
freq[j] = temp1;
}
}
}
int element=0;
for(int i=0; i<n; i++){
for(int j=0; j<freq[i]; j++){
element++;
int value = val[i];
if(element==k){
System.out.print(value);
break;
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two arrays - value and frequency both containing N elements.
There is also a third array C which is currently empty. Then you perform N insertion operation in the array. For ith operation you insert value[i] to the end of the array frequency[i] number of times.
Finally you have to tell the kth smallest element in the array C.First line of input contains N.
Second line contains N integers denoting array - value
Third line contains N integers denoting array - frequency
Fourth line contains single integer K.
Constraints
1 <= N, value[i], frequency[i] <= 100000
1 <= k <= frequency[1] + frequency[2] +frequency[3] +........ + frequency[N]
Output a single integer which is the kth smallest element of the array C.Sample input 1
5
1 2 3 4 5
1 1 1 2 2
3
Sample output 1
3
Explanation 1:
Array C constructed is 1 2 3 4 4 5 5
Third smallest element is 3
Sample input 2
3
2 1 3
3 3 2
2
sample output 2
1
Explanation 2:
Array C constructed is 2 2 2 1 1 1 3 3
Second smallest element is 1, I have written this Solution Code: def myFun():
n = int(input())
arr1 = list(map(int,input().strip().split()))
arr2 = list(map(int,input().strip().split()))
k = int(input())
arr = []
for i in range(n):
arr.append((arr1[i], arr2[i]))
arr.sort()
c = 0
for i in arr:
k -= i[1]
if k <= 0:
print(i[0])
return
myFun()
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two arrays - value and frequency both containing N elements.
There is also a third array C which is currently empty. Then you perform N insertion operation in the array. For ith operation you insert value[i] to the end of the array frequency[i] number of times.
Finally you have to tell the kth smallest element in the array C.First line of input contains N.
Second line contains N integers denoting array - value
Third line contains N integers denoting array - frequency
Fourth line contains single integer K.
Constraints
1 <= N, value[i], frequency[i] <= 100000
1 <= k <= frequency[1] + frequency[2] +frequency[3] +........ + frequency[N]
Output a single integer which is the kth smallest element of the array C.Sample input 1
5
1 2 3 4 5
1 1 1 2 2
3
Sample output 1
3
Explanation 1:
Array C constructed is 1 2 3 4 4 5 5
Third smallest element is 3
Sample input 2
3
2 1 3
3 3 2
2
sample output 2
1
Explanation 2:
Array C constructed is 2 2 2 1 1 1 3 3
Second smallest element is 1, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define inf 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int N;ll K;
cin>>N;
int c=0;
pair<int, ll> A[N];
for(int i=0;i<N;++i){
cin >> A[i].first ;
}
for(int i=0;i<N;++i){
cin >> A[i].second ;
}
cin>>K;
sort(A, A+N);
for(int i=0;i<N;++i){
K -= A[i].second;
if(K <= 0){
cout << A[i].first << endl;;
break;
}
}
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given N nodes and M undirected edges. These nodes and edges represent some trees (graphs with no loops). Now, your task is to merge the trees into a single tree. In one operation of merging, two disconnected trees are merged using an edge from one tree to the other tree. The cost of this merging operation is <b>total number of nodes in the final merged tree</b>. Once merged the trees lose their original existence.
So, if there are T trees in the original state, you'll have to perform the merge operation T-1 times to merge all the trees in a single tree.
All you need to do is find the minimum cost of merging the trees (sum of costs of all T-1 merge operations). See sample for better understanding.The first line of input contains two integers N and M, the total number of nodes and the total number of edges.
The next M lines contain two integers U and V, representing an undirected edge between the two nodes.
Constraints
1 <= N <= 200000
0 <= M < N
1 <= U, V <= N
The given nodes and edges represent some trees (maybe 1)Output a single integer, the cost of merging these trees.Sample Input
7 3
1 2
3 4
4 5
Sample Output
13
Explanation:
The four trees are as follows:
1 --- 2 ; 3 --- 4 --- 5 ; 6 ; 7
We will first add an edge from 6 to 7. (Cost of merging = 2)
The three remaining trees are as follows:
1 --- 2 ; 3 --- 4 --- 5 ; 6 --- 7
We will add an edge from 2 to 6. (Cost of merging = 4)
The two remaining trees are as follows:
1 --- 2 --- 6 --- 7 ; 3 --- 4 --- 5
We will add an edge from 7 to 3. (Cost of merging = 7)
Total cost of merging = 2 + 4 + 7 = 13.
Sample Input
2 1
1 2
Sample Output
0, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int find(int []parent, int x){
if(parent[x] == x){
return x;
}
parent[x] = find(parent, parent[x]);
return parent[x];
}
static void union(int []parent, int x, int y){
int parentX = find(parent, x);
int parentY = find(parent, y);
parent[parentY] = parentX;
}
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] str = br.readLine().trim().split(" ");
int N = Integer.parseInt(str[0]);
int M = Integer.parseInt(str[1]);
int totalCost = 0;
int[] parent = new int[N+1];
for(int i=1; i<=N; i++){
parent[i] = i;
}
for (int i=0; i<M; i++) {
int x, y;
String[] str1 = br.readLine().trim().split(" ");
x = Integer.parseInt(str1[0]);
y = Integer.parseInt(str1[1]);
union(parent, x, y);
}
for(int i=1; i<=N; i++){
find(parent, i);
}
PriorityQueue<Integer> pQueue = new PriorityQueue<Integer>();
HashMap<Integer, Integer> counts = new HashMap<>();
for(int i=1; i<=N; i++){
counts.put(parent[i], counts.getOrDefault(parent[i], 0) + 1);
}
for (Map.Entry<Integer,Integer> entry : counts.entrySet()) {
pQueue.add(entry.getValue());
}
while(pQueue.size() > 1) {
int s1 = pQueue.poll();
int s2 = pQueue.poll();
totalCost += s1 + s2;
pQueue.add(s1 + s2);
}
System.out.println(totalCost);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given N nodes and M undirected edges. These nodes and edges represent some trees (graphs with no loops). Now, your task is to merge the trees into a single tree. In one operation of merging, two disconnected trees are merged using an edge from one tree to the other tree. The cost of this merging operation is <b>total number of nodes in the final merged tree</b>. Once merged the trees lose their original existence.
So, if there are T trees in the original state, you'll have to perform the merge operation T-1 times to merge all the trees in a single tree.
All you need to do is find the minimum cost of merging the trees (sum of costs of all T-1 merge operations). See sample for better understanding.The first line of input contains two integers N and M, the total number of nodes and the total number of edges.
The next M lines contain two integers U and V, representing an undirected edge between the two nodes.
Constraints
1 <= N <= 200000
0 <= M < N
1 <= U, V <= N
The given nodes and edges represent some trees (maybe 1)Output a single integer, the cost of merging these trees.Sample Input
7 3
1 2
3 4
4 5
Sample Output
13
Explanation:
The four trees are as follows:
1 --- 2 ; 3 --- 4 --- 5 ; 6 ; 7
We will first add an edge from 6 to 7. (Cost of merging = 2)
The three remaining trees are as follows:
1 --- 2 ; 3 --- 4 --- 5 ; 6 --- 7
We will add an edge from 2 to 6. (Cost of merging = 4)
The two remaining trees are as follows:
1 --- 2 --- 6 --- 7 ; 3 --- 4 --- 5
We will add an edge from 7 to 3. (Cost of merging = 7)
Total cost of merging = 2 + 4 + 7 = 13.
Sample Input
2 1
1 2
Sample Output
0, I have written this Solution Code: import heapq
def dfs(adjlist,vis,node):
global temparr
temparr.append(node)
vis[node] = True
for i in adjlist[node]:
if not vis[i]:
dfs(adjlist,vis,i)
ne=list(map(int,input().rstrip().split()))
n=ne[0]
e=ne[1]
adjlist = [[] for i in range(n+1)]
for i in range(e):
edge = list(map(int,input().rstrip().split()))
adjlist[edge[0]].append(edge[1])
adjlist[edge[1]].append(edge[0])
vis = [False for i in range(n+1)]
temparr=[]
mx = 0
arr=[]
for x in range(1,n+1):
if vis[x] == False:
temparr = []
dfs(adjlist,vis,x)
arr.append(len(temparr))
heapq.heapify(arr)
while len(arr)>=2:
a=heapq.heappop(arr)
b=heapq.heappop(arr)
mx+=a+b
heapq.heappush(arr,a+b)
print(mx), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given N nodes and M undirected edges. These nodes and edges represent some trees (graphs with no loops). Now, your task is to merge the trees into a single tree. In one operation of merging, two disconnected trees are merged using an edge from one tree to the other tree. The cost of this merging operation is <b>total number of nodes in the final merged tree</b>. Once merged the trees lose their original existence.
So, if there are T trees in the original state, you'll have to perform the merge operation T-1 times to merge all the trees in a single tree.
All you need to do is find the minimum cost of merging the trees (sum of costs of all T-1 merge operations). See sample for better understanding.The first line of input contains two integers N and M, the total number of nodes and the total number of edges.
The next M lines contain two integers U and V, representing an undirected edge between the two nodes.
Constraints
1 <= N <= 200000
0 <= M < N
1 <= U, V <= N
The given nodes and edges represent some trees (maybe 1)Output a single integer, the cost of merging these trees.Sample Input
7 3
1 2
3 4
4 5
Sample Output
13
Explanation:
The four trees are as follows:
1 --- 2 ; 3 --- 4 --- 5 ; 6 ; 7
We will first add an edge from 6 to 7. (Cost of merging = 2)
The three remaining trees are as follows:
1 --- 2 ; 3 --- 4 --- 5 ; 6 --- 7
We will add an edge from 2 to 6. (Cost of merging = 4)
The two remaining trees are as follows:
1 --- 2 --- 6 --- 7 ; 3 --- 4 --- 5
We will add an edge from 7 to 3. (Cost of merging = 7)
Total cost of merging = 2 + 4 + 7 = 13.
Sample Input
2 1
1 2
Sample Output
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
vector<int> adj[N];
priority_queue<int, vector<int>, greater<int> > pq;
int ct = 0;
bool vis[N];
void dfs(int u, int p){
ct++;
vis[u]=true;
for(int v: adj[u]){
if(v != p){
dfs(v, u);
}
}
}
void solve(){
int n, m; cin>>n>>m;
For(i, 0, m){
int u, v; cin>>u>>v;
adj[u].pb(v);
adj[v].pb(u);
}
For(i, 1, n+1){
if(!vis[i]){
ct = 0;
dfs(i, 0);
pq.push(ct);
}
}
int ans = 0;
while(sz(pq) > 1){
int c1 = pq.top();
pq.pop();
int c2 = pq.top();
pq.pop();
ans += (c1+c2);
pq.push(c1+c2);
}
assert(pq.top() == n);
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the first 2 terms A and B of an Arithmetic Series, tell the Nth term of the series.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>NthAP()</b> that takes the integer A, B, and N as a parameter.
<b>Constraints:</b>
-10<sup>3</sup> ≤ A ≤ 10<sup>3</sup>
-10<sup>3</sup> ≤ B ≤ 10<sup>3</sup>
1 ≤ N ≤ 10<sup>4</sup>Return the Nth term of AP series.Sample Input 1:
2 3 4
Sample Output 1:
5
Sample Input 2:
1 2 10
Sample output 2:
10, I have written this Solution Code: class Solution {
public static int NthAP(int a, int b, int n){
return a+(n-1)*(b-a);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
InputStreamReader is = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(is);
int n=Integer.parseInt(br.readLine().trim());
if(n%3==1)
System.out.println("Dead");
else
System.out.println("Alive");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: n = int(input().strip())
if n%3 == 1:
print("Dead")
else:
print("Alive"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
if(n%3==1)
cout<<"Dead";
else
cout<<"Alive";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Separate chaining technique in hashing allows us to use a linked list at each hash slot to handle the problem of collisions. That is, every slot of the hash table is a linked list, so whenever a collision occurs, the element can be appened as a node to the linked list at the slot.
In this question, we'll learn how to fill up the hash table using Separate chaining technique. You are given hash table size which you'll use to insert elements into their correct position in the hash table i.e.(arr[i]%hashSize). You are also given an array arr. The size of the array is denoted by sizeOfArray. You need to fill up the hash table using Separate chaining and print the resultant hash table.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase contains 2 lines of input. The first line contains size of the hashtable and the size of the array. The next line contains elements of the array
Constraints:
1 <= T <= 100
2 <= hashSize <= 10^3
1 <= sizeOfArray <= 10^3
0 <= arr[i] <= 10^7For each testcase, in a new line, print the hash table. You need to print the hash table as represented in the example.
Note: Please print tilde ( '~') character at the end of every testcase for separation of list from one another. Given below in the example represent to how to separate each testcase by tilde character.Sample Input:
2
10 6
92 4 14 24 44 91
10 5
12 45 36 87 11
Sample Output:
1->91
2->92
4->4->14->24->44
~
1->11
2->12
5->45
6->36
7->87
~
Explanation:
Testcase1: 92%10=2 so 92 goes to slot 2. 4%10=4 so 4 goes to slot 4. 14%10=4. But 4 is already occupied so we make a linked list at this position and add 14 after 4 in slot 4 and so on.
Testcase2: 12%10=2 so 12 goes to slot 2. 45%10=5 goes to slot 5. 36%10=6 goes to slot 6. 87%10=7 goes to slot 7 and finally 11%10=1 goes to slot 1., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] str=br.readLine().split(" ");
int t=Integer.parseInt(str[0]);
while(t-->0){
str=br.readLine().split(" ");
int k=Integer.parseInt(str[0]);
int n=Integer.parseInt(str[1]);
int[] arr=new int[n];
str=br.readLine().split(" ");
for(int i=0;i<n;i++){
arr[i]=Integer.parseInt(str[i]);
}
String[] st=new String[k];
for(int i=0;i<k;i++){
st[i]="";
}
for(int i=0;i<n;i++){
st[arr[i]%k]+="->"+str[i];
}
for(int i=0;i<k;i++){
if(st[i]!=""){
System.out.println(i+st[i]);
}
}
System.out.println("~");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Separate chaining technique in hashing allows us to use a linked list at each hash slot to handle the problem of collisions. That is, every slot of the hash table is a linked list, so whenever a collision occurs, the element can be appened as a node to the linked list at the slot.
In this question, we'll learn how to fill up the hash table using Separate chaining technique. You are given hash table size which you'll use to insert elements into their correct position in the hash table i.e.(arr[i]%hashSize). You are also given an array arr. The size of the array is denoted by sizeOfArray. You need to fill up the hash table using Separate chaining and print the resultant hash table.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase contains 2 lines of input. The first line contains size of the hashtable and the size of the array. The next line contains elements of the array
Constraints:
1 <= T <= 100
2 <= hashSize <= 10^3
1 <= sizeOfArray <= 10^3
0 <= arr[i] <= 10^7For each testcase, in a new line, print the hash table. You need to print the hash table as represented in the example.
Note: Please print tilde ( '~') character at the end of every testcase for separation of list from one another. Given below in the example represent to how to separate each testcase by tilde character.Sample Input:
2
10 6
92 4 14 24 44 91
10 5
12 45 36 87 11
Sample Output:
1->91
2->92
4->4->14->24->44
~
1->11
2->12
5->45
6->36
7->87
~
Explanation:
Testcase1: 92%10=2 so 92 goes to slot 2. 4%10=4 so 4 goes to slot 4. 14%10=4. But 4 is already occupied so we make a linked list at this position and add 14 after 4 in slot 4 and so on.
Testcase2: 12%10=2 so 12 goes to slot 2. 45%10=5 goes to slot 5. 36%10=6 goes to slot 6. 87%10=7 goes to slot 7 and finally 11%10=1 goes to slot 1., I have written this Solution Code: t=int(input())
for i in range(t):
rowNcol=input().split()
els=input().split()
ds=[]
for i in range(int(rowNcol[0])):
col=[]
ds.append(col)
for i in range(int(rowNcol[1])):
ds[int(int(els[i])%int(rowNcol[0]))].append(els[i])
for i in range(len(ds)):
if(len(ds[i])==0):
continue
else:
print(i,end="")
for j in range(len(ds[i])):
print("->"+ds[i][j],end="")
print()
print("~"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Separate chaining technique in hashing allows us to use a linked list at each hash slot to handle the problem of collisions. That is, every slot of the hash table is a linked list, so whenever a collision occurs, the element can be appened as a node to the linked list at the slot.
In this question, we'll learn how to fill up the hash table using Separate chaining technique. You are given hash table size which you'll use to insert elements into their correct position in the hash table i.e.(arr[i]%hashSize). You are also given an array arr. The size of the array is denoted by sizeOfArray. You need to fill up the hash table using Separate chaining and print the resultant hash table.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase contains 2 lines of input. The first line contains size of the hashtable and the size of the array. The next line contains elements of the array
Constraints:
1 <= T <= 100
2 <= hashSize <= 10^3
1 <= sizeOfArray <= 10^3
0 <= arr[i] <= 10^7For each testcase, in a new line, print the hash table. You need to print the hash table as represented in the example.
Note: Please print tilde ( '~') character at the end of every testcase for separation of list from one another. Given below in the example represent to how to separate each testcase by tilde character.Sample Input:
2
10 6
92 4 14 24 44 91
10 5
12 45 36 87 11
Sample Output:
1->91
2->92
4->4->14->24->44
~
1->11
2->12
5->45
6->36
7->87
~
Explanation:
Testcase1: 92%10=2 so 92 goes to slot 2. 4%10=4 so 4 goes to slot 4. 14%10=4. But 4 is already occupied so we make a linked list at this position and add 14 after 4 in slot 4 and so on.
Testcase2: 12%10=2 so 12 goes to slot 2. 45%10=5 goes to slot 5. 36%10=6 goes to slot 6. 87%10=7 goes to slot 7 and finally 11%10=1 goes to slot 1., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 1000001
int main(){
int t;
cin>>t;
while(t--){
int n,m,x;
cin>>m>>n;
vector<int> a[m];
for(int i=0;i<n;i++){
cin>>x;
a[x%m].push_back(x);
}
for(int i=0;i<m;i++){
if(a[i].size()==0){continue;}
cout<<i<<"->";
for(int j=0;j<a[i].size()-1;j++){
cout<<a[i][j]<<"->";
}
cout<<a[i][a[i].size()-1]<<endl;
}
cout<<"~"<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2X2 square matrix. You need to find the determinant of the matrix.The input contains two-line, each line contain two integers separated by spaces.
Each element of the matrix is from 1 to 100.Output a single integer, the determinant of the matrix.Sample Input
4 5
2 3
Sample Output
2
Sample Input
2 10
10 40
Sample Output
-20, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
int[][]arr=new int[2][2];
for(int i=0;i<2;i++){
for(int j=0;j<2;j++){
arr[i][j]=sc.nextInt();
}
}
int determinent=(arr[0][0]*arr[1][1])-(arr[1][0]*arr[0][1]);
System.out.print(determinent);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2X2 square matrix. You need to find the determinant of the matrix.The input contains two-line, each line contain two integers separated by spaces.
Each element of the matrix is from 1 to 100.Output a single integer, the determinant of the matrix.Sample Input
4 5
2 3
Sample Output
2
Sample Input
2 10
10 40
Sample Output
-20, I have written this Solution Code: arr1=list(map(int,input().split()))
arr2=list(map(int,input().split()))
lis=[]
a11=arr1[0]
a12=arr1[1]
a21=arr2[0]
a22=arr2[1]
det=(a11*a22)-(a12*a21)
print(det), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2X2 square matrix. You need to find the determinant of the matrix.The input contains two-line, each line contain two integers separated by spaces.
Each element of the matrix is from 1 to 100.Output a single integer, the determinant of the matrix.Sample Input
4 5
2 3
Sample Output
2
Sample Input
2 10
10 40
Sample Output
-20, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int a,b,c,d;
cin>>a>>b>>c>>d;
cout<<a*d-b*c;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size N*N, your task is to find the sum of the primary and secondary diagonal of the matrix.
For Matrix:-
M<sub>00</sub> M<sub>01</sub> M<sub>02</sub>
M<sub>10</sub> M<sub>11</sub> M<sub>12</sub>
M<sub>20</sub> M<sub>21</sub> M<sub>22</sub>
Primary diagonal:- M<sub>00</sub> M<sub>11</sub> M<sub>22</sub>
Secondary diagonal:- M<sub>02</sub> M<sub>11</sub> M<sub>20</sub>The first line of input contains a single integer N, The next N lines of input contains N space-separated integers depicting the values of the matrix.
Constraints:-
1 <= N <= 500
1 <= Matrix[][] <= 100000Print the sum of primary and secondary diagonal separated by a space.Sample Input:-
2
1 4
2 6
Sample Output:-
7 6
Sample Input:-
3
1 4 2
1 5 7
3 8 1
Sample Output:-
7 10, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1000001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
signed main(){
int n;
cin>>n;
int a[n][n];
FOR(i,n){
FOR(j,n){
cin>>a[i][j];}}
int sum=0,sum1=0;;
FOR(i,n){
sum+=a[i][i];
sum1+=a[n-i-1][i];
}
out1(sum);out(sum1);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size N*N, your task is to find the sum of the primary and secondary diagonal of the matrix.
For Matrix:-
M<sub>00</sub> M<sub>01</sub> M<sub>02</sub>
M<sub>10</sub> M<sub>11</sub> M<sub>12</sub>
M<sub>20</sub> M<sub>21</sub> M<sub>22</sub>
Primary diagonal:- M<sub>00</sub> M<sub>11</sub> M<sub>22</sub>
Secondary diagonal:- M<sub>02</sub> M<sub>11</sub> M<sub>20</sub>The first line of input contains a single integer N, The next N lines of input contains N space-separated integers depicting the values of the matrix.
Constraints:-
1 <= N <= 500
1 <= Matrix[][] <= 100000Print the sum of primary and secondary diagonal separated by a space.Sample Input:-
2
1 4
2 6
Sample Output:-
7 6
Sample Input:-
3
1 4 2
1 5 7
3 8 1
Sample Output:-
7 10, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String args[])throws Exception {
InputStreamReader inr= new InputStreamReader(System.in);
BufferedReader br= new BufferedReader(inr);
String str=br.readLine();
int row = Integer.parseInt(str);
int col=row;
int [][] arr=new int [row][col];
for(int i=0;i<row;i++){
String line =br.readLine();
String[] elements = line.split(" ");
for(int j=0;j<col;j++){
arr[i][j]= Integer.parseInt(elements[j]);
}
}
int sumPrimary=0;
int sumSecondary=0;
for(int i=0;i<row;i++){
sumPrimary=sumPrimary + arr[i][i];
sumSecondary= sumSecondary + arr[i][row-1-i];
}
System.out.println(sumPrimary+ " " +sumSecondary);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size N*N, your task is to find the sum of the primary and secondary diagonal of the matrix.
For Matrix:-
M<sub>00</sub> M<sub>01</sub> M<sub>02</sub>
M<sub>10</sub> M<sub>11</sub> M<sub>12</sub>
M<sub>20</sub> M<sub>21</sub> M<sub>22</sub>
Primary diagonal:- M<sub>00</sub> M<sub>11</sub> M<sub>22</sub>
Secondary diagonal:- M<sub>02</sub> M<sub>11</sub> M<sub>20</sub>The first line of input contains a single integer N, The next N lines of input contains N space-separated integers depicting the values of the matrix.
Constraints:-
1 <= N <= 500
1 <= Matrix[][] <= 100000Print the sum of primary and secondary diagonal separated by a space.Sample Input:-
2
1 4
2 6
Sample Output:-
7 6
Sample Input:-
3
1 4 2
1 5 7
3 8 1
Sample Output:-
7 10, I have written this Solution Code: // mat is the matrix/ 2d array
// the dimensions of array are n * n
function diagonalSum(mat, n) {
// write code here
// console.log the answer as in example
let principal = 0, secondary = 0;
for (let i = 0; i < n; i++) {
principal += mat[i][i];
secondary += mat[i][n - i - 1];
}
console.log(`${principal} ${secondary}`);
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size N*N, your task is to find the sum of the primary and secondary diagonal of the matrix.
For Matrix:-
M<sub>00</sub> M<sub>01</sub> M<sub>02</sub>
M<sub>10</sub> M<sub>11</sub> M<sub>12</sub>
M<sub>20</sub> M<sub>21</sub> M<sub>22</sub>
Primary diagonal:- M<sub>00</sub> M<sub>11</sub> M<sub>22</sub>
Secondary diagonal:- M<sub>02</sub> M<sub>11</sub> M<sub>20</sub>The first line of input contains a single integer N, The next N lines of input contains N space-separated integers depicting the values of the matrix.
Constraints:-
1 <= N <= 500
1 <= Matrix[][] <= 100000Print the sum of primary and secondary diagonal separated by a space.Sample Input:-
2
1 4
2 6
Sample Output:-
7 6
Sample Input:-
3
1 4 2
1 5 7
3 8 1
Sample Output:-
7 10, I have written this Solution Code: n = int(input())
sum1 = 0
sum2 = 0
for i in range(n):
a = [int(j) for j in input().split()]
sum1 = sum1+a[i]
sum2 = sum2+a[n-1-i]
print(sum1,sum2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer K and an array height[] of size N, where height[i] denotes the height of the ith tree in a forest. The task is to make a cut of height X from the ground such that at max K units wood is collected. Find the minimum value of X
<b>If you make a cut of height X from the ground then every tree with a height greater than X will be reduced to X and the remaining part of the wood can be collected</b>The first line contains two integers N and K.
The next line contains N integers denoting the elements of the array height[]
<b>Constraints</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ arr[i] ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>7</sup>Print a single integer with the value of X.Sample Input:
4 2
1 2 1 2
Sample Output:
1
<b>Explanation:</b>
Make a cut at height 1, the updated array will be {1, 1, 1, 1} and
the collected wood will be {0, 1, 0, 1} i. e. 0 + 1 + 0 + 1 = 2., I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
// don't change the name of this class
// you can add inner classes if needed
class Main {
public static void main (String[] args) {
// Your code here
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int k = sc.nextInt();
int arr[] = new int[n];
for(int i = 0; i < n; i++)
arr[i] = sc.nextInt();
System.out.println(minValue(arr, n, k));
}
static int minValue(int arr[], int N, int k)
{
int l = 0, h = N;
while(l+1 < h){
int m = (l + h) >> 1;
if(f(arr, m, N) <= k)
h = m;
else
l = m;
}
return h;
}
static int f(int a[], int x, int n)
{
int sum = 0;
for(int i = 0; i < n; i++)
sum += Math.max(a[i]-x, 0);
return sum;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer K and an array height[] of size N, where height[i] denotes the height of the ith tree in a forest. The task is to make a cut of height X from the ground such that at max K units wood is collected. Find the minimum value of X
<b>If you make a cut of height X from the ground then every tree with a height greater than X will be reduced to X and the remaining part of the wood can be collected</b>The first line contains two integers N and K.
The next line contains N integers denoting the elements of the array height[]
<b>Constraints</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ arr[i] ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>7</sup>Print a single integer with the value of X.Sample Input:
4 2
1 2 1 2
Sample Output:
1
<b>Explanation:</b>
Make a cut at height 1, the updated array will be {1, 1, 1, 1} and
the collected wood will be {0, 1, 0, 1} i. e. 0 + 1 + 0 + 1 = 2., I have written this Solution Code: def minCut(arr,k):
mini = 1
maxi = k
mid = 0
while mini<=maxi:
mid = mini + int((maxi - mini)/2)
wood = 0
for j in range(n):
if(arr[j]-mid>=0):
wood += arr[j] - mid
if wood == k:
break;
elif wood > k:
mini = mid+1
else:
maxi = mid-1
print(mini)
n,k = list(map(int, input().split()))
arr = list(map(int, input().split()))
minCut(arr,k), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer K and an array height[] of size N, where height[i] denotes the height of the ith tree in a forest. The task is to make a cut of height X from the ground such that at max K units wood is collected. Find the minimum value of X
<b>If you make a cut of height X from the ground then every tree with a height greater than X will be reduced to X and the remaining part of the wood can be collected</b>The first line contains two integers N and K.
The next line contains N integers denoting the elements of the array height[]
<b>Constraints</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ arr[i] ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>7</sup>Print a single integer with the value of X.Sample Input:
4 2
1 2 1 2
Sample Output:
1
<b>Explanation:</b>
Make a cut at height 1, the updated array will be {1, 1, 1, 1} and
the collected wood will be {0, 1, 0, 1} i. e. 0 + 1 + 0 + 1 = 2., I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define ll long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N], n, k;
int f(int x){
int sum = 0;
for(int i = 1; i <= n; i++)
sum += max(a[i]-x, 0);
return sum;
}
void solve(){
cin >> n >> k;
for(int i = 1; i <= n; i++)
cin >> a[i];
int l = 0, h = N;
while(l+1 < h){
int m = (l + h) >> 1;
if(f(m) <= k)
h = m;
else
l = m;
}
cout << h;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
testcases();
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, your task is to calculate the largest number that can be formed by appending the numbers from the given array.
For eg:- If the numbers are { 10, 3, 31} then the largest number will be 33110First line of input contains a single integer N, next line contains N space separated integers depicting values of array.
Constraints:-
1 < = N < = 100000
1 < = Arr[i] < = 100000Print the largest number that can be formed using the array elements.Sample Input:-
3
10 3 31
Sample Output:-
33110
Sample Input:-
5
8 9 10 11 12
Sample Output:-
98121110, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static void Largest(ArrayList<String> a)
{
Collections.sort(a, new Comparator<String>(){
@Override
public int compare(String X, String Y) {
String XY=X + Y;
String YX=Y + X;
return XY.compareTo(YX) > 0 ? -1:1;
}
});
for(String value:a)
{
System.out.print(value);
}
}
public static void main (String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
ArrayList<String> al=new ArrayList<>();
String[] nums=br.readLine().split("\\s");
for(int i=0;i<n;i++)
{
al.add(nums[i]);
}
Largest(al);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, your task is to calculate the largest number that can be formed by appending the numbers from the given array.
For eg:- If the numbers are { 10, 3, 31} then the largest number will be 33110First line of input contains a single integer N, next line contains N space separated integers depicting values of array.
Constraints:-
1 < = N < = 100000
1 < = Arr[i] < = 100000Print the largest number that can be formed using the array elements.Sample Input:-
3
10 3 31
Sample Output:-
33110
Sample Input:-
5
8 9 10 11 12
Sample Output:-
98121110, I have written this Solution Code: // C++ program to find sum of all minimum and maximum
// elements Of Sub-array Size k.
#include<bits/stdc++.h>
using namespace std;
#define int long long int
int myCompare(string X, string Y)
{
// first append Y at the end of X
string XY = X.append(Y);
// then append X at the end of Y
string YX = Y.append(X);
// Now see which of the two formed numbers is greater
return XY.compare(YX) > 0 ? 1: 0;
}
// The main function that prints the arrangement with the largest value.
// The function accepts a vector of strings
void printLargest(vector<string> arr)
{
// Sort the numbers using library sort function. The function uses
// our comparison function myCompare() to compare two strings.
// See http://www.cplusplus.com/reference/algorithm/sort/ for details
sort(arr.begin(), arr.end(), myCompare);
for (int i=0; i < arr.size() ; i++ )
cout << arr[i];
}
// Driver program to test above functions
signed main()
{
int n,k;
cin>>n;
vector<string> v(n);
for(int i=0;i<n;i++){
cin>>v[i];
}
printLargest(v);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b your task is to print the summation of these two numbersInput contains a single line containing two space separated integers a and b.
Constraints:-
1 <= a, b <= 10<sup>20000</sup>Print the sum of a and b.Sample Input:-
3 8
Sample Output:-
11
Sample Input:-
15 1
Sample Output:-
16, I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
import java.math.BigInteger;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
BigInteger sum;
String ip1 = sc.next();
String ip2 = sc.next();
BigInteger a = new BigInteger(ip1);
BigInteger b = new BigInteger(ip2);
sum = a.add(b);
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b your task is to print the summation of these two numbersInput contains a single line containing two space separated integers a and b.
Constraints:-
1 <= a, b <= 10<sup>20000</sup>Print the sum of a and b.Sample Input:-
3 8
Sample Output:-
11
Sample Input:-
15 1
Sample Output:-
16, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-02-03 02:46:30
**/
#include <bits/stdc++.h>
#define NX 105
#define MX 3350
using namespace std;
const int mod = 998244353;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
typedef long long INT;
const int pb = 10;
const int base_digits = 9;
const int base = 1000000000;
const int DIV = 100000;
struct bigint {
vector<int> a;
int sign;
bigint() : sign(1) {}
bigint(INT v) { *this = v; }
bigint(const string &s) { read(s); }
void operator=(const bigint &v) { sign = v.sign, a = v.a; }
void operator=(INT v) {
sign = 1;
if (v < 0) sign = -1, v = -v;
for (; v > 0; v = v / base) a.push_back(v % base);
}
bigint operator+(const bigint &v) const {
if (sign == v.sign) {
bigint res = v;
for (int i = 0, carry = 0; i < (int)max(a.size(), v.a.size()) ||
carry;
i++) {
if (i == (int)res.a.size()) res.a.push_back(0);
res.a[i] += carry + (i < (int)a.size() ? a[i] : 0);
carry = res.a[i] >= base;
if (carry) res.a[i] -= base;
}
return res;
}
return *this - (-v);
}
bigint operator-(const bigint &v) const {
if (sign == v.sign) {
if (abs() >= v.abs()) {
bigint res = *this;
for (int i = 0, carry = 0; i < (int)v.a.size() || carry; i++) {
res.a[i] -= carry + (i < (int)v.a.size() ? v.a[i] : 0);
carry = res.a[i] < 0;
if (carry) res.a[i] += base;
}
res.trim();
return res;
}
return -(v - *this);
}
return *this + (-v);
}
void operator*=(int v) {
if (v < 0) sign = -sign, v = -v;
for (int i = 0, carry = 0; i < (int)a.size() || carry; i++) {
if (i == (int)a.size()) a.push_back(0);
INT cur = a[i] * (INT)v + carry;
carry = (int)(cur / base);
a[i] = (int)(cur % base);
}
trim();
}
bigint operator*(int v) const {
bigint res = *this;
res *= v;
return res;
}
friend pair<bigint, bigint> DIVmod(const bigint &a1, const bigint &b1) {
int norm = base / (b1.a.back() + 1);
bigint a = a1.abs() * norm;
bigint b = b1.abs() * norm;
bigint q, r;
q.a.resize(a.a.size());
for (int i = a.a.size() - 1; i >= 0; i--) {
r *= base;
r += a.a[i];
int s1 = r.a.size() <= b.a.size() ? 0 : r.a[b.a.size()];
int s2 = r.a.size() <= b.a.size() - 1 ? 0 : r.a[b.a.size() - 1];
int d = ((INT)base * s1 + s2) / b.a.back();
r -= b * d;
while (r < 0) r += b, --d;
q.a[i] = d;
}
q.sign = a1.sign * b1.sign;
r.sign = a1.sign;
q.trim();
r.trim();
return make_pair(q, r / norm);
}
bigint operator/(const bigint &v) const { return DIVmod(*this, v).first; }
bigint operator%(const bigint &v) const { return DIVmod(*this, v).second; }
void operator/=(int v) {
if (v < 0) sign = -sign, v = -v;
for (int i = (int)a.size() - 1, rem = 0; i >= 0; i--) {
INT cur = a[i] + rem * (INT)base;
a[i] = (int)(cur / v);
rem = (int)(cur % v);
}
trim();
}
bigint operator/(int v) const {
bigint res = *this;
res /= v;
return res;
}
int operator%(int v) const {
if (v < 0) v = -v;
int m = 0;
for (int i = a.size() - 1; i >= 0; --i)
m = (a[i] + m * (INT)base) % v;
return m * sign;
}
void operator+=(const bigint &v) { *this = *this + v; }
void operator-=(const bigint &v) { *this = *this - v; }
void operator*=(const bigint &v) { *this = *this * v; }
void operator/=(const bigint &v) { *this = *this / v; }
bool operator<(const bigint &v) const {
if (sign != v.sign) return sign < v.sign;
if (a.size() != v.a.size()) return a.size() * sign < v.a.size() * v.sign;
for (int i = a.size() - 1; i >= 0; i--)
if (a[i] != v.a[i]) return a[i] * sign < v.a[i] * sign;
return false;
}
bool operator>(const bigint &v) const { return v < *this; }
bool operator<=(const bigint &v) const { return !(v < *this); }
bool operator>=(const bigint &v) const { return !(*this < v); }
bool operator==(const bigint &v) const { return !(*this < v) && !(v < *this); }
bool operator!=(const bigint &v) const { return *this < v || v < *this; }
void trim() {
while (!a.empty() && !a.back()) a.pop_back();
if (a.empty()) sign = 1;
}
bool isZero() const { return a.empty() || (a.size() == 1 && !a[0]); }
bigint operator-() const {
bigint res = *this;
res.sign = -sign;
return res;
}
bigint abs() const {
bigint res = *this;
res.sign *= res.sign;
return res;
}
INT longValue() const {
INT res = 0;
for (int i = a.size() - 1; i >= 0; i--) res = res * base + a[i];
return res * sign;
}
friend bigint gcd(const bigint &a, const bigint &b) { return b.isZero() ? a : gcd(b, a % b); }
friend bigint lcm(const bigint &a, const bigint &b) { return a / gcd(a, b) * b; }
void read(const string &s) {
sign = 1;
a.clear();
int pos = 0;
while (pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
if (s[pos] == '-') sign = -sign;
pos++;
}
for (int i = s.size() - 1; i >= pos; i -= base_digits) {
int x = 0;
for (int j = max(pos, i - base_digits + 1); j <= i; j++) x = x *
pb +
s[j] - '0';
a.push_back(x);
}
trim();
}
friend istream &operator>>(istream &stream, bigint &v) {
string s;
stream >> s;
v.read(s);
return stream;
}
friend ostream &operator<<(ostream &stream, const bigint &v) {
if (v.sign == -1) stream << '-';
stream << (v.a.empty() ? 0 : v.a.back());
for (int i = (int)v.a.size() - 2; i >= 0; --i) stream << setw(base_digits) << setfill('0') << v.a[i];
return stream;
}
static vector<int> convert_base(const vector<int> &a, int old_digits, int new_digits) {
vector<INT> p(max(old_digits, new_digits) + 1);
p[0] = 1;
for (int i = 1; i < (int)p.size(); i++) p[i] = p[i - 1] * pb;
vector<int> res;
INT cur = 0;
int cur_digits = 0;
for (int i = 0; i < (int)a.size(); i++) {
cur += a[i] * p[cur_digits];
cur_digits += old_digits;
while (cur_digits >= new_digits) {
res.push_back(int(cur % p[new_digits]));
cur /= p[new_digits];
cur_digits -= new_digits;
}
}
res.push_back((int)cur);
while (!res.empty() && !res.back()) res.pop_back();
return res;
}
typedef vector<INT> vll;
static vll karatsubaMultiply(const vll &a, const vll &b) {
int n = a.size();
vll res(n + n);
if (n <= 32) {
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
res[i + j] += a[i] * b[j];
return res;
}
int k = n >> 1;
vll a1(a.begin(), a.begin() + k);
vll a2(a.begin() + k, a.end());
vll b1(b.begin(), b.begin() + k);
vll b2(b.begin() + k, b.end());
vll a1b1 = karatsubaMultiply(a1, b1);
vll a2b2 = karatsubaMultiply(a2, b2);
for (int i = 0; i < k; i++) a2[i] += a1[i];
for (int i = 0; i < k; i++) b2[i] += b1[i];
vll r = karatsubaMultiply(a2, b2);
for (int i = 0; i < (int)a1b1.size(); i++) r[i] -= a1b1[i];
for (int i = 0; i < (int)a2b2.size(); i++) r[i] -= a2b2[i];
for (int i = 0; i < (int)r.size(); i++) res[i + k] += r[i];
for (int i = 0; i < (int)a1b1.size(); i++) res[i] += a1b1[i];
for (int i = 0; i < (int)a2b2.size(); i++) res[i + n] += a2b2[i];
return res;
}
bigint operator*(const bigint &v) const {
vector<int> a5 = convert_base(this->a, base_digits, 5);
vector<int> b5 = convert_base(v.a, base_digits, 5);
vll a(a5.begin(), a5.end());
vll b(b5.begin(), b5.end());
while (a.size() < b.size()) a.push_back(0);
while (b.size() < a.size()) b.push_back(0);
while (a.size() & (a.size() - 1)) a.push_back(0), b.push_back(0);
vll c = karatsubaMultiply(a, b);
bigint res;
res.sign = sign * v.sign;
for (int i = 0, carry = 0; i < (int)c.size(); i++) {
INT cur = c[i] + carry;
res.a.push_back((int)(cur % DIV));
carry = (int)(cur / DIV);
}
res.a = convert_base(res.a, 5, base_digits);
res.trim();
return res;
}
inline bool isOdd() { return a[0] & 1; }
};
int main() {
bigint n, m;
cin >> n >> m;
cout << n + m << "\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b your task is to print the summation of these two numbersInput contains a single line containing two space separated integers a and b.
Constraints:-
1 <= a, b <= 10<sup>20000</sup>Print the sum of a and b.Sample Input:-
3 8
Sample Output:-
11
Sample Input:-
15 1
Sample Output:-
16, I have written this Solution Code: n,m = map(int,input().split())
print(n+m) , In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: How many ways are there to place a black and a white knight on an N * M chessboard such that they do not attack each other? The knights have to be placed on different squares. A knight can move two squares horizontally and one square vertically (L shaped), or two squares vertically and one square horizontally (L shaped). The knights attack each other if one can reach the other in one move.The first line contains the number of test cases T. Each of the next T lines contains two integers N and M which is size of matrix.
Constraints:
1 <= T <= 100
1 <= N, M <= 100For each testcase in a new line, print the required answer, i.e, number of possible ways to place knights.Sample Input:
3
2 2
2 3
4 5
Sample Output:
12
26
312
Explanation:
Test Case 1: We can place a black and a white knight in 12 possible ways such that none of them attacks each other., I have written this Solution Code: import java.io.*;
import java.util.*;
import java.math.BigInteger;
class Main{
public static void main (String[] args) throws IOException {
StringBuilder output = new StringBuilder();
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
String s = reader.readLine();
int T = Integer.parseInt(s);
for (int t=0; t<T; t++) {
s = reader.readLine();
int index = s.indexOf(' ');
long N = Integer.parseInt(s.substring(0,index));
long M = Integer.parseInt(s.substring(index+1));
if (N < M) {
long temp = N;
N = M;
M = temp;
}
long answer;
if (M == 1) {
answer = 0;
} else if (M == 2) {
if (N == 2) {
answer = 0;
} else if (N == 3) {
answer = 4;
} else {
answer = 2*N*M-8;
}
} else if (M == 3) {
if (N == 3) {
answer = 16;
} else {
answer = 4*N*M-20;
}
} else {
answer = (N-4)*(8*M-12)+2*(4*M-6)+2*(6*M-10);
}
BigInteger all = BigInteger.valueOf(N*M);
BigInteger result = all.multiply(all.add(BigInteger.ONE.negate())).add(BigInteger.valueOf(-answer));
output.append(result).append("\n");
}
System.out.print(output);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: How many ways are there to place a black and a white knight on an N * M chessboard such that they do not attack each other? The knights have to be placed on different squares. A knight can move two squares horizontally and one square vertically (L shaped), or two squares vertically and one square horizontally (L shaped). The knights attack each other if one can reach the other in one move.The first line contains the number of test cases T. Each of the next T lines contains two integers N and M which is size of matrix.
Constraints:
1 <= T <= 100
1 <= N, M <= 100For each testcase in a new line, print the required answer, i.e, number of possible ways to place knights.Sample Input:
3
2 2
2 3
4 5
Sample Output:
12
26
312
Explanation:
Test Case 1: We can place a black and a white knight in 12 possible ways such that none of them attacks each other., I have written this Solution Code: t = int(input())
for i in range(t):
n, m = input().split()
n = int(n)
m = int(m)
if n < m:
n, m = m, n
if n == 1 or m == 1:
print(n * (n - 1))
else:
b = n * m
print((b * (b - 1)) - (4 * (((n - 1) * (m - 2)) + ((m - 1) * (n - 2))))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: How many ways are there to place a black and a white knight on an N * M chessboard such that they do not attack each other? The knights have to be placed on different squares. A knight can move two squares horizontally and one square vertically (L shaped), or two squares vertically and one square horizontally (L shaped). The knights attack each other if one can reach the other in one move.The first line contains the number of test cases T. Each of the next T lines contains two integers N and M which is size of matrix.
Constraints:
1 <= T <= 100
1 <= N, M <= 100For each testcase in a new line, print the required answer, i.e, number of possible ways to place knights.Sample Input:
3
2 2
2 3
4 5
Sample Output:
12
26
312
Explanation:
Test Case 1: We can place a black and a white knight in 12 possible ways such that none of them attacks each other., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
int t;cin>>t;while(t--){
long n,m,c,count=0,len=0;
cin>>n>>m;
c=n*m;
count=c*(c-1);
if(m>1 && n>1){
len+=(2*(m-2))*(n-1);
len+=(2*(n-2))*(m-1);
len*=2;}
/*for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(i+1<n && j+2<m)
len++;
if(i+1<n && j-2>=0)
len++;
if(i-1>=0 && j+2<m)
len++;
if(i-1>=0 && j-2>=0)
len++;
if(i+2<n && j+1<m)
len++;
if(i+2<n && j-1>=0)
len++;
if(i-2>0 && j+1<m)
len++;
if(i-2>=0 && j-1>=0)
len++;
}
}*/
count-=len;
cout<<count<<endl;
}
//code
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string S of length N. Find the character which occurs the most number of times in this string. If there are mutliple such characters present, print -1.The first line of the input contains a single integer N.
The second line of the input contains a string S.
<b>Constraints:</b>
2 <b>≤</b> N <b>≤</b> 10<sup>5</sup>
All characters in string S are lowercase english letters.Print the character which occurs the most number of times in this string. If there are mutliple such characters present, print -1.Sample Input:
7
abadcar
Sample Output:
a, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
String s = sc.next();
int arr[]= new int[26];
for(int i=0;i<n;i++){
char ch = s.charAt(i);
int j = ch - 'a';
arr[j]++;
}
int max = 0;
int sec = 0;
int fr = 0;
int j =0;
for(int i=0;i<26; i++){
if(arr[i]> max){
sec = max;
max = arr[i];
j=i;
}else if(arr[i]>sec){
sec = arr[i];
}
}
if(sec == max){
System.out.print(-1);
}else{
for(int i=0;i<n; i++){
char c = s.charAt(i);
if(c-'a' == j){
System.out.print(c);
break;
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string S of length N. Find the character which occurs the most number of times in this string. If there are mutliple such characters present, print -1.The first line of the input contains a single integer N.
The second line of the input contains a string S.
<b>Constraints:</b>
2 <b>≤</b> N <b>≤</b> 10<sup>5</sup>
All characters in string S are lowercase english letters.Print the character which occurs the most number of times in this string. If there are mutliple such characters present, print -1.Sample Input:
7
abadcar
Sample Output:
a, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n;
cin >> n;
string s;
cin >> s;
map<char, int> mp;
for(auto i : s) mp[i]++;
int mx = 0, f = 0;
char ch;
for(auto i : mp){
if(i.second > mx){
mx = i.second;
f = 1;
ch = i.first;
}
else if(i.second == mx){
f++;
}
}
if(f == 1){
cout << ch;
}
else cout << -1;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: static int isPresent(long arr[], int n, long k)
{
int left = 0;
int right = n-1;
int res = -1;
while(left<=right){
int mid = (left+right)/2;
if(arr[mid] == k){
res = 1;
break;
}else if(arr[mid] < k){
left = mid + 1;
}else{
right = mid - 1;
}
}
return res;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
unordered_map<long long,int> m;
long k;
cin>>k;
long long a;
for(int i=0;i<n;i++){
cin>>a;
m[a]++;
}
if(m.find(k)!=m.end()){
cout<<1<<endl;
}
else{
cout<<-1<<endl;
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: def binary_search(arr, low, high, x):
if high >= low:
mid = (high + low) // 2
if arr[mid] == x:
return 1
elif arr[mid] > x:
return binary_search(arr, low, mid - 1, x)
else:
return binary_search(arr, mid + 1, high, x)
else:
return -1
def position(n,arr,x):
return binary_search(arr,0,n-1,x)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: // arr is they array to search from
// x is target
function binSearch(arr, x) {
// write code here
// do not console.log
// return the 1 or -1
let l = 0;
let r = arr.length - 1;
let mid;
while (r >= l) {
mid = l + Math.floor((r - l) / 2);
// If the element is present at the middle
// itself
if (arr[mid] == x)
return 1;
// If element is smaller than mid, then
// it can only be present in left subarray
if (arr[mid] > x)
r = mid - 1;
// Else the element can only be present
// in right subarray
else
l = mid + 1;
}
// We reach here when element is not
// present in array
return -1;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; i++)
cout << a[i] << " ";
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: // arr is unsorted array
// n is the number of elements in the array
function bubbleSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => a - b)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int temp;
for(int i=1;i<n;i++){
if(a[i]<a[i-1]){
for(int j=i;j>0;j--){
if(a[j]<a[j-1]){
temp=a[j];
a[j]=a[j-1];
a[j-1]=temp;
}
else{
break;
}
}
}
}
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: def bubbleSort(arr):
arr.sort()
return arr
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, your task is to update every element with multiplication of previous and next elements with following exceptions:-
a) First element is replaced by multiplication of first and second.
b) Last element is replaced by multiplication of last and second last.
See example for more clarityFirst line of input contains the size of array N, next line contains N space separated integers denoting values of array.
Constraints:-
2 < = N < = 100000
1 < = arr[i] < = 100000Print the modified arraySample Input :-
5
2 3 4 5 6
Sample Output:-
6 8 15 24 30
Explanation:-
{2*3, 2*4, 3*5, 4*6, 5*6}
Sample Input:-
2
3 4
Sample Output:-
12 12, I have written this Solution Code: // arr is the array of numbers, n is the number fo elements
function replaceArray(arr, n) {
// write code here
// do not console.log
// return the new array
const newArr = []
newArr[0] = arr[0] * arr[1]
newArr[n-1] = arr[n-1] * arr[n-2]
for(let i= 1;i<n-1;i++){
newArr[i] = arr[i-1] * arr[i+1]
}
return newArr
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, your task is to update every element with multiplication of previous and next elements with following exceptions:-
a) First element is replaced by multiplication of first and second.
b) Last element is replaced by multiplication of last and second last.
See example for more clarityFirst line of input contains the size of array N, next line contains N space separated integers denoting values of array.
Constraints:-
2 < = N < = 100000
1 < = arr[i] < = 100000Print the modified arraySample Input :-
5
2 3 4 5 6
Sample Output:-
6 8 15 24 30
Explanation:-
{2*3, 2*4, 3*5, 4*6, 5*6}
Sample Input:-
2
3 4
Sample Output:-
12 12, I have written this Solution Code: n = int(input())
X = [int(x) for x in input().split()]
lst = []
for i in range(len(X)):
if i == 0:
lst.append(X[i]*X[i+1])
elif i == (len(X) - 1):
lst.append(X[i-1]*X[i])
else:
lst.append(X[i-1]*X[i+1])
for i in lst:
print(i,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, your task is to update every element with multiplication of previous and next elements with following exceptions:-
a) First element is replaced by multiplication of first and second.
b) Last element is replaced by multiplication of last and second last.
See example for more clarityFirst line of input contains the size of array N, next line contains N space separated integers denoting values of array.
Constraints:-
2 < = N < = 100000
1 < = arr[i] < = 100000Print the modified arraySample Input :-
5
2 3 4 5 6
Sample Output:-
6 8 15 24 30
Explanation:-
{2*3, 2*4, 3*5, 4*6, 5*6}
Sample Input:-
2
3 4
Sample Output:-
12 12, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin>>n;
long long b[n],a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=1;i<n-1;i++){
b[i]=a[i-1]*a[i+1];
}
b[0]=a[0]*a[1];
b[n-1]=a[n-1]*a[n-2];
for(int i=0;i<n;i++){
cout<<b[i]<<" ";}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, your task is to update every element with multiplication of previous and next elements with following exceptions:-
a) First element is replaced by multiplication of first and second.
b) Last element is replaced by multiplication of last and second last.
See example for more clarityFirst line of input contains the size of array N, next line contains N space separated integers denoting values of array.
Constraints:-
2 < = N < = 100000
1 < = arr[i] < = 100000Print the modified arraySample Input :-
5
2 3 4 5 6
Sample Output:-
6 8 15 24 30
Explanation:-
{2*3, 2*4, 3*5, 4*6, 5*6}
Sample Input:-
2
3 4
Sample Output:-
12 12, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n=sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
System.out.print(a[0]*a[1]+" ");
for(int i=1;i<n-1;i++){
System.out.print(a[i-1]*a[i+1]+" ");
}
System.out.print(a[n-1]*a[n-2]);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Each item in Newton’s menu is assigned a spice level from 1 to 10. Based on the spice level, the item is categorized as:
MILD: If the spice level is less than 4.
MEDIUM: If the spice level is greater than or equal to 4 but less than 7.
HOT: If the spice level is greater than or equal to 7.
Given that the spice level of an item is X, find the category it lies in.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You have to complete the function <b>Solve()</b> that takes the integer X as parameter.
<b>Constraints :</b>
1 ≤ X, ≤ 10Output on a new line, the category that the item lies in.Sample Input:-
4
Sample Output:-
MEDIUM
Sample Input:-
1
Sample Output:-
MILD, I have written this Solution Code: class Solution {
public void Solve(int x) {
if(x>=1 && x<4)
System.out.println("MILD");
else if(x>=4 && x<=6)
System.out.println("MEDIUM");
else
System.out.println("HOT");
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A[] of size N. For any subarray size i (1 <= i <= N), you have to do the following:
<ul><li>Write all subarrays of size i from array A on a piece of paper.</li><li>Now, replace each of these subarrays with their minimum element.</li><li>Now, find the maximum value among all these minimums.</li>
Print this maximum value for all subarray sizes k (1 <= k <= N).The first line contains an integer N denoting the size of the array. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array.
Constraint:-
1 <= N <= 10^5
1 <= A[i] <= 10^6Print the array of numbers of size N for each of the considered window size 1, 2, ..., N respectively.Sample Input:
7
10 20 30 50 10 70 30
Sample Output:
70 30 20 10 10 10 10
Explanation:
Testcase 1:
First element in output indicates maximum of minimums of all windows of size 1. Minimums of windows of size 1 are {10}, {20}, {30}, {50}, {10}, {70} and {30}. Maximum of these minimums is 70.
Second element in output indicates maximum of minimums of all windows of size 2. Minimums of windows of size 2 are {10}, {20}, {30}, {10}, {10}, and {30}. Maximum of these minimums is 30.
Third element in output indicates maximum of minimums of all windows of size 3. Minimums of windows of size 3 are {10}, {20}, {10}, {10} and {10}. Maximum of these minimums is 20.
Similarly other elements of output are computed., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader (System.in));
int n = Integer.parseInt(br.readLine());
String s= br.readLine();
String[] str = s.split(" ");
int arr[] = new int[n];
for(int i=0 ; i<n ; i++){
arr[i]=Integer.parseInt(str[i]) ;
}
printMaxOfMin(n,arr);
}
static void printMaxOfMin(int n,int[] arr)
{
Stack<Integer> s = new Stack<>();
int left[] = new int[n+1];
int right[] = new int[n+1];
for (int i=0; i<n; i++)
{
left[i] = -1;
right[i] = n;
}
for (int i=0; i<n; i++)
{
while (!s.empty() && arr[s.peek()] >= arr[i])
s.pop();
if (!s.empty())
left[i] = s.peek();
s.push(i);
}
while (!s.empty())
s.pop();
for (int i = n-1 ; i>=0 ; i-- )
{
while (!s.empty() && arr[s.peek()] >= arr[i])
s.pop();
if(!s.empty())
right[i] = s.peek();
s.push(i);
}
int ans[] = new int[n+1];
for (int i=0; i<=n; i++)
ans[i] = 0;
for (int i=0; i<n; i++)
{
int len = right[i] - left[i] - 1;
ans[len] = Math.max(ans[len], arr[i]);
}
for (int i=n-1; i>=1; i--)
ans[i] = Math.max(ans[i], ans[i+1]);
for (int i=1; i<=n; i++)
System.out.print(ans[i] + " ");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A[] of size N. For any subarray size i (1 <= i <= N), you have to do the following:
<ul><li>Write all subarrays of size i from array A on a piece of paper.</li><li>Now, replace each of these subarrays with their minimum element.</li><li>Now, find the maximum value among all these minimums.</li>
Print this maximum value for all subarray sizes k (1 <= k <= N).The first line contains an integer N denoting the size of the array. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array.
Constraint:-
1 <= N <= 10^5
1 <= A[i] <= 10^6Print the array of numbers of size N for each of the considered window size 1, 2, ..., N respectively.Sample Input:
7
10 20 30 50 10 70 30
Sample Output:
70 30 20 10 10 10 10
Explanation:
Testcase 1:
First element in output indicates maximum of minimums of all windows of size 1. Minimums of windows of size 1 are {10}, {20}, {30}, {50}, {10}, {70} and {30}. Maximum of these minimums is 70.
Second element in output indicates maximum of minimums of all windows of size 2. Minimums of windows of size 2 are {10}, {20}, {30}, {10}, {10}, and {30}. Maximum of these minimums is 30.
Third element in output indicates maximum of minimums of all windows of size 3. Minimums of windows of size 3 are {10}, {20}, {10}, {10} and {10}. Maximum of these minimums is 20.
Similarly other elements of output are computed., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int l[N], r[N], ans[N], a[N];
signed main() {
IOS;
int n; cin >> n;
stack<int> s;
a[0] = a[n+1] = -inf;
s.push(0);
for(int i = 1; i <= n; i++){
cin >> a[i];
while(!s.empty() && a[s.top()] >= a[i])
s.pop();
l[i] = s.top();
s.push(i);
}
while(!s.empty())
s.pop();
s.push(n+1);
for(int i = n; i >= 1; i--){
while(!s.empty() && a[s.top()] >= a[i])
s.pop();
r[i] = s.top();
s.push(i);
}
for(int i = 1; i <= n; i++){
int len = r[i] - l[i] - 1;
ans[len] = max(ans[len], a[i]);
}
for(int i = n-1; i >= 1; i--)
ans[i] = max(ans[i], ans[i+1]);
for(int i = 1; i <= n; i++)
cout << ans[i] << " ";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a circular linked list consisting of N nodes and an integer K, your task is to add the integer K at the end of the list.
<b>Note:
Sample Input and Output just show how a linked list will look depending on the questions. Do not copy-paste as it is in custom input</b><b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Insertion()</b> that takes head node of circular linked list and the integer K as parameter.
Constraints:
1 <=N <= 1000
1 <= Node.data, K<= 1000Return the head node of the modified circular linked list.Sample Input 1:-
3
1- >2- >3
4
Sample Output 1:-
1- >2- >3- >4
Sample Input 2:-
3
1- >3- >2
1
Sample Output 2:-
1- >3- >2- >1, I have written this Solution Code: public static Node Insertion(Node head, int K){
Node node=head;
while ( node.next != head)
{node = node.next; }
Node temp = new Node(K);
node.next=temp;
temp.next=head;
return head;}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number is called Silly if it can be represented as the sum of the square of consecutive natural numbers starting from 1.
For a given number N, find the closest silly number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SillyNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the closest Silly number.
Note:- If more than one answer exists return the minimum one.Sample Input:-
18
Sample Output:-
14
Explanation:-
1*1 + 2*2 + 3*3 = 14
Sample Input:-
2
Sample Output:-
1, I have written this Solution Code: static int SillyNumber(int N){
int sum=0;
int x=1;
while(sum<N){
sum+=x*x;
x++;
}
x--;
if(sum-N < N-(sum-x*x)){
return sum;
}
else{
return sum-x*x;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number is called Silly if it can be represented as the sum of the square of consecutive natural numbers starting from 1.
For a given number N, find the closest silly number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SillyNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the closest Silly number.
Note:- If more than one answer exists return the minimum one.Sample Input:-
18
Sample Output:-
14
Explanation:-
1*1 + 2*2 + 3*3 = 14
Sample Input:-
2
Sample Output:-
1, I have written this Solution Code: int SillyNumber(int N){
int sum=0;
int x=1;
while(sum<N){
sum+=x*x;
x++;
}
x--;
if(sum-N < N-(sum-x*x)){
return sum;
}
else{
return sum-x*x;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number is called Silly if it can be represented as the sum of the square of consecutive natural numbers starting from 1.
For a given number N, find the closest silly number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SillyNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the closest Silly number.
Note:- If more than one answer exists return the minimum one.Sample Input:-
18
Sample Output:-
14
Explanation:-
1*1 + 2*2 + 3*3 = 14
Sample Input:-
2
Sample Output:-
1, I have written this Solution Code: int SillyNumber(int N){
int sum=0;
int x=1;
while(sum<N){
sum+=x*x;
x++;
}
x--;
if(sum-N < N-(sum-x*x)){
return sum;
}
else{
return sum-x*x;
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number is called Silly if it can be represented as the sum of the square of consecutive natural numbers starting from 1.
For a given number N, find the closest silly number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SillyNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the closest Silly number.
Note:- If more than one answer exists return the minimum one.Sample Input:-
18
Sample Output:-
14
Explanation:-
1*1 + 2*2 + 3*3 = 14
Sample Input:-
2
Sample Output:-
1, I have written this Solution Code: def SillyNumber(N):
sum=0
x=1
while sum<N:
sum=sum+x*x
x=x+1
x=x-1
if (sum-N) < (N-(sum-x*x)):
return sum;
else:
return sum - x*x
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: static int equationSum(int n)
{
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: def equationSum(N) :
re=(N*(N+1))//2
re=re*re
re=re+9*((N*(N+1))//2)
re=re+4*N
return re
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: int equationSum(int n){
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: int equationSum(int n){
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono and Solo are one of the best programmers on the planet. Solo knows how to solve the problem on maximum subarray sum using the Kadane algorithm (don't you?). Now, Tono gives Solo a trickier problem to solve. In this problem, Solo can choose any non- empty subarray, the value of the subarray will be the <b>(sum of all the elements of the subarray - largest element of the subarray)</b>. Can you find the maximum value of any subarray?The first line of the input contains a single integer N, the size of the array.
The second line of the input contains N elements A[1], A[2],. , A[N].
Constraints
1 <= N <= 100000
-30 <= A[i] <= 30Output a single integer, the value of the subarray having maximum value.Sample Input
7
5 2 5 3 -30 -30 6
Sample Output
10
Explanation: Solo chooses the subarray [5, 2, 5, 3]. The value of the subarray is 10.
Sample Input
3
-10 6 -15
Sample Output
0, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int size = Integer.parseInt(br.readLine());
String line[] = br.readLine().split(" ");
long a[] = new long[size + 1];
for(int i = 1 ; i <= size; i++)
a[i] = Long.parseLong(line[i - 1]);
long value1 = maxofsubarray(a,size);
long value2 = maxofsubarrayBreakingatNegativeVal(a, size);
System.out.print(Math.max(value1, value2));
}
static long maxofsubarray(long[] a , int size)
{
long currsum = 0 , maxsum = 0, maxEl = 0;
for(int i = 1 ; i <= size ; i++)
{
currsum += a[i];
if(currsum < 0) currsum =0;
if(currsum > maxsum)
{
maxsum = currsum; maxEl = Math.max(maxEl, a[i]);
}
} return (maxsum - maxEl);
}
static long maxofsubarrayBreakingatNegativeVal(long[] a , int size)
{
long currsum = 0 , maxsum = 0, maxEl = 0, maxofall = 0;
for(int i = 1 ; i <= size ; i++)
{
maxEl = Math.max(maxEl, a[i]);
currsum += a[i];
if(currsum - maxEl < 0)
{
currsum =0; maxEl = 0;
}
if(currsum - maxEl > maxsum)
{
maxofall = currsum - maxEl; maxsum = maxofall;
}
}
return (maxofall);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono and Solo are one of the best programmers on the planet. Solo knows how to solve the problem on maximum subarray sum using the Kadane algorithm (don't you?). Now, Tono gives Solo a trickier problem to solve. In this problem, Solo can choose any non- empty subarray, the value of the subarray will be the <b>(sum of all the elements of the subarray - largest element of the subarray)</b>. Can you find the maximum value of any subarray?The first line of the input contains a single integer N, the size of the array.
The second line of the input contains N elements A[1], A[2],. , A[N].
Constraints
1 <= N <= 100000
-30 <= A[i] <= 30Output a single integer, the value of the subarray having maximum value.Sample Input
7
5 2 5 3 -30 -30 6
Sample Output
10
Explanation: Solo chooses the subarray [5, 2, 5, 3]. The value of the subarray is 10.
Sample Input
3
-10 6 -15
Sample Output
0, I have written this Solution Code: def maxSumTill(arr,n):
currSum = 0
maxSum = 0
maxEle = 0
maxi = 0
for i in range(1,n+1):
maxEle = max(maxEle,arr[i])
currSum += arr[i]
if currSum-maxEle < 0:
currSum = 0
maxEle =0
if currSum-maxEle > maxSum:
maxi = currSum - maxEle
maxSum = maxi
return maxi
def maxOfSum(arr,n):
currSum = 0
maxSum = 0
maxEle = 0
for i in range(1,n+1):
currSum += arr[i]
if currSum < 0:
currSum = 0
if currSum > maxSum:
maxSum = currSum
maxEle = max(maxEle, arr[i])
return maxSum - maxEle
n = int(input())
arr = list(map(int,input().split()))
arr = [0]+arr
val1 = maxSumTill(arr,n)
val2 = maxOfSum(arr,n)
print(max(val1,val2))
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono and Solo are one of the best programmers on the planet. Solo knows how to solve the problem on maximum subarray sum using the Kadane algorithm (don't you?). Now, Tono gives Solo a trickier problem to solve. In this problem, Solo can choose any non- empty subarray, the value of the subarray will be the <b>(sum of all the elements of the subarray - largest element of the subarray)</b>. Can you find the maximum value of any subarray?The first line of the input contains a single integer N, the size of the array.
The second line of the input contains N elements A[1], A[2],. , A[N].
Constraints
1 <= N <= 100000
-30 <= A[i] <= 30Output a single integer, the value of the subarray having maximum value.Sample Input
7
5 2 5 3 -30 -30 6
Sample Output
10
Explanation: Solo chooses the subarray [5, 2, 5, 3]. The value of the subarray is 10.
Sample Input
3
-10 6 -15
Sample Output
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 1e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
int arr[N];
int n;
int kadane(){
int mx = 0;
int cur = 0;
For(i, 0, n){
cur += arr[i];
mx = max(mx, cur);
if(cur < 0)
cur = 0;
}
return mx;
}
void play(int x){
For(i, 0, n){
if(arr[i]==x){
arr[i]=-1000000000;
}
}
}
void solve(){
cin>>n;
For(i, 0, n){
cin>>arr[i];
assert(arr[i]>=-30 && arr[i]<=30);
}
int ans = 0;
for(int i=30; i>=1; i--){
ans = max(ans, kadane()-i);
play(i);
}
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice has Q queries for you. If you can solve all her queries, she will invite you to participate in a fireworks display with her.
In each query, she gives you a positive integer N, and you have to find the number of positive integers M such that M × d(M) = N.
Here, d(M) refers to the sum of all digits of M. For example, d(1023) = 6.The first line consists of a single integer Q – the number of queries.
Then Q lines follow, each line containing a single integer N denoting a query.
<b> Constraints: </b>
1 ≤ Q ≤ 1000
1 ≤ N ≤ 10<sup>18</sup>Print Q lines, the i<sup>th</sup> line containing the answer to the i<sup>th</sup> query.Sample Input 1:
3
1
6
36
Sample Output 1:
1
0
2
Sample Explanation 1:
For the first query, the only possibility is 1.
For the third query, the only possibilities are 6 and 12., I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
private static final int START_TEST_CASE = 1;
public static void solveCase(FastIO io, int testCase) {
final long N = io.nextLong();
int count = 0;
for (int i = 1; i <= 200; ++i) {
if (N % i == 0) {
long j = N / i;
if (digitSum(j) == i) {
++count;
}
}
}
io.println(count);
}
private static long digitSum(long x) {
long s = 0;
while (x > 0) {
s += x % 10;
x /= 10;
}
return s;
}
public static void solve(FastIO io) {
final int T = io.nextInt();
for (int t = 0; t < T; ++t) {
solveCase(io, START_TEST_CASE + t);
}
}
public static class FastIO {
private InputStream reader;
private PrintWriter writer;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public FastIO(InputStream r, OutputStream w) {
reader = r;
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(w)));
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = reader.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
public String nextString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public double nextDouble() {
return Double.parseDouble(nextString());
}
public int[] nextIntArray(int n) {
return nextIntArray(n, 0);
}
public int[] nextIntArray(int n, int off) {
int[] arr = new int[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) {
return nextLongArray(n, 0);
}
public long[] nextLongArray(int n, int off) {
long[] arr = new long[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextLong();
}
return arr;
}
private boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void printArray(int[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printArray(long[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printlnArray(int[] arr) {
printArray(arr);
writer.println();
}
public void printlnArray(long[] arr) {
printArray(arr);
writer.println();
}
public void printf(String format, Object... args) {
print(String.format(format, args));
}
public void flush() {
writer.flush();
}
}
private static class Solution implements Runnable {
@Override
public void run() {
FastIO io = new FastIO(System.in, System.out);
solve(io);
io.flush();
}
}
public static void main(String[] args) throws InterruptedException {
Thread t = new Thread(null, new Solution(), "Solution", 1 << 30);
t.start();
t.join();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice has Q queries for you. If you can solve all her queries, she will invite you to participate in a fireworks display with her.
In each query, she gives you a positive integer N, and you have to find the number of positive integers M such that M × d(M) = N.
Here, d(M) refers to the sum of all digits of M. For example, d(1023) = 6.The first line consists of a single integer Q – the number of queries.
Then Q lines follow, each line containing a single integer N denoting a query.
<b> Constraints: </b>
1 ≤ Q ≤ 1000
1 ≤ N ≤ 10<sup>18</sup>Print Q lines, the i<sup>th</sup> line containing the answer to the i<sup>th</sup> query.Sample Input 1:
3
1
6
36
Sample Output 1:
1
0
2
Sample Explanation 1:
For the first query, the only possibility is 1.
For the third query, the only possibilities are 6 and 12., I have written this Solution Code:
q = int(input())
for _ in range(q):
n = int(input())
count = 0
for i in range(1,9*len(str(n))):
if not n % i:
dig = n//i
if sum(map(int,str(dig))) == i:
count += 1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice has Q queries for you. If you can solve all her queries, she will invite you to participate in a fireworks display with her.
In each query, she gives you a positive integer N, and you have to find the number of positive integers M such that M × d(M) = N.
Here, d(M) refers to the sum of all digits of M. For example, d(1023) = 6.The first line consists of a single integer Q – the number of queries.
Then Q lines follow, each line containing a single integer N denoting a query.
<b> Constraints: </b>
1 ≤ Q ≤ 1000
1 ≤ N ≤ 10<sup>18</sup>Print Q lines, the i<sup>th</sup> line containing the answer to the i<sup>th</sup> query.Sample Input 1:
3
1
6
36
Sample Output 1:
1
0
2
Sample Explanation 1:
For the first query, the only possibility is 1.
For the third query, the only possibilities are 6 and 12., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
// #pragma gcc optimize("ofast")
// #pragma gcc target("avx,avx2,fma")
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define endl '\n'
#define fi first
#define se second
// const int mod = 1e9 + 7;
const int mod=998'244'353;
const long long INF = 2e18 + 10;
// const int INF=4e9+10;
#define readv(x, n) \
vector<int> x(n); \
for (auto &i : x) \
cin >> i;
template <typename t>
using v = vector<t>;
template <typename t>
using vv = vector<vector<t>>;
template <typename t>
using vvv = vector<vector<vector<t>>>;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<vector<int>> vvi;
typedef vector<vector<vector<int>>> vvvi;
typedef vector<vector<vector<vector<int>>>> vvvvi;
typedef vector<vector<double>> vvd;
typedef pair<int, int> pii;
int multiply(int a, int b, int in_mod) { return (int)(1ll * a * b % in_mod); }
int mult_identity(int a) { return 1; }
const double pi = acosl(-1);
vector<vector<int> > multiply(vector<vector<int>> a, vector<vector<int>> b, int in_mod)
{
int n = a.size();
int l = b.size();
int m = b[0].size();
vector<vector<int> > result(n,vector<int>(n));
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
for(int k=0;k<l;k++)
{
result[i][j] = (result[i][j] + a[i][k]*b[k][j])%in_mod;
}
}
}
return result;
}
vector<vector<int>> operator%(vector<vector<int>> a, int in_mod)
{
for(auto &i:a)
for(auto &j:i)
j%=in_mod;
return a;
}
vector<vector<int>> mult_identity(vector<vector<int>> a)
{
int n=a.size();
vector<vector<int>> output(n, vector<int> (n));
for(int i=0;i<n;i++)
output[i][i]=1;
return output;
}
vector<int> mat_vector_product(vector<vector<int>> a, vector<int> b, int in_mod)
{
int n =a.size();
vector<int> output(n);
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
output[i]+=a[i][j]*b[j];
output[i]%=in_mod;
}
}
return output;
}
auto power(auto a, auto b, const int in_mod)
{
auto prod = mult_identity(a);
auto mult = a % in_mod;
while (b != 0)
{
if (b % 2)
{
prod = multiply(prod, mult, in_mod);
}
if(b/2)
mult = multiply(mult, mult, in_mod);
b /= 2;
}
return prod;
}
auto mod_inv(auto q, const int in_mod)
{
return power(q, in_mod - 2, in_mod);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define stp cout << fixed << setprecision(20);
int digit_sum(int x){
int sm = 0;
while(x){
sm += x%10;
x/= 10;
}
return sm;
}
void solv(){
int n;
cin>>n;
int cnt = 0;
for(int sm = 1;sm<= 200;sm++){
if(n %sm == 0){
if( digit_sum(n/sm) == sm){
cnt++;
}
}
}
cout<<cnt<<endl;
}
void solve()
{
int t = 1;
cin>>t;
for(int T=1;T<=t;T++)
{
// cout<<"Case #"<<T<<": ";
solv();
}
}
signed main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
auto clk = clock();
// -------------------------------------Code starts here---------------------------------------------------------------------
signed t = 1;
// cin >> t;
for (signed test = 1; test <= t; test++)
{
// cout<<"Case #"<<test<<": ";
solve();
}
// -------------------------------------Code ends here------------------------------------------------------------------
clk = clock() - clk;
#ifndef ONLINE_JUDGE
cerr << fixed << setprecision(6) << "\nTime: " << ((float)clk) / CLOCKS_PER_SEC << "\n";
#endif
return 0;
}
/*
000100
1000100
1 0 -1 -2 -1 -2 -3
*/
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
int[] arr=new int[n];
String[] s=br.readLine().trim().split(" ");
for(int i=0;i<arr.length;i++)
arr[i]=Integer.parseInt(s[i]);
int max=Integer.MIN_VALUE,c=0;
for(int i=0;i<arr.length;i++){
if(max==arr[i])
c++;
if(max<arr[i]){
max=arr[i];
c=1;
}
}
System.out.println(c);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: n=int(input())
a=list(map(int,input().split()))
print(a.count(max(a))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
vector<int> a(101, 0);
For(i, 0, n){
int x; cin>>x;
a[x]++;
}
for(int i=100; i>=1; i--){
if(a[i]){
cout<<a[i];
return;
}
}
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Insertion is a basic but frequently used operation. Arrays in most languages cannnot be dynamically shrinked or expanded. Here, we will work with such arrays and try to insert an element at the end of array.
You are given an array arr. The size of the array is N. You need to insert an element at its end and print this newly modified array.The first line of input contains T denoting the number of testcases.
T testcases follow. Each testcase contains two lines of input.
The first line contains size of the array denoted by N and element to be inserted.
The third line contains N elements separated by spaces.
Constraints:
1 <= T <= 20
2 <= N <= 10000
0 <= element, arri <= 10^6For each testcase, in a new line, print the modified array.Input:
2
5 90
1 2 3 4 5
3 50
1 2 3
Output:
1 2 3 4 5 90
1 2 3 50
Explanation:
Testcase 1: After inserting 90 at end, we have array elements as 1 2 3 4 5 90.
Testcase 2: After inserting 50 at end, we have array elements as 1 2 3 50., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main() {
int t;
cin>>t;
while(t--) {
long int n, el;
cin>>n>>el;
long int arr[n+1];
for(long int i=0; i<n; i++) {
cin>>arr[i];
}
arr[n] = el;
for(long int i=0; i<=n; i++) {
cout<<arr[i];
if (i != n) {
cout<<" ";
}
else if(t != 0) {
cout<<endl;
}
}
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Insertion is a basic but frequently used operation. Arrays in most languages cannnot be dynamically shrinked or expanded. Here, we will work with such arrays and try to insert an element at the end of array.
You are given an array arr. The size of the array is N. You need to insert an element at its end and print this newly modified array.The first line of input contains T denoting the number of testcases.
T testcases follow. Each testcase contains two lines of input.
The first line contains size of the array denoted by N and element to be inserted.
The third line contains N elements separated by spaces.
Constraints:
1 <= T <= 20
2 <= N <= 10000
0 <= element, arri <= 10^6For each testcase, in a new line, print the modified array.Input:
2
5 90
1 2 3 4 5
3 50
1 2 3
Output:
1 2 3 4 5 90
1 2 3 50
Explanation:
Testcase 1: After inserting 90 at end, we have array elements as 1 2 3 4 5 90.
Testcase 2: After inserting 50 at end, we have array elements as 1 2 3 50., I have written this Solution Code: t=int(input())
while t>0:
t-=1
li = list(map(int,input().strip().split()))
n=li[0]
num=li[1]
a= list(map(int,input().strip().split()))
a.insert(len(a),num)
for i in a:
print(i,end=" ")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Insertion is a basic but frequently used operation. Arrays in most languages cannnot be dynamically shrinked or expanded. Here, we will work with such arrays and try to insert an element at the end of array.
You are given an array arr. The size of the array is N. You need to insert an element at its end and print this newly modified array.The first line of input contains T denoting the number of testcases.
T testcases follow. Each testcase contains two lines of input.
The first line contains size of the array denoted by N and element to be inserted.
The third line contains N elements separated by spaces.
Constraints:
1 <= T <= 20
2 <= N <= 10000
0 <= element, arri <= 10^6For each testcase, in a new line, print the modified array.Input:
2
5 90
1 2 3 4 5
3 50
1 2 3
Output:
1 2 3 4 5 90
1 2 3 50
Explanation:
Testcase 1: After inserting 90 at end, we have array elements as 1 2 3 4 5 90.
Testcase 2: After inserting 50 at end, we have array elements as 1 2 3 50., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n =Integer.parseInt(br.readLine().trim());
while(n-->0){
String str[]=br.readLine().trim().split(" ");
String newel =br.readLine().trim()+" "+str[1];
System.out.println(newel);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
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