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For this Question: Sara is trying a new type of game in which she can jump in either the left direction or in the right direction. Also after each jump the range of her jump increases by 1 unit. i.e if starts from 1 in the next jump she has to jump 2 units then 3 units and so on.
Given the number of jumps as N, the range of the first jump to be 1. What will be the minimum distance Sara can be at from the starting point.<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>minDistanceCoveredBySara()</b> that takes integer N as an argument.
Constraints:-
1 <= N <= 1000Return the minimum distance Sara can be at from the starting point.Sample Input:-
3
Sample Output:-
0
Explanation:-
First jump:- right
Second jump:- right
Third jump:- left
Total distance covered = 1+2-3 = 0
Sample Input:-
5
Sample Output:-
1, I have written this Solution Code: int minDistanceCoveredBySara(int N){
if(N%4==1 || N%4==2){return 1;}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new type of game in which she can jump in either the left direction or in the right direction. Also after each jump the range of her jump increases by 1 unit. i.e if starts from 1 in the next jump she has to jump 2 units then 3 units and so on.
Given the number of jumps as N, the range of the first jump to be 1. What will be the minimum distance Sara can be at from the starting point.<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>minDistanceCoveredBySara()</b> that takes integer N as an argument.
Constraints:-
1 <= N <= 1000Return the minimum distance Sara can be at from the starting point.Sample Input:-
3
Sample Output:-
0
Explanation:-
First jump:- right
Second jump:- right
Third jump:- left
Total distance covered = 1+2-3 = 0
Sample Input:-
5
Sample Output:-
1, I have written this Solution Code: def minDistanceCoveredBySara(N):
if N%4==1 or N%4==2:
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new type of game in which she can jump in either the left direction or in the right direction. Also after each jump the range of her jump increases by 1 unit. i.e if starts from 1 in the next jump she has to jump 2 units then 3 units and so on.
Given the number of jumps as N, the range of the first jump to be 1. What will be the minimum distance Sara can be at from the starting point.<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>minDistanceCoveredBySara()</b> that takes integer N as an argument.
Constraints:-
1 <= N <= 1000Return the minimum distance Sara can be at from the starting point.Sample Input:-
3
Sample Output:-
0
Explanation:-
First jump:- right
Second jump:- right
Third jump:- left
Total distance covered = 1+2-3 = 0
Sample Input:-
5
Sample Output:-
1, I have written this Solution Code: static int minDistanceCoveredBySara(int N){
if(N%4==1 || N%4==2){return 1;}
return 0;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a number n. Your task is to print the number of prime numbers before that number.The first line of the number of test cases T.
Next T lines contains the value of N.
<b>Constraints</b>
1 <= T <= 100
1 <= N <= 1000Print the number of primes numbers before that number.Sample Input 1:
3
10
19
4
Sample Output 1:
4
8
2, I have written this Solution Code: n = 1000
arr = [True for i in range(n+1)]
i = 2
while i*i <= n:
if arr[i] == True:
for j in range(i*2, n+1, i):
arr[j] = False
i +=1
arr2 = [0] * (n+1)
for i in range(2,n+1):
if arr[i]:
arr2[i] = arr2[i-1] + 1
else:
arr2[i] = arr2[i-1]
x = int(input())
for i in range(x):
y = int(input())
print(arr2[y]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a number n. Your task is to print the number of prime numbers before that number.The first line of the number of test cases T.
Next T lines contains the value of N.
<b>Constraints</b>
1 <= T <= 100
1 <= N <= 1000Print the number of primes numbers before that number.Sample Input 1:
3
10
19
4
Sample Output 1:
4
8
2, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
vector<bool> sieve(int n) {
vector<bool> is_prime(n + 1, true);
is_prime[0] = is_prime[1] = false;
for (int i = 2; i * i <= n; i++) {
if (is_prime[i]) {
for (int j = i * i; j <= n; j += i)
is_prime[j] = false;
}
}
return is_prime;
}
int main() {
vector<bool> prime = sieve(1e5 + 1);
vector<int> prefix(1e5 + 1, 0);
for (int i = 1; i <= 1e5; i++) {
if (prime[i]) {
prefix[i] = prefix[i - 1] + 1;
} else {
prefix[i] = prefix[i - 1];
}
}
int tt;
cin >> tt;
while (tt--) {
int n;
cin >> n;
cout << prefix[n] << "\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two positive integers X and Y, check if one integer is obtained by rotating bits of other.
Bit Rotation: A rotation (or circular shift) is an operation similar to shift except that the bits that fall off at one end are put back to the other end.The only line of input contains two integers X and Y
0 <= X, Y <= 2^32 - 1
The numbers are in 32-bit formatPrint "Yes" if its possible otherwise print "No"Sample Input:
8 1
Output:
Yes
Explanation :
Represntation of X = 8 : 0000 0000 0000 0000 0000 0000 0000 1000
Represntation of Y = 1 : 0000 0000 0000 0000 0000 0000 0000 0001
If we rotate X by 3 units right we get Y, hence answer is Yes, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static boolean isRotation(long x, long y) {
int i = 32;
while (i-- > 0) {
long lastBit = (x & 1);
x >>= 1;
x |= (lastBit << 31);
if(x == y){
return true;
}
}
return false;
}
public static void main(String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str[] = br.readLine().strip().split(" ");
long x = Long.parseLong(str[0]);
long y = Long.parseLong(str[1]);
if (isRotation(x, y)) {
System.out.println("Yes");
} else {
System.out.println("No");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two positive integers X and Y, check if one integer is obtained by rotating bits of other.
Bit Rotation: A rotation (or circular shift) is an operation similar to shift except that the bits that fall off at one end are put back to the other end.The only line of input contains two integers X and Y
0 <= X, Y <= 2^32 - 1
The numbers are in 32-bit formatPrint "Yes" if its possible otherwise print "No"Sample Input:
8 1
Output:
Yes
Explanation :
Represntation of X = 8 : 0000 0000 0000 0000 0000 0000 0000 1000
Represntation of Y = 1 : 0000 0000 0000 0000 0000 0000 0000 0001
If we rotate X by 3 units right we get Y, hence answer is Yes, I have written this Solution Code: inp = input().split(" ")
x = int(inp[0])
y = int(inp[1])
counter = 32
while(counter != 0):
z = x & 1
x = x >> 1
x = x + (z << 31)
if(x == y):
print("Yes")
exit()
counter -= 1
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two positive integers X and Y, check if one integer is obtained by rotating bits of other.
Bit Rotation: A rotation (or circular shift) is an operation similar to shift except that the bits that fall off at one end are put back to the other end.The only line of input contains two integers X and Y
0 <= X, Y <= 2^32 - 1
The numbers are in 32-bit formatPrint "Yes" if its possible otherwise print "No"Sample Input:
8 1
Output:
Yes
Explanation :
Represntation of X = 8 : 0000 0000 0000 0000 0000 0000 0000 1000
Represntation of Y = 1 : 0000 0000 0000 0000 0000 0000 0000 0001
If we rotate X by 3 units right we get Y, hence answer is Yes, I have written this Solution Code: #include "bits/stdc++.h"
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 = 500 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N][N];
void bin(int x){
for(int i = 0; i < 32; i++){
cout << ((x >> i) & 1);
}
cout << endl;
}
void solve(){
int x = 32;
int a, b;
cin >> a >> b;
while(x--){
int x = (a&1);
a >>= 1;
a += x*(1LL << 31);
/*bin(a); bin(b);
cout << endl;*/
if(a == b){
cout << "Yes";
return;
}
}
cout << "No";
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N and a number P, your task is to calculate N<sup>P</sup>. Since the ans can be very long calculate your ans as N<sup>P</sup>%M where M = 10<sup>9</sup>+7Input contains only two integers N and P separated by spaces.
Constraints:-
1 < = N, P <= 1000000000Print N<sup>P</sup>%M.Sample Input:-
2 5
Sample Output:-
32
Sample Input:-
2 100
Sample Output:-
976371285, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException{
BufferedReader rd=new BufferedReader(new InputStreamReader(System.in));
String y[]=rd.readLine().split(" ");
long n=Long.parseLong(y[0]);
long p=Long.parseLong(y[1]);
long v=1;
while(p>0){
if((p&1L)==1L)
v=(v*n)%1000000007;
p/=2;
n=(n*n)%1000000007;
}
System.out.print(v);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N and a number P, your task is to calculate N<sup>P</sup>. Since the ans can be very long calculate your ans as N<sup>P</sup>%M where M = 10<sup>9</sup>+7Input contains only two integers N and P separated by spaces.
Constraints:-
1 < = N, P <= 1000000000Print N<sup>P</sup>%M.Sample Input:-
2 5
Sample Output:-
32
Sample Input:-
2 100
Sample Output:-
976371285, I have written this Solution Code: n, p =input().split()
n, p =int(n), int(p)
def FastModularExponentiation(b, k, m):
res = 1
b = b % m
while (k > 0):
if ((k & 1) == 1):
res = (res * b) % m
k = k >> 1
b = (b * b) % m
return res
m=pow(10,9)+7
print(FastModularExponentiation(n, p, m)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N and a number P, your task is to calculate N<sup>P</sup>. Since the ans can be very long calculate your ans as N<sup>P</sup>%M where M = 10<sup>9</sup>+7Input contains only two integers N and P separated by spaces.
Constraints:-
1 < = N, P <= 1000000000Print N<sup>P</sup>%M.Sample Input:-
2 5
Sample Output:-
32
Sample Input:-
2 100
Sample Output:-
976371285, I have written this Solution Code:
// author-Shivam gupta
#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 MOD 1000000007
#define read(type) readInt<type>()
#define max1 100001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int power(int x, unsigned int y, int p)
{
int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
if (x == 0) return 0; // In case x is divisible by p;
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
// Driver code
signed main()
{
int x ;
int y;
cin>>x>>y;
int p = 1e9+7;
cout<< power(x, y, p);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are a substitute teacher of a merged class of students of
different sections.
A few minutes before the period gets over you circulated a sheet of paper for attendance.
You told the students to first write their section(A, B, C, D..., Z) and then their roll number(1, 2, 3....) on the sheet of paper.
Some students were naughty so they wrote their section with roll numbers more than once on the attendance sheet.
Now, Ms. Rekha, the head teacher of the section <code>sec</code> wants the roll numbers of her class students to mark attendance in her register.
She requested you to give the attendance in ascending order with non-repeating roll numbers of her section's students only.
Your task is to complete the function <code> getStudentRollNumber() </code> which outputs an array of non-repeating roll numbers of a particular section.The function will take two arguments
1) 1st argument will be a string of one character
2) 2nd argument will be the section and roll numbers separated by spaceThe function will return an array of a string of unique roll numbers, arranged in ascending order<pre>
<code>
const section='B'
const arr=[ 'B1', 'B5', 'A12', 'B45', 'C22', 'B1', 'A1' ]
const answer=getStudentRollNumber(section,arr)
//returns an array of unique string roll numbers in ascending order
console.log(answer)
//[ '1', '5', '45' ]
const section='A'
const arr=[ 'B1', 'B5', 'A12', 'B45', 'C22', 'B1', 'A100', 'D2', 'D2', 'A1', 'A100', 'A100', 'A22' ]
const answer=getStudentRollNumber(section,arr)
//returns an array of unique string roll numbers in ascending order
console.log(answer)
//[ '1', '12', '22', '100' ]
</code>
</pre>
, I have written this Solution Code:
function getStudentRollNumber(section,array){
array=array.filter(e=>e.includes(section)) ;
//filtering the students and taking the students of required section only
array=array.map(w=>w.split(section).pop()) ;
//removing Alphabet section and taking only roll numbers
array=[...new Set(array)];
//removing roll numberes coming more than once
array=array.sort(function(a, b){return a-b})
//sorting the array ,the function defines the sort order
return array;
}, 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 (N is always even), you need to find exactly (N/2) numbers where each number represents the pair-wise sum of consecutive elements of the array A.
In simple terms print (A[1]+A[2]), (A[3]+A[4]), ..., (A[N-1]+A[N]).The first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N(even number) denoting the number of elements in the array. The next line contains N (white-space separated) integers.
Constraints
1 <= N <= 100
1 <= A[I] <= 1000000000For each test case, output N/2 elements representing the pairwise sum of adjacent elements in the array.Input:-1
4
1 2 6 4
output-1
3 10
input-2
10
1 2 3 4 5 6 0 7 8 9
output-2
3 7 11 7 17
Explanation(might now be the optimal solution):
Testcase 1:
Follow the below steps:-
Step 1: [1 2 6 4]
Step 2: (1 2) and (6 4)
Step 3: 3 10, I have written this Solution Code: n = int(input())
all_no = input().split(' ')
i = 0
joined_str = ''
while(i < n-1):
if(i == 0):
joined_str = str(int(all_no[i]) + int(all_no[i+1]))
else:
joined_str = joined_str + ' ' + str(int(all_no[i]) + int(all_no[i+1]))
i = i + 2
print(joined_str), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N (N is always even), you need to find exactly (N/2) numbers where each number represents the pair-wise sum of consecutive elements of the array A.
In simple terms print (A[1]+A[2]), (A[3]+A[4]), ..., (A[N-1]+A[N]).The first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N(even number) denoting the number of elements in the array. The next line contains N (white-space separated) integers.
Constraints
1 <= N <= 100
1 <= A[I] <= 1000000000For each test case, output N/2 elements representing the pairwise sum of adjacent elements in the array.Input:-1
4
1 2 6 4
output-1
3 10
input-2
10
1 2 3 4 5 6 0 7 8 9
output-2
3 7 11 7 17
Explanation(might now be the optimal solution):
Testcase 1:
Follow the below steps:-
Step 1: [1 2 6 4]
Step 2: (1 2) and (6 4)
Step 3: 3 10, 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();
}
int t;
for(int i=0;i<n;i+=2){
System.out.print(a[i]+a[i+1]+" ");
}
}
}, 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 (N is always even), you need to find exactly (N/2) numbers where each number represents the pair-wise sum of consecutive elements of the array A.
In simple terms print (A[1]+A[2]), (A[3]+A[4]), ..., (A[N-1]+A[N]).The first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N(even number) denoting the number of elements in the array. The next line contains N (white-space separated) integers.
Constraints
1 <= N <= 100
1 <= A[I] <= 1000000000For each test case, output N/2 elements representing the pairwise sum of adjacent elements in the array.Input:-1
4
1 2 6 4
output-1
3 10
input-2
10
1 2 3 4 5 6 0 7 8 9
output-2
3 7 11 7 17
Explanation(might now be the optimal solution):
Testcase 1:
Follow the below steps:-
Step 1: [1 2 6 4]
Step 2: (1 2) and (6 4)
Step 3: 3 10, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
long a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<n;i+=2){
cout<<a[i]+a[i+1]<<" ";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Olivia likes triangles. Hence, John decided to give her three integers and ask whether a triangle with edge length as given three integers is possible.
Help Olivia in answering it.The input consists of a single line containing three space-separated integers A, B, and C.
<b>Constraints </b>
1 <= A, B, C <= 100Output "Yes" if the triangle is possible otherwise, "No" (without quotes).Sample Input 1:
5 3 4
Sample Output 1:
Yes
Sample Explanation 1:
The possible triangle is a right-angled triangle with a hypotenuse of 5., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int a, b, c;
cin >> a >> b >> c;
if(a<(b+c) && b<(c+a) && c<(a+b)){
cout << "Yes\n";
}else{
cout << "No\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, 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(" ");
long l=Long.parseLong(str[0]);
long r=Long.parseLong(str[1]);
long k=Long.parseLong(str[2]);
long count=(r/k)-((l-1)/k);
System.out.print(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code: l,r,k=[int(x)for x in input().split()]
print(r//k - (l-1)//k), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
unsigned long long x,l,r,k;
cin>>l>>r>>k;
x=l/k;
if(l%k==0){x--;}
cout<<r/k-x;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers x and y, check if x can be converted to y by adding F(n) to x where F(n){n>=1} is defined as:- 1+3+9+27+81+. .3^n.<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>Ifpossible()</b> that takes the integer x and y as parameter.
Constraints:-
1<= x < y <= 10^9Return 1 if it is possible to convert x into y else return 0.Sample Input:-
5 7
Sample Output:-
0
Sample Input:-
3 16
Sample Output:-
1
Explanation:-
F(3) = 1 + 3 + 9 = 13
3 + 13 = 16, I have written this Solution Code:
public static int Ifpossible(long x, long y){
long sum=y-x;
long ans=0;
long ch = 1;
while(ans<sum){
ans+=ch;
if(ans==sum){return 1;}
ch*=3;
}
return 0;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers x and y, check if x can be converted to y by adding F(n) to x where F(n){n>=1} is defined as:- 1+3+9+27+81+. .3^n.<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>Ifpossible()</b> that takes the integer x and y as parameter.
Constraints:-
1<= x < y <= 10^9Return 1 if it is possible to convert x into y else return 0.Sample Input:-
5 7
Sample Output:-
0
Sample Input:-
3 16
Sample Output:-
1
Explanation:-
F(3) = 1 + 3 + 9 = 13
3 + 13 = 16, I have written this Solution Code:
int Ifpossible(long x, long y){
long sum=y-x;
long ans=0;
long ch = 1;
while(ans<sum){
ans+=ch;
if(ans==sum){return 1;break;}
ch*=(long)3;
}
return 0;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers x and y, check if x can be converted to y by adding F(n) to x where F(n){n>=1} is defined as:- 1+3+9+27+81+. .3^n.<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>Ifpossible()</b> that takes the integer x and y as parameter.
Constraints:-
1<= x < y <= 10^9Return 1 if it is possible to convert x into y else return 0.Sample Input:-
5 7
Sample Output:-
0
Sample Input:-
3 16
Sample Output:-
1
Explanation:-
F(3) = 1 + 3 + 9 = 13
3 + 13 = 16, I have written this Solution Code:
int Ifpossible(long x, long y){
long sum=y-x;
long ans=0;
long ch = 1;
while(ans<sum){
ans+=ch;
if(ans==sum){return 1;}
ch*=3;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers x and y, check if x can be converted to y by adding F(n) to x where F(n){n>=1} is defined as:- 1+3+9+27+81+. .3^n.<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>Ifpossible()</b> that takes the integer x and y as parameter.
Constraints:-
1<= x < y <= 10^9Return 1 if it is possible to convert x into y else return 0.Sample Input:-
5 7
Sample Output:-
0
Sample Input:-
3 16
Sample Output:-
1
Explanation:-
F(3) = 1 + 3 + 9 = 13
3 + 13 = 16, I have written this Solution Code:
def Ifpossible(x,y) :
result = y-x
ans = 0
ch = 1
while ans<result :
ans+=ch
if ans==result:
return 1;
ch*=3
return 0;
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>nums</b> of N integers where each element is in between 1 to N. Your task is to print all the integers from 1 to N which are not present in the given array.First line contains n the length of input
Second line of input contains the array of integers in range from (1 to n)
<b> Constraints </b>
1 <= n <= 100000
1 <= arr[i] <= nPrint all the integers in range [1, n] that dose not appear in array.Sample Input :
5
1 2 3 4 4
Sample Output :
5
Sample Input:
7
4 1 3 2 5 5 4
Sample Output:
6 7
Explaination:
In the first test, only 5 is the integer from 1 to n, which is missing from the array.
In the second test, both 6 and 7 are missing., 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();
int[] arr=new int[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
}
List<Integer> ans=missing(arr);
for(int i=0;i<ans.size();i++){
System.out.print(ans.get(i)+" ");
}
System.out.println();
}
static List<Integer> missing(int[] arr){
int i=0;
while (i< arr.length){
int correct=arr[i]-1;
if (arr[i]!=arr[correct]){
swap(arr,i,correct);
}else {
i++;
}
}
List<Integer> ans=new ArrayList<>();
for (int j = 0; j < arr.length; j++) {
if (arr[j] != j+1) {
ans.add(j + 1);
}
}
return ans;
}
static void swap(int[] arr,int first,int second){
int temp=arr[first];
arr[first]=arr[second];
arr[second]=temp;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>nums</b> of N integers where each element is in between 1 to N. Your task is to print all the integers from 1 to N which are not present in the given array.First line contains n the length of input
Second line of input contains the array of integers in range from (1 to n)
<b> Constraints </b>
1 <= n <= 100000
1 <= arr[i] <= nPrint all the integers in range [1, n] that dose not appear in array.Sample Input :
5
1 2 3 4 4
Sample Output :
5
Sample Input:
7
4 1 3 2 5 5 4
Sample Output:
6 7
Explaination:
In the first test, only 5 is the integer from 1 to n, which is missing from the array.
In the second test, both 6 and 7 are missing., I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-01-08 12:34:09
**/
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
vector<int> findDisappearedNumbers(vector<int>& nums) {
int i = 0, size = nums.size();
while (i < size) {
if (nums[i] != i + 1 && nums[nums[i] - 1] != nums[i])
swap(nums[i], nums[nums[i] - 1]);
else
i++;
}
vector<int> ans;
for (i = 0; i < size; i++)
if (nums[i] != i + 1) ans.push_back(i + 1);
return ans;
}
int main() {
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
vector<int> x = findDisappearedNumbers(a);
for (auto& it : x) {
cout << it << " ";
}
cout << "\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The head of the doubly linked list will be given to you, and you must sort it using merge sort.You will be given the head of doubly Linked List and its length.
Constraints:
0<= n <=10000Return the HEAD of the sorted linked list.Sample Input:
5
3 2 1 3 2
Output:
1 2 2 3 3, I have written this Solution Code: class Node:
def __init__(self, data, next=None, prev=None):
self.data = data
self.next = next
self.prev = prev
def push(head, key):
node = Node(key, head)
if head:
head.prev = node
return node
def printDDL(head):
while head:
print(head.data, end=' ')
head = head.next
def split(head):
slow = head
fast = head.next
while fast:
fast = fast.next
if fast:
slow = slow.next
fast = fast.next
return slow
def merge(a, b):
if a is None:
return b
if b is None:
return a
if a.data <= b.data:
a.next = merge(a.next, b)
a.next.prev = a
a.prev = None
return a
else:
b.next = merge(a, b.next)
b.next.prev = b
b.prev = None
return b
def mergesort(head):
if head is None or head.next is None:
return head
a = head
slow = split(head)
b = slow.next
slow.next = None
a = mergesort(a)
b = mergesort(b)
head = merge(a, b)
return head
n=int(input())
keys=list(map(int,input().split()))
head = None
for key in keys:
head = push(head, key)
head = mergesort(head)
printDDL(head), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The head of the doubly linked list will be given to you, and you must sort it using merge sort.You will be given the head of doubly Linked List and its length.
Constraints:
0<= n <=10000Return the HEAD of the sorted linked list.Sample Input:
5
3 2 1 3 2
Output:
1 2 2 3 3, I have written this Solution Code:
Node split(Node head) {
Node fast = head, slow = head;
while (fast.next != null && fast.next.next != null) {
fast = fast.next.next;
slow = slow.next;
}
Node temp = slow.next;
slow.next = null;
return temp;
}
Node mergeSort(Node node, int n) {
if (node == null || node.next == null) {
return node;
}
Node second = split(node);
node = mergeSort(node,n);
second = mergeSort(second,n);
return merge(node, second);
}
Node merge(Node first, Node second) {
if (first == null) {
return second;
}
if (second == null) {
return first;
}
if (first.data < second.data) {
first.next = merge(first.next, second);
first.next.prev = first;
first.prev = null;
return first;
} else {
second.next = merge(first, second.next);
second.next.prev = second;
second.prev = null;
return second;
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<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>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code:
int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<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>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code:
int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<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>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code: static int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<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>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code: def mythOrFact(N):
prime=1
cnt=N+1
for i in range (2,N):
if N%i==0:
prime=0
cnt=cnt+i
x = 3*N
x=x//2
if(cnt <= x and prime==1) or (cnt>x and prime==0):
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list, the task is to move all 0’s to the front of the linked list. The order of all another element except 0 should be same after rearrangement.
Note: Avoid use of any type of Java Collection frameworks.
Note: For custom input/output, enter the list in reverse order, and the output will come in right order.<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>moveZeroes()</b> that takes head node as parameter.
Constraints:
1 <= T <= 100
1 <= N <= 100000
0<=Node.data<=100000
Note:- Sum of all test cases doesn't exceed 10^5
Return the head of the modified linked list.Input:
2
10
0 4 0 5 0 2 1 0 1 0
7
1 1 2 3 0 0 0
Output:
0 0 0 0 0 4 5 2 1 1
0 0 0 1 1 2 3
Explanation:
Testcase 1:
Original list was 0->4->0->5->0->2->1->0->1->0->NULL.
After processing list becomes 0->0->0->0->0->4->5->2->1->1->NULL.
Testcase 2:
Original list was 1->1->2->3->0->0->0->NULL.
After processing list becomes 0->0->0->1->1->2->3->NULL., I have written this Solution Code: static public Node moveZeroes(Node head){
ArrayList<Integer> a=new ArrayList<>();
int c=0;
while(head!=null){
if(head.data==0){
c++;
}
else{
a.add(head.data);
}
head=head.next;
}
head=null;
for(int i=a.size()-1;i>=0;i--){
if(head==null){
head=new Node(a.get(i));
}
else{
Node temp=new Node(a.get(i));
temp.next=head;
head=temp;
}
}
while(c-->0){
Node temp=new Node(0);
temp.next=head;
head=temp;
}
return head;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of N integers arranged in a circular fashion. Your task is to find the maximum contiguous subarray sum containing at least 1 element.The first line of input contains a single integer T which denotes the number of test cases. The first line of each test case contains a single integer N which denotes the total number of elements. The second line of each test case contains N space-separated integers denoting the elements of the array.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^5
-10^6 <= Arr[i] <= 10^6
The Sum of N over all test cases is less than equal to 10^6.For each test case print the maximum sum obtained by adding the consecutive elements.<b>Input:</b>
4
7
8 -8 9 -9 10 -11 12
8
10 -3 -4 7 6 5 -4 -1
8
-1 40 -14 7 6 5 -4 -1
4
-1 -2 -3 -4
<b>Output:</b>
22
23
52
-1
Explanation:
Testcase 1: Starting from the last element of the array, i.e, 12, and moving in a circular fashion, we have max subarray as 12, 8, -8, 9, -9, 10, which gives the maximum sum as 22., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
public static long maxSubarraySumCircular(long[] array) {
long currentSum = 0, maxSum = Long.MIN_VALUE;
for(int i = 0; i < array.length; i++) {
currentSum = Math.max(currentSum + array[i], array[i]);
maxSum = Math.max(maxSum, currentSum);
}
if(maxSum < 0) return maxSum;
currentSum = 0;
long minSum = Long.MAX_VALUE;
for(int i = 0; i < array.length; i++) {
currentSum = Math.min(currentSum + array[i], array[i]);
minSum = Math.min(minSum, currentSum);
}
long totalSum = 0;
for(long element : array) totalSum += element;
return Math.max(maxSum, totalSum - minSum);
}
public static void main (String[] args) {
FastReader sc = new FastReader();
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
long a[] = new long[n];
long sum = 0l;
for(int i=0;i<n;i++){
a[i] = sc.nextLong();
}
System.out.println(maxSubarraySumCircular(a));
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of N integers arranged in a circular fashion. Your task is to find the maximum contiguous subarray sum containing at least 1 element.The first line of input contains a single integer T which denotes the number of test cases. The first line of each test case contains a single integer N which denotes the total number of elements. The second line of each test case contains N space-separated integers denoting the elements of the array.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^5
-10^6 <= Arr[i] <= 10^6
The Sum of N over all test cases is less than equal to 10^6.For each test case print the maximum sum obtained by adding the consecutive elements.<b>Input:</b>
4
7
8 -8 9 -9 10 -11 12
8
10 -3 -4 7 6 5 -4 -1
8
-1 40 -14 7 6 5 -4 -1
4
-1 -2 -3 -4
<b>Output:</b>
22
23
52
-1
Explanation:
Testcase 1: Starting from the last element of the array, i.e, 12, and moving in a circular fashion, we have max subarray as 12, 8, -8, 9, -9, 10, which gives the maximum sum as 22., I have written this Solution Code: def kadane(a):
Max = a[0]
temp = Max
for i in range(1,len(a)):
temp += a[i]
if temp < a[i]:
temp = a[i]
Max = max(Max,temp)
return Max
def maxCircularSum(a):
n = len(a)
max_kadane = kadane(a)
neg_a = [-1*x for x in a]
max_neg_kadane = kadane(neg_a)
max_wrap = -(sum(neg_a)-max_neg_kadane)
res = max(max_wrap,max_kadane)
return res if res != 0 else max_kadane
for _ in range(int(input())):
s=int(input())
a=list(map(int,input().split()))
print(maxCircularSum(a)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of N integers arranged in a circular fashion. Your task is to find the maximum contiguous subarray sum containing at least 1 element.The first line of input contains a single integer T which denotes the number of test cases. The first line of each test case contains a single integer N which denotes the total number of elements. The second line of each test case contains N space-separated integers denoting the elements of the array.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^5
-10^6 <= Arr[i] <= 10^6
The Sum of N over all test cases is less than equal to 10^6.For each test case print the maximum sum obtained by adding the consecutive elements.<b>Input:</b>
4
7
8 -8 9 -9 10 -11 12
8
10 -3 -4 7 6 5 -4 -1
8
-1 40 -14 7 6 5 -4 -1
4
-1 -2 -3 -4
<b>Output:</b>
22
23
52
-1
Explanation:
Testcase 1: Starting from the last element of the array, i.e, 12, and moving in a circular fashion, we have max subarray as 12, 8, -8, 9, -9, 10, which gives the maximum sum as 22., 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 EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 10001
#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
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int INF = 4557430888798830399ll;
signed main()
{
fast();
int t;
cin>>t;
while(t--){
int n;
cin>>n;
vector<int> A(n);
FOR(i,n){
cin>>A[i];
}
int maxTillNow = -INF;
int maxEndingHere = -INF;
int start = 0;
while (start < A.size()) {
maxEndingHere = max(maxEndingHere + A[start], (int)A[start]);
maxTillNow = max(maxTillNow, maxEndingHere);
start++;
}
vector<int> prefix(n, 0), suffix(n, 0), maxSuffTill(n, 0);
for (int i = 0; i < n; ++i) {
prefix[i] = A[i];
if (i != 0) prefix[i] += prefix[i - 1];
}
for (int i = n - 1; i >= 0; --i) {
suffix[i] = A[i];
maxSuffTill[i] = max(A[i],(int) 0);
if (i != n - 1) {
suffix[i] += suffix[i + 1];
maxSuffTill[i] = max(suffix[i], maxSuffTill[i + 1]);
}
}
for (int i = 0; i < n; ++i) {
int sum = prefix[i];
if (i != n - 1) sum += maxSuffTill[i + 1];
maxTillNow = max(maxTillNow, sum);
}
out(maxTillNow);
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array A of size N of non-negative integers, find a continuous sub-array which adds to a given number S.Each test case consists of two lines. The first line of each test case is N and S, where N is the size of the array and S is the sum. The second line of each test case contains N space-separated integers denoting the array elements.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ Ai ≤ 10<sup>5</sup>Print the starting and ending positions (1 indexing) of first such occurring subarray from the left if sum equals to subarray, else print -1.Sample Input
5 12
1 2 3 7 5
Sample Output
2 4
Explanation:
subarray starting from index 2 and ending at index 4 => {2 , 3 , 7}
sum = 2 + 3 + 7 = 12
Sample Input
10 15
1 2 3 4 5 6 7 8 9 10
Sample Output
1 5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException
{
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String StrInput[] = br.readLine().trim().split(" ");
int n = Integer.parseInt(StrInput[0]);
int s = Integer.parseInt(StrInput[1]);
int arr[] = new int[n];
String StrInput2[] = br.readLine().trim().split(" ");
for(int i=0;i<n;i++)
{
arr[i] = Integer.parseInt(StrInput2[i]);
}
int sum = arr[0];
int startingindex = 0;
int endingindex = 1;
int j = 0;
int i;
for(i=1;i<=n;i++)
{
if(sum < s && arr[i] != 0)
{
sum += arr[i];
}
while(sum > s && startingindex < i-1)
{
sum -= arr[startingindex];
startingindex++;
}
if(sum == s)
{
endingindex = i+1;
if(arr[0] == 0)
{
System.out.print(startingindex+2 + " " + endingindex);
}
else
{
System.out.print(startingindex+1 + " "+ endingindex);
}
break;
}
if(i == n && sum < s)
{
System.out.print(-1);
break;
}
}
}
catch(Exception e)
{
System.out.print(-1);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array A of size N of non-negative integers, find a continuous sub-array which adds to a given number S.Each test case consists of two lines. The first line of each test case is N and S, where N is the size of the array and S is the sum. The second line of each test case contains N space-separated integers denoting the array elements.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ Ai ≤ 10<sup>5</sup>Print the starting and ending positions (1 indexing) of first such occurring subarray from the left if sum equals to subarray, else print -1.Sample Input
5 12
1 2 3 7 5
Sample Output
2 4
Explanation:
subarray starting from index 2 and ending at index 4 => {2 , 3 , 7}
sum = 2 + 3 + 7 = 12
Sample Input
10 15
1 2 3 4 5 6 7 8 9 10
Sample Output
1 5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n,k;
cin>>n>>k;
int a[n];
for(int i=0;i<n;i++)
{
cin>>a[i];
}
int sum=0;
unordered_map<int,int> m;
for(int i=0;i<n;i++){
sum+=a[i];
if(sum==k){cout<<1<<" "<<i+1;return 0;}
if(m.find(sum-k)!=m.end()){
cout<<m[sum-k]+2<<" "<<i+1;
return 0;
}
m[sum]=i;
}
cout<<-1;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array A of size N of non-negative integers, find a continuous sub-array which adds to a given number S.Each test case consists of two lines. The first line of each test case is N and S, where N is the size of the array and S is the sum. The second line of each test case contains N space-separated integers denoting the array elements.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ Ai ≤ 10<sup>5</sup>Print the starting and ending positions (1 indexing) of first such occurring subarray from the left if sum equals to subarray, else print -1.Sample Input
5 12
1 2 3 7 5
Sample Output
2 4
Explanation:
subarray starting from index 2 and ending at index 4 => {2 , 3 , 7}
sum = 2 + 3 + 7 = 12
Sample Input
10 15
1 2 3 4 5 6 7 8 9 10
Sample Output
1 5, I have written this Solution Code: def sumFinder(N,S,a):
currentSum = a[0]
start = 0
i = 1
while i <= N:
while currentSum > S and start < i-1:
currentSum = currentSum - a[start]
start += 1
if currentSum == S:
return (start+1,i)
if i < N:
currentSum = currentSum + a[i]
i += 1
return(-1)
N, S = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
ans = sumFinder(N,S,a)
if(ans==-1):
print(ans)
else:
print(ans[0],ans[1]), In this Programming Language: Python, 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: Given an integer array <strong>arr[]</strong> of size <strong>N</strong> and an element <strong>X</strong>. The task is to find and print the indices of the given element if it is present in array if not then print “<strong>Not found</strong>” without quotes.
<strong>Note</strong>: The elements may be present more than once.The first line of input contains T, denoting the number of test cases.
The first line of each test case contains N and X, N is the size of array and X is an element. Second line contains elements of array space separated. If not present then print "Not found" without quotes
<strong>Constraints</strong>:
1 <= T <= 100
1 <= N, X <= 10000
1 <= arr[i] <= 100000For each test case in new line you need to print all the positions where you find the X separated by space.
Assume 0-indexingInput:
2
5 6
2 3 6 5 6
4 3
2 4 6 5
Output:
2 4
Not found, 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;
signed main() {
IOS;
int t; cin >> t;
while(t--){
vector<int> v;
int n, x; cin >> n >> x;
for(int i = 1; i <= n; i++){
int p; cin >> p;
if(p == x)
v.push_back(i-1);
}
if(v.size() == 0)
cout << "Not found\n";
else{
for(auto i: v)
cout << 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 integer array <strong>arr[]</strong> of size <strong>N</strong> and an element <strong>X</strong>. The task is to find and print the indices of the given element if it is present in array if not then print “<strong>Not found</strong>” without quotes.
<strong>Note</strong>: The elements may be present more than once.The first line of input contains T, denoting the number of test cases.
The first line of each test case contains N and X, N is the size of array and X is an element. Second line contains elements of array space separated. If not present then print "Not found" without quotes
<strong>Constraints</strong>:
1 <= T <= 100
1 <= N, X <= 10000
1 <= arr[i] <= 100000For each test case in new line you need to print all the positions where you find the X separated by space.
Assume 0-indexingInput:
2
5 6
2 3 6 5 6
4 3
2 4 6 5
Output:
2 4
Not found, I have written this Solution Code: def position(n,arr,x):
res = []
cnt = 0
for i in arr:
if(i == x):
res.append(cnt)
cnt += 1
return res
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array <strong>arr[]</strong> of size <strong>N</strong> and an element <strong>X</strong>. The task is to find and print the indices of the given element if it is present in array if not then print “<strong>Not found</strong>” without quotes.
<strong>Note</strong>: The elements may be present more than once.The first line of input contains T, denoting the number of test cases.
The first line of each test case contains N and X, N is the size of array and X is an element. Second line contains elements of array space separated. If not present then print "Not found" without quotes
<strong>Constraints</strong>:
1 <= T <= 100
1 <= N, X <= 10000
1 <= arr[i] <= 100000For each test case in new line you need to print all the positions where you find the X separated by space.
Assume 0-indexingInput:
2
5 6
2 3 6 5 6
4 3
2 4 6 5
Output:
2 4
Not found, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main (String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t =Integer.parseInt(read.readLine());
while(t-- > 0)
{
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
int x = Integer.parseInt(str[1]);
int arr[] = new int[n];
str = read.readLine().trim().split(" ");
for(int i = 0; i < n; i++)
arr[i] = Integer.parseInt(str[i]);
findPositions(arr, n, x);
}
}
static void findPositions(int arr[], int n, int x)
{
boolean flag = false;
StringBuffer sb = new StringBuffer();
for(int i = 0; i < n; i++)
{
if(arr[i] == x)
{
sb.append(i + " ");
flag = true;
}
}
if(flag ==true)
System.out.println(sb.toString());
else System.out.println("Not found");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of N positive elements. The task is to find the Maximum AND Value generated by any pair of element from the array.
Note: AND is bitwise '&' operator.The first line of the input contains a single integer T, denoting the number of test cases. Then T test cases follow. Each test- case has two lines of the input, the first line contains an integer denoting the size of an array N and the second line of input contains N positive integers
Constraints:
1 <= T <= 50
1 <= N <= 10000
1 <= arr[i] <= 10000For each testcase, print Maximum AND Value of a pair in a separate line.Input:
2
4
4 8 12 16
4
4 8 16 2
Output:
8
0
Explanation:
Testcase 1: Pair (8, 12) has the Maximum AND Value i. e. 8.
Testcase 2: Maximum AND Value is 0., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
// Utility function to check number of elements
// having set msb as of pattern
int checkBit(int pattern, int arr[], int n)
{
int count = 0;
for (int i = 0; i < n; i++)
if ((pattern & arr[i]) == pattern)
count++;
return count;
}
// Function for finding maximum and value pair
int maxAND (int arr[], int n)
{
int res = 0, count;
// iterate over total of 30bits from msb to lsb
for (int bit = 31; bit >= 0; bit--)
{
// find the count of element having set msb
count = checkBit(res | (1 << bit),arr,n);
// if count >= 2 set particular bit in result
if ( count >= 2 )
res |= (1 << bit);
}
return res;
}
#define qw1 freopen("input1.txt", "r", stdin); freopen("output1.txt", "w", stdout);
#define qw2 freopen("input2.txt", "r", stdin); freopen("output2.txt", "w", stdout);
#define qw3 freopen("input3.txt", "r", stdin); freopen("output3.txt", "w", stdout);
#define qw4 freopen("input4.txt", "r", stdin); freopen("output4.txt", "w", stdout);
#define qw5 freopen("input5.txt", "r", stdin); freopen("output5.txt", "w", stdout);
#define qw6 freopen("input6.txt", "r", stdin); freopen("output6.txt", "w", stdout);
#define qw freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
// Driver function
int main()
{
#ifndef ONLINE_JUDGE
#endif
int t;
cin>>t;
while(t>0)
{
t--;
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
{
cin>>arr[i];
}
cout<<maxAND(arr,n)<<endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In this season of love, everyone wants to surprise each other.
You are also super excited and you wish to buy roses of 3 different colors. You always buy roses in order, white, yellow, red.
So if you buy 7 roses, they will be "white, yellow, red, white, yellow, red, white".
You need to find the number of yellow roses that you will buy?The only line of input contains a single integer, N, the number of roses that you will buy.
Constraints
1 <= N <= 1000Output a single integer, the number of yellow roses.Sample Input 1
2
Sample Output 1
1
Sample Input 2
8
Sample Ouput 2
3
Explanation;-
testcase1;- 2 flower will be white,yellow
so number of yellow flower is 1, I have written this Solution Code: n=int(input())
x=n/3
if n%3==2:
x+=1
print(int(x)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In this season of love, everyone wants to surprise each other.
You are also super excited and you wish to buy roses of 3 different colors. You always buy roses in order, white, yellow, red.
So if you buy 7 roses, they will be "white, yellow, red, white, yellow, red, white".
You need to find the number of yellow roses that you will buy?The only line of input contains a single integer, N, the number of roses that you will buy.
Constraints
1 <= N <= 1000Output a single integer, the number of yellow roses.Sample Input 1
2
Sample Output 1
1
Sample Input 2
8
Sample Ouput 2
3
Explanation;-
testcase1;- 2 flower will be white,yellow
so number of yellow flower is 1, 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 ans = n/3;
if(n%3==2){ans++;}
System.out.print(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In this season of love, everyone wants to surprise each other.
You are also super excited and you wish to buy roses of 3 different colors. You always buy roses in order, white, yellow, red.
So if you buy 7 roses, they will be "white, yellow, red, white, yellow, red, white".
You need to find the number of yellow roses that you will buy?The only line of input contains a single integer, N, the number of roses that you will buy.
Constraints
1 <= N <= 1000Output a single integer, the number of yellow roses.Sample Input 1
2
Sample Output 1
1
Sample Input 2
8
Sample Ouput 2
3
Explanation;-
testcase1;- 2 flower will be white,yellow
so number of yellow flower is 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int x=n/3;
if(n%3==2){
x++;}
cout<<x;}, In this Programming Language: C++, 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: N numbers are arranged in Circle. Find the Sum of all K contiguous Sub-arrays.The first line of the input contains an integer N, the length of the array, and K. The next line contains N integers which are elements of the array.
<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 kCircleSum(arr, N, K) which contains arr(array) and N(size of the array), and K as a parameter
<b>Constraints</b>
1 <= N <= 100000
1 <= arr[I] <= 100000
1 <= K <= NYou need to print N space-separated integers ith integer denoting Sum of sub-array of length K starting at index i.Sample Input
3 1
1 2 3
Sample Output
1 2 3
Explanation : k=1 so ans is 1, 2, and 3.
Sample Input
5 2
6 4 3 4 1
Sample Output
10 7 7 5 7, I have written this Solution Code: void kCircleSum(int arr[],int n,int k){
long long ans=0;
for(int i=0;i<k;i++){
ans+=arr[i];
}
for(int i=0;i<n;i++){
printf("%lli ",ans);
ans+=arr[(i+k)%n]-arr[i];
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N numbers are arranged in Circle. Find the Sum of all K contiguous Sub-arrays.The first line of the input contains an integer N, the length of the array, and K. The next line contains N integers which are elements of the array.
<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 kCircleSum(arr, N, K) which contains arr(array) and N(size of the array), and K as a parameter
<b>Constraints</b>
1 <= N <= 100000
1 <= arr[I] <= 100000
1 <= K <= NYou need to print N space-separated integers ith integer denoting Sum of sub-array of length K starting at index i.Sample Input
3 1
1 2 3
Sample Output
1 2 3
Explanation : k=1 so ans is 1, 2, and 3.
Sample Input
5 2
6 4 3 4 1
Sample Output
10 7 7 5 7, I have written this Solution Code: function kCircleSum(arr, arrSize, k)
{
var list = new Array(2*arrSize + 5)
for(var i = 0; i < arrSize; i++)
{
list[i+1] = arr[i]
list[i+arrSize+1] = list[i+1]
}
for(var i = 0; i < 2*arrSize; i++)
dp[i] = 0
for(var i=1;i<=2*arrSize;i++)
{
dp[i] = dp[i-1]+list[i]
}
var ans = ""
for(var i = 1; i <= arrSize; i++)
{
ans += (dp[i+k-1]-dp[i-1]) + " "
}
console.log(ans)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N numbers are arranged in Circle. Find the Sum of all K contiguous Sub-arrays.The first line of the input contains an integer N, the length of the array, and K. The next line contains N integers which are elements of the array.
<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 kCircleSum(arr, N, K) which contains arr(array) and N(size of the array), and K as a parameter
<b>Constraints</b>
1 <= N <= 100000
1 <= arr[I] <= 100000
1 <= K <= NYou need to print N space-separated integers ith integer denoting Sum of sub-array of length K starting at index i.Sample Input
3 1
1 2 3
Sample Output
1 2 3
Explanation : k=1 so ans is 1, 2, and 3.
Sample Input
5 2
6 4 3 4 1
Sample Output
10 7 7 5 7, I have written this Solution Code: static void kCircleSum(int arr[],int n,int k){
long ans=0;
for(int i=0;i<k;i++){
ans+=arr[i];
}
for(int i=0;i<n;i++){
System.out.print(ans+" ");
ans+=arr[(i+k)%n]-arr[i];
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N numbers are arranged in Circle. Find the Sum of all K contiguous Sub-arrays.The first line of the input contains an integer N, the length of the array, and K. The next line contains N integers which are elements of the array.
<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 kCircleSum(arr, N, K) which contains arr(array) and N(size of the array), and K as a parameter
<b>Constraints</b>
1 <= N <= 100000
1 <= arr[I] <= 100000
1 <= K <= NYou need to print N space-separated integers ith integer denoting Sum of sub-array of length K starting at index i.Sample Input
3 1
1 2 3
Sample Output
1 2 3
Explanation : k=1 so ans is 1, 2, and 3.
Sample Input
5 2
6 4 3 4 1
Sample Output
10 7 7 5 7, I have written this Solution Code: def kCircleSum(arr,n,k):
ans=0
for i in range (0,k):
ans=ans+arr[i]
for i in range (0,n):
print(ans,end=" ")
ans=ans+arr[int((i+k)%n)]-arr[i]
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N numbers are arranged in Circle. Find the Sum of all K contiguous Sub-arrays.The first line of the input contains an integer N, the length of the array, and K. The next line contains N integers which are elements of the array.
<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 kCircleSum(arr, N, K) which contains arr(array) and N(size of the array), and K as a parameter
<b>Constraints</b>
1 <= N <= 100000
1 <= arr[I] <= 100000
1 <= K <= NYou need to print N space-separated integers ith integer denoting Sum of sub-array of length K starting at index i.Sample Input
3 1
1 2 3
Sample Output
1 2 3
Explanation : k=1 so ans is 1, 2, and 3.
Sample Input
5 2
6 4 3 4 1
Sample Output
10 7 7 5 7, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
void kCircleSum(int arr[],int n,int k){
long long ans=0;
for(int i=0;i<k;i++){
ans+=arr[i];
}
for(int i=0;i<n;i++){
printf("%lli ",ans);
ans+=arr[(i+k)%n]-arr[i];
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given array A of size n. You need to find the maximum-sum sub-array with the condition that you are allowed to skip at most one element.The first line of input contains an integer T denoting the number of test cases. The description of T test cases follows. The first line of each test case contains a single integer n denoting the size of array. The second line contains n space-separated integers A1, A2, ..., An denoting the elements of the array.
Constraints:
1 <= T <= 10
1 <= n <= 10^5
-10^3 <= Ai<= 10^3For each test case, in a new line, print the maximum sum.Input:
2
5
1 2 3 -4 5
8
-2 -3 4 -1 -2 1 5 -3
Output:
11
9
Explanation:
Testcase1:
Input : A[] = {1, 2, 3, -4, 5}
Output : 11
We can get maximum sum subarray by skipping -4.
Testcase2:
Input : A[] = [-2, -3, 4, -1, -2, 1, 5, -3]
Output : 9
We can get maximum sum subarray by skipping -2 as [4, -1, 1, 5] sums to 9, which is the maximum achievable sum., I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main(String[] args)throws IOException
{
// Scanner sc = new Scanner(System.in);
// int t = sc.nextInt();
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine());
while(t-->0)
{
int n = Integer.parseInt(read.readLine());
int[] a = new int[n];
String str[] = read.readLine().trim().split(" ");
for(int i = 0;i < n; i++)
a[i] = Integer.parseInt(str[i]);
NS g = new NS();
System.out.println(g.maxSumSubarray(a , n));
}
}
}
class NS
{
public static int maxSumSubarray(int A[], int n)
{
int[] fw=new int[n]; // 2 pointer approach, L[i] and R[i] will store Left and Right max sum at ith pos.
int[] bw=new int[n];
int cur_max = A[0], max_so_far = A[0];
fw[0] = cur_max;
for (int i = 1; i < n; i++)
{
cur_max = Math.max(A[i], cur_max + A[i]); // either adding previous best ans to the new element makes it max or current element is max
max_so_far = Math.max(max_so_far, cur_max);
fw[i] = cur_max;
}
cur_max = max_so_far = bw[n-1] = A[n-1];
for (int i = n-2; i >= 0; i--)
{
cur_max = Math.max(A[i], cur_max + A[i]); // either adding previous best ans to the new element makes it max or current element is max
max_so_far = Math.max(max_so_far, cur_max);
bw[i] = cur_max;
}
int fans = max_so_far;
for (int i = 1; i < n - 1; i++)
fans = Math.max(fans, fw[i - 1] + bw[i + 1]); // Skipping the ith value and getting the best answer from L[i-1] and R[i+1]
return fans;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given array A of size n. You need to find the maximum-sum sub-array with the condition that you are allowed to skip at most one element.The first line of input contains an integer T denoting the number of test cases. The description of T test cases follows. The first line of each test case contains a single integer n denoting the size of array. The second line contains n space-separated integers A1, A2, ..., An denoting the elements of the array.
Constraints:
1 <= T <= 10
1 <= n <= 10^5
-10^3 <= Ai<= 10^3For each test case, in a new line, print the maximum sum.Input:
2
5
1 2 3 -4 5
8
-2 -3 4 -1 -2 1 5 -3
Output:
11
9
Explanation:
Testcase1:
Input : A[] = {1, 2, 3, -4, 5}
Output : 11
We can get maximum sum subarray by skipping -4.
Testcase2:
Input : A[] = [-2, -3, 4, -1, -2, 1, 5, -3]
Output : 9
We can get maximum sum subarray by skipping -2 as [4, -1, 1, 5] sums to 9, which is the maximum achievable sum., 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], l[N], r[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n, ans = 0; cin >> n;
for(int i = 1; i <= n; i++){
cin >> a[i];
l[i] = max(a[i], a[i] + l[i-1]);
ans = max(ans, l[i]);
}
r[n+1] = 0;
for(int i = n; i >= 2; i--){
r[i] = max(a[i], a[i] + r[i+1]);
ans = max(ans, l[i-2] + r[i]);
}
cout << ans << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether character is an alphabet or not using conditional operator.The first line of the input contains a character ch.
<b>Constraints</b>
'a', 'A' <= ch <= 'z', 'Z'Print "YES" if it's alphabet otherwise "NO".Sample Input :
a
Sample Output
YES, 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);
char c = sc.next().charAt(0);
String output =(c>='a' && c<='z') || (c>='A' && c<='Z')
? "YES" : "NO";
System.out.println(output);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether character is an alphabet or not using conditional operator.The first line of the input contains a character ch.
<b>Constraints</b>
'a', 'A' <= ch <= 'z', 'Z'Print "YES" if it's alphabet otherwise "NO".Sample Input :
a
Sample Output
YES, I have written this Solution Code: c = input()
if(c.isalpha()):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether character is an alphabet or not using conditional operator.The first line of the input contains a character ch.
<b>Constraints</b>
'a', 'A' <= ch <= 'z', 'Z'Print "YES" if it's alphabet otherwise "NO".Sample Input :
a
Sample Output
YES, I have written this Solution Code: #include <stdio.h>
int main()
{
char ch;
scanf("%c", &ch);
(ch>='a' && ch<='z') || (ch>='A' && ch<='Z')
? printf("YES")
: printf("NO");
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: You are given a tree rooted at 1 (root is at level 1). The entire tree can be created in the following fashion:-
<li> Nodes at odd level have 2 children. So node 1 has two children, nodes 2 and 3.
<li> Nodes at even level have 4 children. So node 2 has 4 children, nodes 4, 5, 6, and 7. Similarly node 3 has 4 children, nodes 8, 9, 10, and 11.
<li> This process goes on infinitely. Node 4 has 2 children, nodes 12 and 13, node 5 has 2 children, nodes 14 and 15,... , and so on.
Now you are given an integer N, you need to report the path from the root to the node.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 10<sup>16</sup>Output the path from the root to the node N in the form of space separated integers.Sample Input
14
Sample Output
1 2 5 14
Explanation: The path to node 14 from root is 1 -> 2 -> 5 -> 14 (Explained in problem text)
Sample Input
999
Sample Output
1 2 6 16 44 125 353 999
, I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
InputStream is;
PrintWriter out;
String INPUT = "";
void solve(int TC) throws Exception {
ArrayList<Long> al = new ArrayList<>();
ArrayList<Long> al2 = new ArrayList<>();
long last = 1L, lastP = 1L;
al.add(1L);
al2.add(1L);
for(int i=0;i<40;i++) {
if(i%2==0) {
last *= 2L;
al.add(last);
al2.add(last + al2.get(al2.size() - 1));
}else{
last *= 4L;
al.add(last);
al2.add(last + al2.get(al2.size() - 1));
}
}
long n = nl();
ArrayList<Long> ans = new ArrayList<>();
ans.add(n);
long cur = n;
while(cur > 1) {
int pos = 0;
while(al2.get(pos) < cur) {
++pos;
}
if(pos == 0) {
ans.add(1L);
break;
}
else if(pos == 1) {
if(cur>=8 && cur<=11) ans.add(3L);
else if(cur>=4 && cur<=7) ans.add(2L);
ans.add(1L);
break;
}
if(pos%2 == 1) {
cur = al2.get(pos - 2) + ((cur - al2.get(pos - 1) + 1) / 2);
ans.add(cur);
}else{
cur = al2.get(pos - 2) + (cur - al2.get(pos - 1) + 3) / 4;
ans.add(cur);
}
}
int pp = ans.size()-1;
while(pp >= 0) {
p(ans.get(pp) + " ");
--pp;
}
pn("");
}
boolean TestCases = false;
public static void main(String[] args) throws Exception { new Main().run(); }
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
static void dbg(Object... o){System.err.println(Arrays.deepToString(o));}
void run() throws Exception {
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
int T = TestCases ? ni() : 1;
for(int t=1;t<=T;t++) solve(t);
out.flush();
if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
}
void p(Object o) { out.print(o); }
void pn(Object o) { out.println(o); }
void pni(Object o) { out.println(o);out.flush(); }
int ni() {
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-') {
minus = true;
b = readByte();
}
while(true) {
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
long nl() {
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-') {
minus = true;
b = readByte();
}
while(true) {
if(b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
double nd() { return Double.parseDouble(ns()); }
char nc() { return (char)skip(); }
int BUF_SIZE = 1024 * 8;
byte[] inbuf = new byte[BUF_SIZE];
int lenbuf = 0, ptrbuf = 0;
int readByte() {
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
} return inbuf[ptrbuf++];
}
boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
String ns() {
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))) {
sb.appendCodePoint(b); b = readByte();
} return sb.toString();
}
char[] ns(int n) {
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
} return n == p ? buf : Arrays.copyOf(buf, p);
}
void tr(Object... o) { if(INPUT.length() > 0)System.out.println(Arrays.deepToString(o)); }
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a tree rooted at 1 (root is at level 1). The entire tree can be created in the following fashion:-
<li> Nodes at odd level have 2 children. So node 1 has two children, nodes 2 and 3.
<li> Nodes at even level have 4 children. So node 2 has 4 children, nodes 4, 5, 6, and 7. Similarly node 3 has 4 children, nodes 8, 9, 10, and 11.
<li> This process goes on infinitely. Node 4 has 2 children, nodes 12 and 13, node 5 has 2 children, nodes 14 and 15,... , and so on.
Now you are given an integer N, you need to report the path from the root to the node.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 10<sup>16</sup>Output the path from the root to the node N in the form of space separated integers.Sample Input
14
Sample Output
1 2 5 14
Explanation: The path to node 14 from root is 1 -> 2 -> 5 -> 14 (Explained in problem text)
Sample Input
999
Sample Output
1 2 6 16 44 125 353 999
, I have written this Solution Code: n = int(input())
import math
check=1
temp=1
ans=1
stk=[]
if n == 1:
print('1')
else:
while ans<n:
if check:
temp*=2
stk.append(ans+1)
ans+=temp
check=0
else:
temp*=4
stk.append(ans+1)
ans+=temp
check=1
path=[]
pos = n - stk[-1]+1
while stk:
path.append(stk[-1]+pos-1)
if not check:
pos = math.ceil(pos/2)
check=1
else:
pos = math.ceil(pos/4)
check=0
stk.pop()
path.append(1)
while stk:
path.append(stk[-1]+pos-1)
if not check:
pos = math.ceil(pos/2)
check=1
else:
pos = math.ceil(pos/4)
check=0
path=path[::-1]
ans=''
for i in path:
ans+=str(i)+' '
print(ans[:-1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a tree rooted at 1 (root is at level 1). The entire tree can be created in the following fashion:-
<li> Nodes at odd level have 2 children. So node 1 has two children, nodes 2 and 3.
<li> Nodes at even level have 4 children. So node 2 has 4 children, nodes 4, 5, 6, and 7. Similarly node 3 has 4 children, nodes 8, 9, 10, and 11.
<li> This process goes on infinitely. Node 4 has 2 children, nodes 12 and 13, node 5 has 2 children, nodes 14 and 15,... , and so on.
Now you are given an integer N, you need to report the path from the root to the node.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 10<sup>16</sup>Output the path from the root to the node N in the form of space separated integers.Sample Input
14
Sample Output
1 2 5 14
Explanation: The path to node 14 from root is 1 -> 2 -> 5 -> 14 (Explained in problem text)
Sample Input
999
Sample Output
1 2 6 16 44 125 353 999
, 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;
int s[50], e[50];
s[1]=1, e[1]=1;
For(i, 2, 50){
int m = 2;
if(i%2){
m = 4;
}
s[i]=e[i-1]+1;
e[i]=e[i-1]+(e[i-1]-s[i-1]+1)*m;
if(s[i] > n){
break;
}
}
int j;
vector<int> vect;
for(int i=1; i<50; i++){
if(s[i] > n){
j = i-1;
break;
}
}
int cur = n;
while(j > 1){
vect.pb(cur);
int cl = e[j]-s[j]+1;
int pl = e[j-1]-s[j-1]+1;
int par = cl/pl;
int pos = cur-s[j];
pos = pos/par;
cur = s[j-1]+pos;
j--;
}
vect.pb(1);
reverse(all(vect));
for(int i: vect){
cout<<i<<" ";
}
}
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 a number N, find the sum of all the digits of the number
Note: Use a recursive method to solve this problem.<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>Sum()</b> that takes the integer N as a parameter.
Constraints:
1 ≤ T ≤ 100
0 ≤ N ≤ 1000000000Return sum of digits.Sample Input
2
25
28
Sample Output
7
10, I have written this Solution Code: // n is the input number
function recSum(n) {
// write code here
// do not console.log
// return the answer as a number
if (n < 10) return n;
return n % 10 + recSum(Math.floor(n / 10))
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the sum of all the digits of the number
Note: Use a recursive method to solve this problem.<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>Sum()</b> that takes the integer N as a parameter.
Constraints:
1 ≤ T ≤ 100
0 ≤ N ≤ 1000000000Return sum of digits.Sample Input
2
25
28
Sample Output
7
10, I have written this Solution Code: static long Sum(long n)
{ if(n==0){return 0;}
return n%10+Sum(n/10);
} , In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a directed graph, detect the presence of a cycle in the graph.The first line of input contains two integers N and M which denotes the no of vertices and no of edges in the graph respectively.
Next M lines contain space-separated integers u and v denoting that there is a directed edge from u to v.
Constraints:
1 <= N, M <= 1000
0 <= u, v <= N-1, u != v
There are no self loops or multiple edges.The method should return 1 if there is a cycle else it should return 0.Sample Input 1:
4 5
0 1
1 2
2 3
3 0
0 2
Sample Output 1:
Yes
Explanation:
There is a cycle with nodes 0, 1, 2, 3
Sample Input 2:
4 4
0 1
1 2
2 3
0 3
Sample Output 2:
No
Explanation:
There is no cycle in this graph, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static ArrayList<ArrayList<Integer>> adj;
static void graph(int v){
adj =new ArrayList<>();
for(int i=0;i<=v;i++) adj.add(new ArrayList<Integer>());
}
static void addedge(int u, int v){
adj.get(u).add(v);
}
static boolean dfs(int i, boolean [] vis, int parent[]){
vis[i] = true; parent[i] = 1;
for(Integer it: adj.get(i)){
if(vis[it]==false) {
if(dfs(it, vis, parent)==true) return true;
}
else if(parent[it]==1) return true;
}
parent[i] = 0;
return false;
}
static boolean helper(int n){
boolean vis[] = new boolean [n+1];
int parent[] = new int[n+1];
for(int i=0;i<=n;i++){
if(vis[i]==false){
if(dfs(i, vis, parent)) return true;
}
}
return false;
}
public static void main (String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String t[] = br.readLine().split(" ");
int n = Integer.parseInt(t[0]);
graph(n);
int e = Integer.parseInt(t[1]);
for(int i=0;i<e;i++) {
String num[] = br.readLine().split(" ");
int u = Integer.parseInt(num[0]);
int v = Integer.parseInt(num[1]);
addedge(u, v);
}
if(helper(n)) System.out.println("Yes");
else System.out.println("No");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a directed graph, detect the presence of a cycle in the graph.The first line of input contains two integers N and M which denotes the no of vertices and no of edges in the graph respectively.
Next M lines contain space-separated integers u and v denoting that there is a directed edge from u to v.
Constraints:
1 <= N, M <= 1000
0 <= u, v <= N-1, u != v
There are no self loops or multiple edges.The method should return 1 if there is a cycle else it should return 0.Sample Input 1:
4 5
0 1
1 2
2 3
3 0
0 2
Sample Output 1:
Yes
Explanation:
There is a cycle with nodes 0, 1, 2, 3
Sample Input 2:
4 4
0 1
1 2
2 3
0 3
Sample Output 2:
No
Explanation:
There is no cycle in this graph, I have written this Solution Code: from collections import defaultdict
class Graph():
def __init__(self,vertices):
self.graph = defaultdict(list)
self.V = vertices
def addEdge(self,u,v):
self.graph[u].append(v)
def isCyclicUtil(self, v, visited, recStack):
visited[v] = True
recStack[v] = True
for neighbour in self.graph[v]:
if visited[neighbour] == False:
if self.isCyclicUtil(neighbour, visited, recStack) == True:
return True
elif recStack[neighbour] == True:
return True
recStack[v] = False
return False
def isCyclic(self):
visited = [False] * (self.V + 1)
recStack = [False] * (self.V + 1)
for node in range(self.V):
if visited[node] == False:
if self.isCyclicUtil(node,visited,recStack) == True:
return True
return False
V,edges=map(int,input().split())
g = Graph(V)
for i in range(edges):
u,v=map(int,input().split())
g.addEdge(u,v)
if g.isCyclic() == 1:
print ("Yes")
else:
print ("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a directed graph, detect the presence of a cycle in the graph.The first line of input contains two integers N and M which denotes the no of vertices and no of edges in the graph respectively.
Next M lines contain space-separated integers u and v denoting that there is a directed edge from u to v.
Constraints:
1 <= N, M <= 1000
0 <= u, v <= N-1, u != v
There are no self loops or multiple edges.The method should return 1 if there is a cycle else it should return 0.Sample Input 1:
4 5
0 1
1 2
2 3
3 0
0 2
Sample Output 1:
Yes
Explanation:
There is a cycle with nodes 0, 1, 2, 3
Sample Input 2:
4 4
0 1
1 2
2 3
0 3
Sample Output 2:
No
Explanation:
There is no cycle in this graph, 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;
vector<int> g[N];
int vis[N];
bool flag = 0;
void dfs(int u){
vis[u] = 1;
for(auto i: g[u]){
if(vis[i] == 1)
flag = 1;
if(vis[i] == 0) dfs(i);
}
vis[u] = 2;
}
signed main() {
IOS;
int n, m;
cin >> n >> m;
for(int i = 1; i <= m; i++){
int u, v;
cin >> u >> v;
g[u].push_back(v);
}
for(int i = 0; i < n; i++){
if(vis[i]) continue;
dfs(i);
}
if(flag)
cout << "Yes";
else
cout << "No";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a queue of integers and N queries. Your the task is to perform these operations:-
enqueue:-this operation will add an element to your current queue.
dequeue:-this operation will delete the element from the starting of the queue
displayfront:-this operation will print the element presented at front
Note:-if queue is empty than dequeue operation will do nothing, and 0 will be printed as a front element of queue if it is empty.User task:
Since this will be a functional problem, you don't have to take input. You just have to complete the functions:
<b>enqueue()</b>:- that takes integer to be added as a parameter.
<b>dequeue()</b>:- that takes no parameter.
<b>displayfront()</b> :- that takes no parameter.
Constraints:
1 <= N(Number of queries) <= 10<sup>3</sup>You don't need to print anything else other than in displayfront() in which you require to print the element at front of your queue in a new line, if the queue is empty you just need to print 0.Sample Input:-
7
displayfront
enqueue 2
displayfront
enqueue 4
displayfront
dequeue
displayfront
Sample Output:-
0
2
2
4
Sample input:
5
enqueue 4
enqueue 5
displayfront
dequeue
displayfront
Sample output:-
4
5, I have written this Solution Code: class Queue
{
private Node front, rear;
private int currentSize;
class Node {
Node next;
int val;
Node(int val) {
this.val = val;
next = null;
}
}
public Queue()
{
front = null;
rear = null;
currentSize = 0;
}
public boolean isEmpty()
{
return (currentSize <= 0);
}
public void dequeue()
{
if (isEmpty())
{
}
else{
front = front.next;
currentSize--;
}
}
//Add data to the end of the list.
public void enqueue(int data)
{
Node oldRear = rear;
rear = new Node(data);
if (isEmpty())
{
front = rear;
}
else
{
oldRear.next = rear;
}
currentSize++;
}
public void displayfront(){
if(isEmpty()){
System.out.println("0");
}
else{
System.out.println(front.val);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is <b>Palindrome</b> or not. A number is said to be Palindrome when it reads the same from backward as forward.User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isPalindrome()</b> which contains N as a parameter.
<b>Constraints:</b>
1 <= N <= 9999You need to return "true" is the number is palindrome otherwise "false".Sample Input:
5
Sample Output:
true
Sample Input:
121
Sample Output:
true, I have written this Solution Code: static boolean isPalindrome(int N)
{
int sum = 0;
int rev = N;
while(N > 0)
{
int digit = N%10;
sum = sum*10+digit;
N = N/10;
}
if(rev == sum)
return true;
else return false;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is <b>Palindrome</b> or not. A number is said to be Palindrome when it reads the same from backward as forward.User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isPalindrome()</b> which contains N as a parameter.
<b>Constraints:</b>
1 <= N <= 9999You need to return "true" is the number is palindrome otherwise "false".Sample Input:
5
Sample Output:
true
Sample Input:
121
Sample Output:
true, I have written this Solution Code: def isPalindrome(N):
sum1 = 0
rev = N
while(N > 0):
digit = N%10
sum1 = sum1*10+digit
N = N//10
if(rev == sum1):
return True
return False, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given n integers a<sub>1</sub>, a<sub>2</sub>, …, a<sub>n</sub>. Find the maximum value of <b>max(a<sub>l</sub>, a<sub>l+1</sub>, …, a<sub>r</sub>) * min(a<sub>l</sub>, a<sub>l+1</sub>, …, a<sub>r</sub>)</b> over all pairs (l, r) of integers for which 1 ≤ l < r ≤ n.The first line of input contains a single integer n. The second line of each test case contains n integers a<sub>1</sub>, a<sub>2</sub>, …, a<sub>n</sub>
<b>Constraints</b>
2 ≤ n ≤ 10<sup>5</sup>
1 ≤ a<sub>i</sub> ≤ 10<sup>6</sup>Print a single integer denoting the maximum possible value of the product from the statement.<b>Sample Input 1</b>
3
2 4 3
<b>Sample Output 1</b>
12
<b>Sample Input 2</b>
4
3 2 3 1
<b>Sample Output 2</b>
6, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
int32_t main()
{
int t=1;
// cin>>t;
while(t--)
{
int n;
cin>>n;
int a[n];
int ans=0;
for(int i=0;i<n;++i) {cin>>a[i];if(i!=0) ans=max(ans,a[i]*a[i-1]);}
cout<<ans<<'\n';
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Quadratic probing is a collision handling technique in hashing. Quadratic probing says that whenever a collision occurs, search for i^2 position.
In this question, we'll learn how to fill up the hash table using Quadratic probing 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[] of size N. You need to fill up the hash table using Quadratic probing and print the resultant hash table.
Note:
1. All the positions that are unoccupied are denoted by -1 in the hash table.
2. An empty slot can only be found if load factor < 0.5 and hash table size is a prime number.
3. The given testcases satisfy the above condition so you can assume that an empty slot is always reachable.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 size of array. The third line contains elements of the array.
Constraints:
1 <= T <= 100
2 <= hashSize (prime) <= 97
1 <= sizeOfArray < hashSize*0.5
0 <= Array[i] <= 10^5For each testcase, in a new line, print the hash table.Input:
2
11 4
21 10 32 43
11 4
880 995 647 172
Output:
10 -1 -1 32 -1 -1 -1 -1 43 -1 21
880 -1 -1 -1 -1 995 -1 172 -1 647 -1
Explanation:
Testcase1: 21%11=10 so 21 goes into hashTable[10] position. 10%11=10. hashTable[10] is already filled so we try for (10+1)%11=0 position. hashTable[0] is empty so we put 10 there. 32%11=10. hashTable[10] is filled. We try (32+1)%11=0. But hashTable[0] is also already filled. We try (32+4)%11=3. hashTable[3] is empty so we put 32 in hashTable[3] position. 43 uses (43+9)%11=8. We put it in hashTable[8].
Testcase2: Using the similar approach as used in above explanation we will get the output like 880 -1 -1 -1 -1 995 -1 172 -1 647 -1 ., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
InputStreamReader in = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(in);
int t = Integer.parseInt(br.readLine().trim());
while(t-- > 0){
String str[] = br.readLine().trim().split(" ");
int capacity = Integer.parseInt(str[0]);
int n = Integer.parseInt(str[1]);
str = br.readLine().trim().split(" ");
ArrayList<Integer> ht = new ArrayList<>(capacity);
for(int i = 0; i < capacity; i++){
ht.add(-1);
}
for(int i = 0; i < n; i++){
int key = Integer.parseInt(str[i]);
int count =0;
int hashKey = key % capacity;
if (ht.get(hashKey)==-1){
ht.set(hashKey, key);
}
else
{
for(int j=0; j<capacity;j++){
hashKey = (key + j*j) %capacity;
if(ht.get(hashKey)==-1){
ht.set(hashKey, key);
break;
}
}
}
}
for(int i = 0; i < capacity; i++){
System.out.print(ht.get(i)+ " ");
}
System.out.println();
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Quadratic probing is a collision handling technique in hashing. Quadratic probing says that whenever a collision occurs, search for i^2 position.
In this question, we'll learn how to fill up the hash table using Quadratic probing 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[] of size N. You need to fill up the hash table using Quadratic probing and print the resultant hash table.
Note:
1. All the positions that are unoccupied are denoted by -1 in the hash table.
2. An empty slot can only be found if load factor < 0.5 and hash table size is a prime number.
3. The given testcases satisfy the above condition so you can assume that an empty slot is always reachable.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 size of array. The third line contains elements of the array.
Constraints:
1 <= T <= 100
2 <= hashSize (prime) <= 97
1 <= sizeOfArray < hashSize*0.5
0 <= Array[i] <= 10^5For each testcase, in a new line, print the hash table.Input:
2
11 4
21 10 32 43
11 4
880 995 647 172
Output:
10 -1 -1 32 -1 -1 -1 -1 43 -1 21
880 -1 -1 -1 -1 995 -1 172 -1 647 -1
Explanation:
Testcase1: 21%11=10 so 21 goes into hashTable[10] position. 10%11=10. hashTable[10] is already filled so we try for (10+1)%11=0 position. hashTable[0] is empty so we put 10 there. 32%11=10. hashTable[10] is filled. We try (32+1)%11=0. But hashTable[0] is also already filled. We try (32+4)%11=3. hashTable[3] is empty so we put 32 in hashTable[3] position. 43 uses (43+9)%11=8. We put it in hashTable[8].
Testcase2: Using the similar approach as used in above explanation we will get the output like 880 -1 -1 -1 -1 995 -1 172 -1 647 -1 ., I have written this Solution Code: for _ in range(int(input())):
hs,n = map(int,input().split())
l = list(map(int,input().split()))
ht = [-1]*hs
lf = 0
for i in l:
if(lf/hs >= 0.5):
break
if(ht[i%hs] == -1):
ht[i%hs] = i
else:
k = 1
j = (i%hs + k**2) % hs
while(ht[j] != -1):
k += 1
j = (i%hs + k**2) % hs
ht[j] = i
lf += 1
for i in ht:
print(i,end=" ")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Quadratic probing is a collision handling technique in hashing. Quadratic probing says that whenever a collision occurs, search for i^2 position.
In this question, we'll learn how to fill up the hash table using Quadratic probing 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[] of size N. You need to fill up the hash table using Quadratic probing and print the resultant hash table.
Note:
1. All the positions that are unoccupied are denoted by -1 in the hash table.
2. An empty slot can only be found if load factor < 0.5 and hash table size is a prime number.
3. The given testcases satisfy the above condition so you can assume that an empty slot is always reachable.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 size of array. The third line contains elements of the array.
Constraints:
1 <= T <= 100
2 <= hashSize (prime) <= 97
1 <= sizeOfArray < hashSize*0.5
0 <= Array[i] <= 10^5For each testcase, in a new line, print the hash table.Input:
2
11 4
21 10 32 43
11 4
880 995 647 172
Output:
10 -1 -1 32 -1 -1 -1 -1 43 -1 21
880 -1 -1 -1 -1 995 -1 172 -1 647 -1
Explanation:
Testcase1: 21%11=10 so 21 goes into hashTable[10] position. 10%11=10. hashTable[10] is already filled so we try for (10+1)%11=0 position. hashTable[0] is empty so we put 10 there. 32%11=10. hashTable[10] is filled. We try (32+1)%11=0. But hashTable[0] is also already filled. We try (32+4)%11=3. hashTable[3] is empty so we put 32 in hashTable[3] position. 43 uses (43+9)%11=8. We put it in hashTable[8].
Testcase2: Using the similar approach as used in above explanation we will get the output like 880 -1 -1 -1 -1 995 -1 172 -1 647 -1 ., I have written this Solution Code:
// author-Shivam gupta
#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 MOD 1000000007
#define read(type) readInt<type>()
#define max1 1000001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x && (!(x&(x-1)));
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
// Returns n^(-1) mod p
ll modInverse(ll n, ll p)
{
return power(n, p-2, p);
}
// Returns nCr % p using Fermat's little
// theorem.
int main(){
int t;
cin>>t;
while(t--){
int n,p;
cin>>p>>n;
int a[n];
li sum;
ll cur=0;
int b[p];
FOR(i,p){
b[i]=-1;}
unordered_map<ll,int> m;
ll cnt=0;
for(int i=0;i<n;i++){
cin>>a[i];
cur=a[i]%p;
int j=0;
while(b[(cur+j*j)%p]!=-1){
j++;
}
b[(cur+j*j)%p]=a[i];
}
FOR(i,p){
out1(b[i]);
}
END;
}
}
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T bug T \ `. _____ / ; /
` \ : "==="""""""""==="" : /
`: ;'
"-. .-"
""--.. ..--""
*/
, 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: Write a program to check whether a number is even or odd using switch case.First Line of the input contains the number n.
<b>Constraints</b>
1 <= n <= 1e9If the number is even print "Even" otherwise "Odd"Sample Input :
23
Sample Output :
Odd
Sample Input :
24
Sample Output :
Even, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t= Integer.parseInt(br.readLine());
if(t%2==0)
System.out.println("Even");
else
System.out.println("Odd");
}
catch (Exception e){
return;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is even or odd using switch case.First Line of the input contains the number n.
<b>Constraints</b>
1 <= n <= 1e9If the number is even print "Even" otherwise "Odd"Sample Input :
23
Sample Output :
Odd
Sample Input :
24
Sample Output :
Even, I have written this Solution Code: n = int(input())
if n % 2 == 0:
print("Even")
else:
print("Odd"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is even or odd using switch case.First Line of the input contains the number n.
<b>Constraints</b>
1 <= n <= 1e9If the number is even print "Even" otherwise "Odd"Sample Input :
23
Sample Output :
Odd
Sample Input :
24
Sample Output :
Even, I have written this Solution Code: #include <stdio.h>
int main()
{
int num;
scanf("%d", &num);
switch(num % 2)
{
case 0:
printf("Even");
break;
/* Else if n%2 == 1 */
case 1:
printf("Odd");
break;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list consisting of N nodes, your task is to check if the given list is palindrome or not.
A palindrome is a string/number that mirrors itself, for example, 21312 reverse is also 21312.User task:
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>Ispalindrome()</b> that takes the head of the linked list as parameter.
Constraints:-
1<=N<=1000
1<=Node.data<=1000Return true if given list is palindrome else return false.Sample Input 1:-
5
1 2 3 2 1
Sample Output 1:-
Yes
Sample Input 2:-
3
1 2 2
Sample Output 2:-
No, I have written this Solution Code: n=int(input())
l=input().split()
l=[int(i) for i in l]
if l[::-1]==l:
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list consisting of N nodes, your task is to check if the given list is palindrome or not.
A palindrome is a string/number that mirrors itself, for example, 21312 reverse is also 21312.User task:
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>Ispalindrome()</b> that takes the head of the linked list as parameter.
Constraints:-
1<=N<=1000
1<=Node.data<=1000Return true if given list is palindrome else return false.Sample Input 1:-
5
1 2 3 2 1
Sample Output 1:-
Yes
Sample Input 2:-
3
1 2 2
Sample Output 2:-
No, I have written this Solution Code: public static boolean IsPalindrome(Node head)
{
Node slow = head;
boolean ispalin = true;
Stack<Integer> stack = new Stack<Integer>();
while (slow != null) {
stack.push(slow.val);
slow = slow.next;
}
while (head != null) {
int i = stack.pop();
if (head.val == i) {
ispalin = true;
}
else {
ispalin = false;
break;
}
head = head.next;
}
return ispalin;
} , In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number (n) is represented in Linked List such that each digit corresponds to a node in linked list. Add 1 to it.
<b>Note:-</b> Linked list representation of a number is from left to right i.e if the number is 123 than in linked list it is represented as 3->2->1<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>addOne()</b> that takes head node of the linked list as parameter.
Constraints:
1 <=length of n<= 1000Return the head of the modified linked list.Input 1:
456
Output 1:
457
Input 2:
999
Output 2:
1000, I have written this Solution Code:
static Node addOne(Node head)
{
Node res = head;
Node temp = null, prev = null;
int carry = 1, sum;
while (head != null) //while both lists exist
{
sum = carry + head.data;
carry = (sum >= 10)? 1 : 0;
sum = sum % 10;
head.data = sum;
temp = head;
head = head.next;
}
if (carry > 0) {
Node x=new Node(carry);
temp.next=x;}
return res;
} , In this Programming Language: Java, 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: Our Tono has got N cute strings. Cute strings are very loving, and they possess a beautiful property. Whenever two cute strings are placed next to each other, they get concatenated in the following manner:
"tono"+"nolo"="tonolo" (suffix-prefix match occurs once). See sample for better understanding.
Now, Tono wants to concatenate these N cute strings from left to right. Can you help her find the merged string.The first line contains an integer N, the number of strings.
The next line contains N strings consisting of only lowercase english letters.
Constraints
1 <= N <= 100000
1 <= sum of lengths of all strings <= 500000
(String matching takes O(N) time, try to analyze the time complexity of your solution)Output a single string, the answer to Tono's problem.Sample Input
5
i want to eat truffle
Sample Output
iwantoeatruffle
Explanation
i + want = iwant
iwant + to = iwanto
iwanto + eat = iwantoeat
iwantoeat + truffle = iwantoeatruffle, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll 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 int long long
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
#define MOD 1000000007
#define INF 1000000000000000007
#define MAXN 300005
// it's swapnil07 ;)
void getZ(string str, int Z[])
{
int n = str.length();
int L, R, k;
L = R = 0;
for (int i = 1; i < n; ++i)
{
if (i > R)
{
L = R = i;
while (R<n && str[R-L] == str[R])
R++;
Z[i] = R-L;
R--;
}
else
{
k = i-L;
if (Z[k] < R-i+1)
Z[i] = Z[k];
else
{
L = i;
while (R<n && str[R-L] == str[R])
R++;
Z[i] = R-L;
R--;
}
}
}
}
signed main(){
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int n; cin>>n;
string ans;
cin>>ans;
string s1, s2;
For(i, 0, n-1){
int v = ans.length();
cin>>s2;
int len = min(ans.length(), s2.length());
s1 = ans.substr(v-len);
string s = s2+"$"+s1;
int Z[s.length()];
getZ(s, Z);
int av = 0;
for(int j=0; j<len; j++){
int x = j+1;
if(Z[s.length()-x]==x){
av = x;
}
}
s2 = s2.substr(av);
ans += s2;
}
cout<<ans;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When Alexa writes an integer, he uses a comma for every third digit from the right. How many commas will be used in total when she writes each integer from 1 to N (inclusive) once?The input consists of a single integer N.
<b>Constraints</b>
1 ≤ N ≤ 10<sup>15</sup>
N is an integer.Print the total number of commas.<b>Sample Input 1</b>
1010
<b>Sample Output 1</b>
11
<b>Explanation</b>
For 1 to 999, the number of commas used is 0.
For 1000 there is 1 comma --> 1,000.
For 1001 there is 1 comma --> 1,001.
For 1002 there is 1 comma --> 1,002.
For 1003 there is 1 comma --> 1,003.
For 1004 there is 1 comma --> 1,004.
For 1005 there is 1 comma --> 1,005.
For 1006 there is 1 comma --> 1,006.
For 1007 there is 1 comma --> 1,007.
For 1008 there is 1 comma --> 1,008.
For 1009 there is 1 comma --> 1,009.
For 1010 there is 1 comma --> 1,010.
Hence the total number of commas required is 11., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
using lint = long long;
void solve() {
lint n;
cin >> n;
lint ans = 0, ten = 1;
for (int k = 1; k <= 5; ++k) {
ten *= 1000;
ans += max(n - ten + 1, 0LL);
}
cout << ans << "\n";
}
int main() {
solve();
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given an array of N numbers. Your task is to first reverse the array (the first number becomes last, 2nd number becomes 2nd from the last, and so on) and then print the sum of the numbers at even indices and print the product of the numbers at odd indices.
<b>Note :</b> 1 based indexing is followed .The first line contains a single integer N: the number of elements
The second line contains N different integers separated by spaces
<b>constraints:-</b>
1 <= N <= 35
1 <= a[i] <= 10Two space separated integers representing sum of the numbers at even places and the product of the numbers at odd places.
Sample Input 1:
6
1 2 3 4 5 6
Sample Output 1:
9 48
Sample Input 2:
3
1 2 3
Sample Output 2:
2 3
<b>Explanation 1:-</b>
After reversing 1 2 3 4 5 6 becomes 6 5 4 3 2 1
Hence the sum of the numbers at even indices: 5+3+1=9
product of the numbers at odd indices: 6*4*2=48
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
long x=1,y=0;
if(n&1){
for(int i=0;i<n;i++){
if(i%2==0){
x*=a[i];
}
else{
y+=a[i];
}
}
}
else{
for(int i=0;i<n;i++){
if(i%2==0){
y+=a[i];
}
else{
x*=a[i];
}
}
}
cout<<y<<" "<<x;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given an array of N numbers. Your task is to first reverse the array (the first number becomes last, 2nd number becomes 2nd from the last, and so on) and then print the sum of the numbers at even indices and print the product of the numbers at odd indices.
<b>Note :</b> 1 based indexing is followed .The first line contains a single integer N: the number of elements
The second line contains N different integers separated by spaces
<b>constraints:-</b>
1 <= N <= 35
1 <= a[i] <= 10Two space separated integers representing sum of the numbers at even places and the product of the numbers at odd places.
Sample Input 1:
6
1 2 3 4 5 6
Sample Output 1:
9 48
Sample Input 2:
3
1 2 3
Sample Output 2:
2 3
<b>Explanation 1:-</b>
After reversing 1 2 3 4 5 6 becomes 6 5 4 3 2 1
Hence the sum of the numbers at even indices: 5+3+1=9
product of the numbers at odd indices: 6*4*2=48
, I have written this Solution Code: n = int(input())
lst = list(map(int, input().strip().split()))[:n]
lst.reverse()
lst_1 = 0
for i in range(1, n, 2):
lst_1 += lst[i]
lst_2 = 1
for i in range(0, n, 2):
lst_2 *= lst[i]
print(lst_1,lst_2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given an array of N numbers. Your task is to first reverse the array (the first number becomes last, 2nd number becomes 2nd from the last, and so on) and then print the sum of the numbers at even indices and print the product of the numbers at odd indices.
<b>Note :</b> 1 based indexing is followed .The first line contains a single integer N: the number of elements
The second line contains N different integers separated by spaces
<b>constraints:-</b>
1 <= N <= 35
1 <= a[i] <= 10Two space separated integers representing sum of the numbers at even places and the product of the numbers at odd places.
Sample Input 1:
6
1 2 3 4 5 6
Sample Output 1:
9 48
Sample Input 2:
3
1 2 3
Sample Output 2:
2 3
<b>Explanation 1:-</b>
After reversing 1 2 3 4 5 6 becomes 6 5 4 3 2 1
Hence the sum of the numbers at even indices: 5+3+1=9
product of the numbers at odd indices: 6*4*2=48
, 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 Arr[] = new int[n];
for(int i=0;i<n;i++){
Arr[i] = sc.nextInt();
}
if(n%2==1){
long sum=0;
long product=1;
for(int i=0;i<n;i++){
if(i%2==0){
product*=Arr[i];
}
else{
sum+=Arr[i];
}
}
System.out.print(sum+" "+product);
}
else{
long sum=0;
long product=1;
for(int i=0;i<n;i++){
if(i%2==1){
product*=Arr[i];
}
else{
sum+=Arr[i];
}
}
System.out.print(sum+" "+product);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement the power(a, n) function which return the value a<sup>n</sup> in log(n) time complexity.
The value can be very large so return the answer with modulo M=1e9+7.You don't have to input anything, just implement the function.
Constraints :
-10^9 <= a <= 10^9, a!=0
0 <= n <= 10^9You don't have to output anything, just return the required value.Input:
2 0
Output:
1
Input:
3 4
Output:
81
Input:
-5 3
Output:
-125, I have written this Solution Code:
class Solution{
public long power(long a,long n){
if(n==0)return 1;
long M=(long)1e9+7;
long p=power(a,n/2);
p=(p*p)%M;
if(n%2 == 1){
p=(p*a)%M;
}
return p;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The question is super small and super simple.
You are given an integer n. Initially you have an empty string. You need to construct the string of length n with the following rules:
1. Insert the first character in the beginning of string, the second in the end, the third in the beginning, fourth in the end, and so on.
2. The first character should be 'a', followed by 'b', 'c', and so on. 'z' will be followed by 'a'.The first and the only line of input contains a single number n.
Constraints
1 <= n <= 500000Output the generated string.Sample Input
4
Sample Output
cabd
Sample Input
30
Sample Output
caywusqomkigecabdfhjlnprtvxzbd
Explanation
In the first case the string transforms as follows: "" -> "a" -> "ab" -> "cab" -> "cabd", I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader scan=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(scan.readLine());
char ch='a';
boolean isFirst = true;
StringBuilder sb = new StringBuilder();
for(int i=0;i<n;i++)
{
if(isFirst) {
sb.insert(0, (char)(ch + i%26));
} else {
sb.append((char) (ch + i%26));
}
isFirst = !isFirst;
}
System.out.print(sb.toString());
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The question is super small and super simple.
You are given an integer n. Initially you have an empty string. You need to construct the string of length n with the following rules:
1. Insert the first character in the beginning of string, the second in the end, the third in the beginning, fourth in the end, and so on.
2. The first character should be 'a', followed by 'b', 'c', and so on. 'z' will be followed by 'a'.The first and the only line of input contains a single number n.
Constraints
1 <= n <= 500000Output the generated string.Sample Input
4
Sample Output
cabd
Sample Input
30
Sample Output
caywusqomkigecabdfhjlnprtvxzbd
Explanation
In the first case the string transforms as follows: "" -> "a" -> "ab" -> "cab" -> "cabd", 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(ll 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 int long long
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define MOD 1000000007
#define INF 1000000000000000007LL
const int N = 100005;
// it's swapnil07 ;)
#ifdef SWAPNIL07
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define end_routine()
#endif
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int n; cin>>n;
deque<char> vect;
int cur = 0;
bool fl = false;
for(int i=0; i<n; i++){
if(!fl)
vect.push_front('a'+cur);
else
vect.push_back('a'+cur);
cur = (cur+1)%26;
fl = !fl;
}
while(!vect.empty()){
char c = vect.front();
vect.pop_front();
cout<<c;
}
// end_routine();
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You must be familiar with various types of operators. One of the most commonly used operators in any language are increment and decrement operators. Given two numbers X and Y. Your task is to print the value of X decremented by 1 and value of Y after incremented by 1.The first line of the input contains two integers X and Y.
<b>Constraints:</b>
1 ≤ X, Y ≤ 10000You need to perform the task as mentioned in the question, and finally, print the result separated by a space.Sample Input 1:
4 5
Sample Output 1:
3 6
Sample Input 2:
5 6
Sample Output 2:
4 7, 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 x = sc.nextInt();
int y = sc.nextInt();
x--;
y++;
System.out.print(x);
System.out.print(" ");
System.out.print(y);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You must be familiar with various types of operators. One of the most commonly used operators in any language are increment and decrement operators. Given two numbers X and Y. Your task is to print the value of X decremented by 1 and value of Y after incremented by 1.The first line of the input contains two integers X and Y.
<b>Constraints:</b>
1 ≤ X, Y ≤ 10000You need to perform the task as mentioned in the question, and finally, print the result separated by a space.Sample Input 1:
4 5
Sample Output 1:
3 6
Sample Input 2:
5 6
Sample Output 2:
4 7, I have written this Solution Code: def incremental_decremental(x, y):
x -= 1
y += 1
print(x, y, end =' ')
def main():
input1 = input().split()
x = int(input1[0])
y = int(input1[1])
#z = int(input1[2])
incremental_decremental(x, y)
if __name__ == '__main__':
main(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers A and a number K, find maximum sum of a subarray of size K.The first line of input contains two integers N and K, denoting the number of elements in the array and the subarray size respectively.
The next line contains N integers denoting the elements of the array respectively.
1 <= K <= N <= 200000
-200000 <= A[i] <= 200000Print a single integer denoting the maximum sum of subarray of size K.Sample Input:
4 2
-1 5 2 -3
Sample Output:
7
Explanation:
Three subarrays of size 2, their sum are 4, 7, -1, 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);
String s[]=sc.nextLine().split(" ");
int n=Integer.parseInt(s[0]);
int k=Integer.parseInt(s[1]);
s=sc.nextLine().split(" ");
int ar[]=new int[n];
for(int i=0;i<n;i++){
ar[i]=Integer.parseInt(s[i]);
}
int max=0;
for(int i=0;i<=n-k;i++){
int t=0;
for(int j=0;j<k;j++){
t+=ar[i+j];
}
if(t>max)
max=t;
}
System.out.println(max);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers A and a number K, find maximum sum of a subarray of size K.The first line of input contains two integers N and K, denoting the number of elements in the array and the subarray size respectively.
The next line contains N integers denoting the elements of the array respectively.
1 <= K <= N <= 200000
-200000 <= A[i] <= 200000Print a single integer denoting the maximum sum of subarray of size K.Sample Input:
4 2
-1 5 2 -3
Sample Output:
7
Explanation:
Three subarrays of size 2, their sum are 4, 7, -1, I have written this Solution Code: n1,k1=input().split()
n,k=int(n1),int(k1)
l=list(map(int,input().strip().split()))
curr_sum=0
for x in range(k):
curr_sum=curr_sum+l[x]
max_sum=curr_sum
i=0
j=k
while i<len(l)-k:
curr_sum=curr_sum-l[i]+l[j]
max_sum=max(max_sum,curr_sum)
i+=1
j+=1
print(max_sum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers A and a number K, find maximum sum of a subarray of size K.The first line of input contains two integers N and K, denoting the number of elements in the array and the subarray size respectively.
The next line contains N integers denoting the elements of the array respectively.
1 <= K <= N <= 200000
-200000 <= A[i] <= 200000Print a single integer denoting the maximum sum of subarray of size K.Sample Input:
4 2
-1 5 2 -3
Sample Output:
7
Explanation:
Three subarrays of size 2, their sum are 4, 7, -1, I have written this Solution Code: #include "bits/stdc++.h"
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];
void solve(){
int n, k, ans = 0;
cin >> n >> k;
for(int i = 1; i <= n; i++)
cin >> a[i];
int sum = 0;
for(int i = 1; i <= k; i++)
sum += a[i];
ans = sum;
for(int i = k+1; i <= n; i++){
sum += a[i] - a[i-k];
ans = max(ans, sum);
}
cout << ans;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ready for a super simple challenge!
You are given two integers A and B. You will create a line of boys and girls in the following manner:
First A girls, then B boys, then A girls, then B boys, and so on.
You will also be given an integer N. Find the number of girls in the first N people of the row.The first and the only line of input contains 3 integers N, A, B.
Constraints
1 < N, A, B <= 10^18 (1000000000000000000)
A+B <= 10^18Output a singer integer, the number of girls in the first N people in the queue.Sample Input
8 3 4
Sample Output
4
Explanation
The arrangement will look like GGGBBBBG, where G represents girl and B represents boy., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
String[] nab = rd.readLine().split(" ");
long n = Long.parseLong(nab[0]);
long girl = Long.parseLong(nab[1]);
long boy = Long.parseLong(nab[2]);
long total = 0;
if(n>1 && girl>1 && boy>1){
long temp = n/(girl+boy);
total = (girl*temp);
if(n%(girl+boy)<girl)
total+=n%(girl+boy);
else
total+=girl;
System.out.print(total);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ready for a super simple challenge!
You are given two integers A and B. You will create a line of boys and girls in the following manner:
First A girls, then B boys, then A girls, then B boys, and so on.
You will also be given an integer N. Find the number of girls in the first N people of the row.The first and the only line of input contains 3 integers N, A, B.
Constraints
1 < N, A, B <= 10^18 (1000000000000000000)
A+B <= 10^18Output a singer integer, the number of girls in the first N people in the queue.Sample Input
8 3 4
Sample Output
4
Explanation
The arrangement will look like GGGBBBBG, where G represents girl and B represents boy., I have written this Solution Code: n,a,b=map(int,input().split())
if n%(a+b)== 0:
print(n//(a+b)*a)
else:
k=n-(n//(a+b))*(a+b)
if k >= a:
print((n//(a+b))*a+a)
else:
print((n//(a+b))*a+k), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ready for a super simple challenge!
You are given two integers A and B. You will create a line of boys and girls in the following manner:
First A girls, then B boys, then A girls, then B boys, and so on.
You will also be given an integer N. Find the number of girls in the first N people of the row.The first and the only line of input contains 3 integers N, A, B.
Constraints
1 < N, A, B <= 10^18 (1000000000000000000)
A+B <= 10^18Output a singer integer, the number of girls in the first N people in the queue.Sample Input
8 3 4
Sample Output
4
Explanation
The arrangement will look like GGGBBBBG, where G represents girl and B represents boy., 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(ll 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 int long long
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define MOD 1000000007
#define INF 1000000000000000007LL
const int N = 100005;
// it's swapnil07 ;)
#ifdef SWAPNIL07
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define end_routine()
#endif
signed main()
{
fast
int n, a, b; cin>>n>>a>>b;
int x = a+b;
int y = (n/x);
n -= (y*x);
int ans = y*a;
if(n > a)
ans += a;
else
ans += n;
cout<<ans;
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: #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 |
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