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For this Question: Find factorial of a given number N.
<b>Note: </b>
The Factorial of a number is the product of an integer and all the integers below it;
e.g. factorial four ( 4! ) is equal to 24 (4*3*2*1).<b>User Task</b>
Since this is a functional problem, you don't have to worry about the input. You just have to complete the function <i>Factorial()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 15
Return the factorial of the given number.Sample Input:-
5
Sample Output:-
120
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code: def factorial(n):
if(n == 1):
return 1
return n * factorial(n-1)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number N.
<b>Note: </b>
The Factorial of a number is the product of an integer and all the integers below it;
e.g. factorial four ( 4! ) is equal to 24 (4*3*2*1).<b>User Task</b>
Since this is a functional problem, you don't have to worry about the input. You just have to complete the function <i>Factorial()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 15
Return the factorial of the given number.Sample Input:-
5
Sample Output:-
120
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code: static int Factorial(int N)
{
if(N==0){
return 1;}
return N*Factorial(N-1);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number N.
<b>Note: </b>
The Factorial of a number is the product of an integer and all the integers below it;
e.g. factorial four ( 4! ) is equal to 24 (4*3*2*1).<b>User Task</b>
Since this is a functional problem, you don't have to worry about the input. You just have to complete the function <i>Factorial()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 15
Return the factorial of the given number.Sample Input:-
5
Sample Output:-
120
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code: // n is the input number
function factorial(n) {
// write code here
// do not console.log
// return the answer as a number
if (n == 1 ) return 1;
return n * factorial(n-1)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: How would you add your own method to the Array object so
the following code would work?
const arr = [1, 2, 3]
console. log(arr.average())
// 2input will be an array, run like this
const anyArray = [5,6...]
anyArray.average should return average of the arraysAverage of the given arrayconst myArray = [1,2,3,4,5]
console.log(myArray.average())
// 3, I have written this Solution Code: Array.prototype.average = function() {
// calculate sum
var sum = this.reduce(function(prev, cur) { return prev + cur; });
// return sum divided by number of elements
return sum / this.length;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary tree of N nodes, with root 1 and Q queries with nodes u and v. For each query find the sum of nodes on the shortest path from u to v, where node v is the ancestor of node u.First line contains the integer N and Q, denoting the number of nodes in the binary tree and the number of queries respectively.
Next N lines contains two integers denoting the left and right child of the i'th node respectively.
If the node doesn't have a left or right child, it is denoted by '-1'
Next Q lines contain two integers u and v
1 <= N <= 10000
1 <= Q <= 100000
1 <= u, v <= NPrint Q lines denoting the sum of nodes on the shortest path from u to vSample Input 1:
6 3
2 4
5 3
-1 -1
-1 -1
6 -1
-1 -1
6 2
3 1
5 5
Sample output 1:
13
6
5
Explanation: Given binary tree
1
/ \
2 4
/ \
5 3
/
6
Query 1: 6+5+2 = 13
Query 2: 3+2+1 = 6
Query 3: 5 , 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 q=sc.nextInt();
int arr[]=new int[n+1];
arr[1]=1;
for(int i=1;i<=n;i++){
int l=sc.nextInt();
int r=sc.nextInt();
if(l!=-1){
arr[l]=l+arr[i];
}
if(r!=-1){
arr[r]=r+arr[i];
}
}
for(int i=0;i<q;i++){
int u=sc.nextInt();
int v=sc.nextInt();
System.out.println(arr[u]-arr[v]+v);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary tree of N nodes, with root 1 and Q queries with nodes u and v. For each query find the sum of nodes on the shortest path from u to v, where node v is the ancestor of node u.First line contains the integer N and Q, denoting the number of nodes in the binary tree and the number of queries respectively.
Next N lines contains two integers denoting the left and right child of the i'th node respectively.
If the node doesn't have a left or right child, it is denoted by '-1'
Next Q lines contain two integers u and v
1 <= N <= 10000
1 <= Q <= 100000
1 <= u, v <= NPrint Q lines denoting the sum of nodes on the shortest path from u to vSample Input 1:
6 3
2 4
5 3
-1 -1
-1 -1
6 -1
-1 -1
6 2
3 1
5 5
Sample output 1:
13
6
5
Explanation: Given binary tree
1
/ \
2 4
/ \
5 3
/
6
Query 1: 6+5+2 = 13
Query 2: 3+2+1 = 6
Query 3: 5 , I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define ll long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int l[N], r[N], p[N], s[N];
void dfs(int u, int p = 0){
if(u == -1) return;
s[u] = s[p] + u;
dfs(l[u], u);
dfs(r[u], u);
}
signed main() {
IOS;
clock_t start = clock();
int n, q; cin >> n >> q;
for(int i = 1; i <= n; i++){
cin >> l[i] >> r[i];
p[l[i]] = p[r[i]] = i;
}
dfs(1);
while(q--){
int u, v;
cin >> u >> v;
cout << s[u] - s[p[v]] << endl;
}
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 year (an integer value, as input), determine whether it is a leap year or not. If it is a leap year, return the Boolean value True, otherwise, return False.
For a year to be Leap Year either of the following conditions should be true:
It is divisible by 400.
It is divisible by 4 but it is not divisible by 100.
If none of the above conditions is true, then it is not a leap year.The first line contains the year to be checked.
<b>Constraints:</b>
1900 ≤ year ≤10^5The function must return a Boolean value (True/False).Sample Input:
1900
Sample Output:
False
Explanation:
1900 is not divisible by 400 .
1900 is a multiple of 4 , but it is also divisible by 100.
, I have written this Solution Code: def isleap(year):
return year % 4 == 0 and (year % 400 == 0 or year % 100 != 0)
n = int(input())
LeapYear=isleap(n)
print (LeapYear), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given number N, Consider a table of N X N in which elements at the intersection of ith row and jth column are the product of i and j, i. e i x j.
Also given a positive integer X. Your task is to count the number of elements in the table that contain the integer X.<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>OccurenceOfX()</b> that takes the integer N and the integer X as parameter.
Constraints:-
1 <= N <= 10^5
1 <= X <= 10^9Return the count of X.Sample Input:-
5 5
Sample Output:-
2
Explanation:-
table :-
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
Count of occurrence of X is :- 2
Sample Input:-
10 13
Sample Output:-
0, I have written this Solution Code: def OccurenceOfX(N,X):
cnt=0
for i in range(1, N+1):
if(X%i==0 and X/i<=N):
cnt=cnt+1
return cnt, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given number N, Consider a table of N X N in which elements at the intersection of ith row and jth column are the product of i and j, i. e i x j.
Also given a positive integer X. Your task is to count the number of elements in the table that contain the integer X.<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>OccurenceOfX()</b> that takes the integer N and the integer X as parameter.
Constraints:-
1 <= N <= 10^5
1 <= X <= 10^9Return the count of X.Sample Input:-
5 5
Sample Output:-
2
Explanation:-
table :-
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
Count of occurrence of X is :- 2
Sample Input:-
10 13
Sample Output:-
0, I have written this Solution Code:
int OccurenceOfX(int N,long X){
int cnt=0,i;
for( i=1;i<=N;i++){
if(X%i==0 && X/i<=N){cnt++;}}
return cnt;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given number N, Consider a table of N X N in which elements at the intersection of ith row and jth column are the product of i and j, i. e i x j.
Also given a positive integer X. Your task is to count the number of elements in the table that contain the integer X.<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>OccurenceOfX()</b> that takes the integer N and the integer X as parameter.
Constraints:-
1 <= N <= 10^5
1 <= X <= 10^9Return the count of X.Sample Input:-
5 5
Sample Output:-
2
Explanation:-
table :-
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
Count of occurrence of X is :- 2
Sample Input:-
10 13
Sample Output:-
0, I have written this Solution Code:
int OccurenceOfX(int N,long X){
int cnt=0,i;
for( i=1;i<=N;i++){
if(X%i==0 && X/i<=N){cnt++;}}
return cnt;
}
int main()
{
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given number N, Consider a table of N X N in which elements at the intersection of ith row and jth column are the product of i and j, i. e i x j.
Also given a positive integer X. Your task is to count the number of elements in the table that contain the integer X.<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>OccurenceOfX()</b> that takes the integer N and the integer X as parameter.
Constraints:-
1 <= N <= 10^5
1 <= X <= 10^9Return the count of X.Sample Input:-
5 5
Sample Output:-
2
Explanation:-
table :-
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
Count of occurrence of X is :- 2
Sample Input:-
10 13
Sample Output:-
0, I have written this Solution Code: public static int OccurenceOfX(int N,int X){
int cnt=0,i;
for( i=1;i<=N;i++){
if(X%i==0 && X/i<=N){cnt++;}}
return cnt;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <b> Update: The time limit for the problem has been updated to 5s.</b>
Bob is a prodigy. To avoid constant pestering, his professor has given him a non-negative integer N and the following task.
Bob needs to find out the smallest such <b>positive</b> integer X such that the product of the digits of X is exactly N. Obviously, this is not possible for many values of N. However, if the product is taken modulo 10<sup>9</sup>+7, the professor claims the answer exists for any value of N (in the range 0 to 10^9+6). Further, he claims that <b>the answer will not exceed 1000 digits</b>.
Alas! Bob solved it within twenty minutes. Can you?The first line of the input contains a single integer T, the number of test cases. (1 <= T <= 10)
The next T lines contain a single integer each, i<sup>th</sup> one being the value of N for the ith test case. (0 <= N <= 10<sup>9</sup>+6)Print the smallest positive number for which the product of digits modulo 10<sup>9</sup>+7 equals N. It is guaranteed that answer exists and contains <= 1000 digits.Sample Input:
2
12
0
Sample Output:
26
10
Explanation:
26 is the smallest positive integer with product of digits = 12.
For the second case, note that we need a positive integer and hence, answer is 10 and not 0., I have written this Solution Code: #include<bits/stdc++.h>
#define ll long long
#define ull unsigned ll
#define uint ungigned
#define db double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pli pair<ll,int>
#define vi vector<int>
#define vpi vector<pii >
#define IT iterator
#define PB push_back
#define MK make_pair
#define LB lower_bound
#define UB upper_bound
#define y1 wzpakking
#define fi first
#define se second
#define BG begin
#define ED end
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define UPD(x,y) (((x)+=(y))>=mo?(x)-=mo:233)
#define CLR(a,v) memset(a,v,sizeof(a))
#define CPY(a,b) memcpy(a,b,sizeof(a))
#define sqr(x) (1ll*x*x)
#define LS3 k*2,l,mid
#define RS3 k*2+1,mid+1,r
#define LS5 k*2,l,mid,x,y
#define RS5 k*2+1,mid+1,r,x,y
#define GET pushdown(k);int mid=(l+r)/2
#define INF (1ll<<60)
using namespace std;
const int mo=1e9+7;
int pri[4]={2,3,5,7};
int P[4][2333];
map<int,pii > mp;
int power(int x,int y){
int s=1;
for (;y;y/=2,x=1ll*x*x%mo)
if (y&1) s=1ll*s*x%mo;
return s;
}
void init(){
For(i,0,3){
P[i][0]=1;
For(j,1,2100) P[i][j]=1ll*P[i][j-1]*pri[i]%mo;
if (i<=1)
For(j,0,2100) P[i][j]=power(P[i][j],mo-2);
}
Rep(i,2100,0) Rep(j,1400,0)
if ((i+2)/3+(j+1)/2<=700)
mp[1ll*P[0][i]*P[1][j]%mo]=pii(i,j);
}
int ans[12],tmp[12],v[4],V[4];
bool check(){
if (tmp[0]!=ans[0]) return tmp[0]<ans[0];
For(i,2,9) if (tmp[i]!=ans[i]) return tmp[i]>ans[i];
return 0;
}
void update(int i,int j,int k,int l){
For(i,0,9) tmp[i]=0;
v[3]=i; v[2]=j; v[1]=k; v[0]=l;
Rep(i,9,2){
int x=i;
For(j,0,3) V[j]=0;
For(j,0,3) for (;x%pri[j]==0;x/=pri[j],V[j]++);
int mn=100000;
For(j,0,3) if (V[j]) mn=min(mn,v[j]/V[j]);
tmp[i]=mn; tmp[0]+=mn;
For(j,0,3) v[j]-=mn*V[j];
}
if (check())
CPY(ans,tmp);
}
void solve(){
int v;
scanf("%d",&v);
if (!v){
puts("10");
return;
}
if (v==1){
puts("1");
return;
}
ans[0]=700;
v=power(v,mo-2);
For(i,0,700) For(j,0,700)
if (i+j<=ans[0]){
int val=1ll*v*P[2][i]%mo*P[3][j]%mo;
if (mp.find(val)==mp.end()) continue;
int l=mp[val].fi,k=mp[val].se;
if ((l+2)/3+(k+1)/2+i+j>=ans[0]) continue;
update(j,i,k,l);
}
For(i,2,9) For(j,1,ans[i])
putchar(i+'0');
puts("");
}
int main(){
init();
int T; scanf("%d",&T);
while (T--) solve();
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have to complete <code>printName</code> function
<ol>
<li>Takes 2 arguments which are EventEmitter object , event name </li>
<li>Using the <code>EventEmitter</code> object you need to register an eventListener for <code>personEvent</code> event which takes a callback function. </li>
<li>The callback function itself takes an argument (which would be emitted by test runner) which is of type string. Using that argument print to console.
`My name is ${argument}`.
</li>
</ol>Function will take two arguments
1) 1st argument will be an object of EventEmitter class
2) 2nd argument will be the event name (string)
Function returns a registered Event using an object of EventEmitter class (which then is used to print the name)<pre>
<code>
const emitter=EventEmitter object
const personEvent="event"
function printName(emitter,personEvent)//registers event with event Name "event"
emitter.emit("event","Dev")//emits the event and give the output "My name is Dev"
</code>
</pre>
, I have written this Solution Code: function printName(emitter,personEvent) {
emitter.on(personEvent,(arg)=>{
console.log('My name is ',arg);
});
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, construct a string of length N such that no two adjacent characters are the same. Among all possible strings, find the lexicographically minimum string.
Note: You can use only lowercase characters from 'a' to 'z'.The first and only line of input contains an integer N.
<b>Constraints:</b>
1 <= N <= 10<sup>5</sup>Print the required string.Sample Input 1:
1
Sample Output 1:
a
Sample Input 2:
2
Sample Output 2:
ab, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(isr);
int len = Integer.parseInt(br.readLine());
char[] str = new char[len];
for(int i = 0; i < len; i++){
if(i%2 == 0){
str[i] = 'a';
} else{
str[i] = 'b';
}
}
System.out.println(str);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, construct a string of length N such that no two adjacent characters are the same. Among all possible strings, find the lexicographically minimum string.
Note: You can use only lowercase characters from 'a' to 'z'.The first and only line of input contains an integer N.
<b>Constraints:</b>
1 <= N <= 10<sup>5</sup>Print the required string.Sample Input 1:
1
Sample Output 1:
a
Sample Input 2:
2
Sample Output 2:
ab, 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();
//////////////
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>
#define rep(i,n) for (int i=0; i<(n); i++)
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
string s(n,'a');
for(int i=1;i<n;i+=2)
s[i]='b';
cout<<s;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, construct a string of length N such that no two adjacent characters are the same. Among all possible strings, find the lexicographically minimum string.
Note: You can use only lowercase characters from 'a' to 'z'.The first and only line of input contains an integer N.
<b>Constraints:</b>
1 <= N <= 10<sup>5</sup>Print the required string.Sample Input 1:
1
Sample Output 1:
a
Sample Input 2:
2
Sample Output 2:
ab, I have written this Solution Code: a="ab"
inp = int(input())
print(a*(inp//2)+a[0:inp%2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary array A[] of size N. The task is to arrange an array in increasing order.
(O(N) time complexity solution)Every test case contains two lines, first line contains an integer N (size of array) and second line contains space separated elements of array.
1 <= N <= 20
0 <= A[i] <= 1Print a single line containing space separated elements of sorted arrays.Input:
5
1 0 1 1 0
Output:
0 0 1 1 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void cal(int arr[], int n){
int countZ = 0;
for(int i = 0; i < n; i++) {
if(arr[i] == 0) {
countZ++;
}
}
for(int i = 1; i <= countZ; i++) {
System.out.print("0 ");
}
for(int i = 1; i <= n - countZ; i++) {
System.out.print("1 ");
}
}
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String nD = br.readLine();
String nDArr[] = nD.split(" ");
int n = Integer.parseInt(nDArr[0]);
int arr[]= new int[n];
String input = br.readLine();
String sar[] = input.split(" ");
for(int i = 0; i < n; i++){
arr[i] = Integer.parseInt(sar[i]);
}
cal(arr, n);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary array A[] of size N. The task is to arrange an array in increasing order.
(O(N) time complexity solution)Every test case contains two lines, first line contains an integer N (size of array) and second line contains space separated elements of array.
1 <= N <= 20
0 <= A[i] <= 1Print a single line containing space separated elements of sorted arrays.Input:
5
1 0 1 1 0
Output:
0 0 1 1 1, I have written this Solution Code: n = int(input())
l = list(map(int, input().split()))
l = sorted(l)
for i in l:
print(i, end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary array A[] of size N. The task is to arrange an array in increasing order.
(O(N) time complexity solution)Every test case contains two lines, first line contains an integer N (size of array) and second line contains space separated elements of array.
1 <= N <= 20
0 <= A[i] <= 1Print a single line containing space separated elements of sorted arrays.Input:
5
1 0 1 1 0
Output:
0 0 1 1 1, 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 n; cin >> n;
int a[2] = {0};
for(int i = 1; i <= n; i++){
int p; cin >> p;
a[p]++;
}
for(int i = 0; i <= 1; i++)
for(int j = 0; j < a[i]; j++)
cout << i << " ";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two arrays and an integer x. Calculate the minimum absolute difference between sum of pair of elements (one from each array) and the integer x.First line of input contains the length of the array N
Second line contains first array elements
Third line contains second array elements
Last line contains the value of x
Constraints:-
1<=N<=10000
1<=elements<=100000
1<=x<=100000Output a single line containing the minimum differenceSample Input:-
4
1 4 5 7
10 20 30 40
32
Sample Output:-
1
Explanation:
Required pair is 30,1., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(in.readLine());
int[] a=new int[n];
int[] b=new int[n];
String s[]=in.readLine().split(" ");
for(int i=0;i<n;i++){
a[i]=Integer.parseInt(s[i]);
}
s=in.readLine().split(" ");
for(int i=0;i<n;i++){
b[i]=Integer.parseInt(s[i]);
}
int x=Integer.parseInt(in.readLine());
Arrays.sort(a);
Arrays.sort(b);
int i=0,j=n-1;
int min=Integer.MAX_VALUE;
while(i<n&&j>=0){
int sum=a[i]+b[j];
if(sum==x){
min=0;
break;
}else if(sum>x){
min=Math.min(sum-x,min);
j--;
}else{
min=Math.min(x-sum,min);
i++;
}
}
System.out.print(min);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two arrays and an integer x. Calculate the minimum absolute difference between sum of pair of elements (one from each array) and the integer x.First line of input contains the length of the array N
Second line contains first array elements
Third line contains second array elements
Last line contains the value of x
Constraints:-
1<=N<=10000
1<=elements<=100000
1<=x<=100000Output a single line containing the minimum differenceSample Input:-
4
1 4 5 7
10 20 30 40
32
Sample Output:-
1
Explanation:
Required pair is 30,1., I have written this Solution Code: n=int(input())
array1=list(map(int,input().split()))
array2=list(map(int,input().split()))
array1=sorted(array1)
array2=sorted(array2)
value=int(input())
i=0
n=len(array1)
j=len(array2)-1
closest=None
while(i<n and j>=0):
v=array1[i]+array2[j]
if closest==None:
closest=v
elif abs(v-value)<abs(closest-value):
closest=v
if v>value:
j-=1
else:
i+=1
print(abs(closest-value)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two arrays and an integer x. Calculate the minimum absolute difference between sum of pair of elements (one from each array) and the integer x.First line of input contains the length of the array N
Second line contains first array elements
Third line contains second array elements
Last line contains the value of x
Constraints:-
1<=N<=10000
1<=elements<=100000
1<=x<=100000Output a single line containing the minimum differenceSample Input:-
4
1 4 5 7
10 20 30 40
32
Sample Output:-
1
Explanation:
Required pair is 30,1., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 10000001
int main(){
int n,q;
cin>>n;
int a[n];
int b[n];
long sum=LONG_MAX;
unordered_map<int,int> a1,a2;
for(int i=0;i<n;i++){
cin>>a[i];a1[a[i]]++;
}
for(int i=0;i<n;i++){
cin>>b[i];
a2[b[i]]++;
}
int x;
cin>>x;
sort(a,a+n);
sort(b,b+n);
int c[2*n];
int i=0,j=0,k=0;
while(i!=n && j!=n){
if(a[i]<b[j]){c[k]=a[i];k++;i++;}
else{c[k]=b[j];j++;k++;}
}
while(i!=n){
c[k]=a[i];
k++;i++;
}
while(j!=n){
c[k]=b[j];
k++;j++;
}
i=0,j=2*n-1;
while(i<j){
if((c[i]+c[j])>x){if(a1.find(c[i])==a1.end() && a1.find(c[j])==a1.end()){j--;continue;}
else if(a2.find(c[i])==a2.end() && a2.find(c[j])==a2.end()){j--;continue;}
sum=min(sum,(long)abs(x-(c[i]+c[j])));
j--;}
else{
if(a1.find(c[i])==a1.end() && a1.find(c[j])==a1.end()){i++;continue;}
else if(a2.find(c[i])==a2.end() && a2.find(c[j])==a2.end()){i++;continue;}
sum=min(sum,(long)abs(x-(c[i]+c[j])));
i++;
}
}
cout<<sum<<endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tom is a teacher responsible for a class of N students. The students are given distinct student numbers from 1 to N.
Today, all the students entered the classroom at different times. According to Tom's record, there were Ai students in the classroom when student number i entered the classroom (including student number i).
From these records, reconstruct the order in which the students entered the classroom.Input is given from Standard Input in the following format:
First line contains a single integer N the array length and the second line contains elements of the array
Constraints
1 ≤ N ≤ 10000
1 ≤ Ai ≤ N
Ai ≠ Aj if (i≠j)Print the student numbers of the students in the order the students entered the classroom.Sample Input
3
2 3 1
Sample Output
3 1 2
Explanation:-
First, student number 3 entered the classroom.
Then, student number 1 entered the classroom.
Finally, student number 2 entered the classroom.
Sample Input:
5
5 4 3 2 1
Sample Output:
5 4 3 2 1 , I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
String s[]=br.readLine().split(" ");
int arr1[]=new int[n];
int arr2[]=new int[n];
for(int i=0;i<n;i++)
{
arr1[i]=Integer.parseInt(s[i]);
arr2[arr1[i]-1]=i+1;
}
for(int i=0;i<n;i++)
{
System.out.print(arr2[i]+" ");
}
System.out.println();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tom is a teacher responsible for a class of N students. The students are given distinct student numbers from 1 to N.
Today, all the students entered the classroom at different times. According to Tom's record, there were Ai students in the classroom when student number i entered the classroom (including student number i).
From these records, reconstruct the order in which the students entered the classroom.Input is given from Standard Input in the following format:
First line contains a single integer N the array length and the second line contains elements of the array
Constraints
1 ≤ N ≤ 10000
1 ≤ Ai ≤ N
Ai ≠ Aj if (i≠j)Print the student numbers of the students in the order the students entered the classroom.Sample Input
3
2 3 1
Sample Output
3 1 2
Explanation:-
First, student number 3 entered the classroom.
Then, student number 1 entered the classroom.
Finally, student number 2 entered the classroom.
Sample Input:
5
5 4 3 2 1
Sample Output:
5 4 3 2 1 , I have written this Solution Code: t=int(input())
li=[int(i) for i in input().split()]
ans=[0]*t
for i in range(t):
ans[li[i]-1] = i+1
for i in ans:
print(i,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tom is a teacher responsible for a class of N students. The students are given distinct student numbers from 1 to N.
Today, all the students entered the classroom at different times. According to Tom's record, there were Ai students in the classroom when student number i entered the classroom (including student number i).
From these records, reconstruct the order in which the students entered the classroom.Input is given from Standard Input in the following format:
First line contains a single integer N the array length and the second line contains elements of the array
Constraints
1 ≤ N ≤ 10000
1 ≤ Ai ≤ N
Ai ≠ Aj if (i≠j)Print the student numbers of the students in the order the students entered the classroom.Sample Input
3
2 3 1
Sample Output
3 1 2
Explanation:-
First, student number 3 entered the classroom.
Then, student number 1 entered the classroom.
Finally, student number 2 entered the classroom.
Sample Input:
5
5 4 3 2 1
Sample Output:
5 4 3 2 1 , I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a[n],b;
for(int i=0;i<n;i++){
cin>>b;
b--;
a[b]=i;
}
for(int i=0;i<n;i++){
cout<<a[i]+1<<" ";
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an ArrayList of N lowercase characters. The task is to insert given elements in the list and count frequency of elements present in the list.
You can use some inbuilt functions as:-
add() to append element in the list
contains() to check an element is present or not in the list
collections.frequency() to find the frequency of the element in the list.<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>insert()</b> and <b>freq()</b> that takes the array list and the character c as parameters.
Constraints:
1 <= T <= 100
1 <= N <= 1000
c will be a lowercase english character
You need to print the count of the character c if it is present else you need to print "Not Present" all in a separate line in function freq().Sample Input:
2
6
i n i e i w i t i n f n
4
i c i p i p f f
Sample Output:
2
Not Present
Explanation:
Testcase 1: Inserting n, e, w, t, n into the list. Frequency of n is 2 in the list.
Testcase 2: Inserting c, p, p into the list. Frequency of f is 0 in the list., I have written this Solution Code:
// Function to insert element
public static void insert(ArrayList<Character> clist, char c)
{
clist.add(c);
}
// Function to count frequency of element
public static void freq(ArrayList<Character> clist, char c)
{
if(clist.contains(c) == true)
System.out.println(Collections.frequency(clist, c));
else
System.out.println("Not Present");
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, I have written this Solution Code: m,n=map(int ,input().split())
matrix=[]
for i in range(m):
l1=[eval(x) for x in input().split()]
matrix.append(l1)
l2=[]
for coloumn in range(n):
sum1=0
for row in range(m):
sum1+= matrix[row][coloumn]
l2.append(sum1)
print(max(l2))
'''for row in range(n):
sum2=0
for col in range(m):
sum2 += matrix[row][col]
print(sum2)''', In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, I have written this Solution Code: // mat is the matrix/ 2d array
// the dimensions of array are:- a rows, b columns
function colMaxSum(mat,a,b) {
// write code here
// do not console.log
// return the answer as a number
let idx = -1;
// Variable to store max sum
let maxSum = Number.MIN_VALUE;
// Traverse matrix column wise
for (let i = 0; i < b; i++) {
let sum = 0;
// calculate sum of column
for (let j = 0; j < a; j++) {
sum += mat[j][i];
}
// Update maxSum if it is
// less than current sum
if (sum > maxSum) {
maxSum = sum;
// store index
idx = i;
}
}
let res;
res = [idx, maxSum];
// return result
return maxSum;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
signed main(){
int n,m;
cin>>n>>m;
int a[m];
for(int i=0;i<m;i++){
a[i]=0;
}
int x;
int sum=0;
FOR(i,n){
FOR(j,m){
cin>>x;
a[j]+=x;
sum=max(sum,a[j]);
}
}
out(sum);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, 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 m = sc.nextInt();
int n = sc.nextInt();
int a[][] = new int[m][n];
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
a[i][j]=sc.nextInt();
}
}
int sum=0;
int ans=0;
for(int i=0;i<n;i++){
sum=0;
for(int j=0;j<m;j++){
sum+=a[j][i];
}
if(sum>ans){ans=sum;}
}
System.out.print(ans);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We are given a string. Your task is to compress the consecutive letters of the string
For example, given string is "AAACCCBBD", thus here
A's occurrence 3 times
C's occurrence 3 times
B's occurrence 2 times
D's occurrence 1 time
So after compressing string becomes "A3C3B2D1".The first line of input contains an integer T denoting the number of test cases. Each test case will have a string provided in the new line.
Constraints:
1 <= T <= 10
1 <= sizeof(String) <= 10^6
All characters of String are upper case letters. (A-Z)
Sum of size of Strings over all testcases is <= 10^6For each testcase, in a new line, print the compressed string for each test case in a new line. Input:
2
AAACCCBBD
ABCD
Output:
A3C3B2D1
A1B1C1D1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void compress(String str, int l){
for (int i = 0; i < l; i++) {
int count = 1;
while (i < l - 1 && str.charAt(i) == str.charAt(i + 1)) {
count++;
i++;
}
System.out.print(str.charAt(i));
System.out.print(count);
}
System.out.println();
}
public static void main (String[] args) throws IOException{
BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
int test = Integer.parseInt(rd.readLine());
while(test-->0){
String s = rd.readLine();
int len = s.length();
compress(s,len);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We are given a string. Your task is to compress the consecutive letters of the string
For example, given string is "AAACCCBBD", thus here
A's occurrence 3 times
C's occurrence 3 times
B's occurrence 2 times
D's occurrence 1 time
So after compressing string becomes "A3C3B2D1".The first line of input contains an integer T denoting the number of test cases. Each test case will have a string provided in the new line.
Constraints:
1 <= T <= 10
1 <= sizeof(String) <= 10^6
All characters of String are upper case letters. (A-Z)
Sum of size of Strings over all testcases is <= 10^6For each testcase, in a new line, print the compressed string for each test case in a new line. Input:
2
AAACCCBBD
ABCD
Output:
A3C3B2D1
A1B1C1D1, I have written this Solution Code:
def compress(st):
n = len(st)
i = 0
while i < n:
count = 1
while (i < n-1 and st[i] == st[i + 1]):
count += 1
i += 1
i += 1
print(st[i-1] +str(count),end="")
t=int(input())
for i in range(t):
s=input()
compress(s)
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We are given a string. Your task is to compress the consecutive letters of the string
For example, given string is "AAACCCBBD", thus here
A's occurrence 3 times
C's occurrence 3 times
B's occurrence 2 times
D's occurrence 1 time
So after compressing string becomes "A3C3B2D1".The first line of input contains an integer T denoting the number of test cases. Each test case will have a string provided in the new line.
Constraints:
1 <= T <= 10
1 <= sizeof(String) <= 10^6
All characters of String are upper case letters. (A-Z)
Sum of size of Strings over all testcases is <= 10^6For each testcase, in a new line, print the compressed string for each test case in a new line. Input:
2
AAACCCBBD
ABCD
Output:
A3C3B2D1
A1B1C1D1, 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 = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
void solve(){
string s; cin >> s;
int c = 1;
char p = 0;
int n = s.length();
for(int i = 1; i < n; i++){
if(s[i] != s[i-1]){
cout << s[i-1] << c;
c = 1;
}
else
c++;
}
cout << s[n-1] << c << endl;
}
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 an array of N integers and an integer K, your task is to calculate the count of pairs whose sum is divisible by K.The first line of input contains two integers N and K, the next line contains N space-separated integers depicting values of an array.
Constraints:-
1 < = N < = 100000
1 < = Arr[i] <= 100000
1 <= K <= 100000Print the count of required pairs.Sample Input
5 4
1 2 3 4 5
Sample Output
2
Sample Input
5 3
1 2 3 4 5
Sample Output
4
Explanation:-
In Sample 2,
(1 5), (1 2), (2 4), and (4 5) are the required pairs, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static long subarraysDivByK(int[] A, int k)
{
long ans =0 ;
int rem;
int[] freq = new int[k];
for(int i=0;i<A.length;i++)
{
rem = A[i]%k;
ans += freq[(k - rem)% k] ;
freq[rem]++;
}
return ans;
}
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] str = br.readLine().split(" ");
String[] input = br.readLine().split(" ");
int n = Integer.parseInt(str[0]);
int k = Integer.parseInt(str[1]);
int [] a = new int [n];
for(int i=0; i<n; i++)
a[i] = Integer.parseInt(input[i]);
System.out.println(subarraysDivByK(a, k));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers and an integer K, your task is to calculate the count of pairs whose sum is divisible by K.The first line of input contains two integers N and K, the next line contains N space-separated integers depicting values of an array.
Constraints:-
1 < = N < = 100000
1 < = Arr[i] <= 100000
1 <= K <= 100000Print the count of required pairs.Sample Input
5 4
1 2 3 4 5
Sample Output
2
Sample Input
5 3
1 2 3 4 5
Sample Output
4
Explanation:-
In Sample 2,
(1 5), (1 2), (2 4), and (4 5) are the required pairs, I have written this Solution Code: def countKdivPairs(A, n, K):
freq = [0] * K
for i in range(n):
freq[A[i] % K]+= 1
sum = freq[0] * (freq[0] - 1) / 2;
i = 1
while(i <= K//2 and i != (K - i) ):
sum += freq[i] * freq[K-i]
i+= 1
if( K % 2 == 0 ):
sum += (freq[K//2] * (freq[K//2]-1)/2);
return int(sum)
a,b=input().split()
a=int(a)
b=int(b)
arr=input().split()
for i in range(0,a):
arr[i]=int(arr[i])
print (countKdivPairs(arr,a, b)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers and an integer K, your task is to calculate the count of pairs whose sum is divisible by K.The first line of input contains two integers N and K, the next line contains N space-separated integers depicting values of an array.
Constraints:-
1 < = N < = 100000
1 < = Arr[i] <= 100000
1 <= K <= 100000Print the count of required pairs.Sample Input
5 4
1 2 3 4 5
Sample Output
2
Sample Input
5 3
1 2 3 4 5
Sample Output
4
Explanation:-
In Sample 2,
(1 5), (1 2), (2 4), and (4 5) are the required pairs, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define 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);
}
signed main(){
fast();
int n;
cin>>n;
int a;
int k;
cin>>k;
int fre[k];
FOR(i,k){
fre[i]=0;}
FOR(i,n){
cin>>a;
fre[a%k]++;
}
int ans=(fre[0]*(fre[0]-1))/2;
for(int i=1;i<=(k-1)/2;i++){
ans+=fre[i]*fre[k-i];
}
if(k%2==0){
ans+=(fre[k/2]*(fre[k/2]-1))/2;
}
out(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array a of n integers a<sub>1</sub>, a<sub>2</sub>,... , a<sub>n</sub>. Your task is to response to the queries like : How many number's values are between l and r ?The first line of the input contains a single space separated integer n denoting the length of the array.
The second line of the input contains n space separated integers denoting an array a.
The third line of the input contains a single space separated integer k denoting the number of queries.
The following k lines contain a pair of integers l and r - query, described above.
<b>Constraints</b>
1 ≤ N ≤ 10<sup>5</sup>
-10<sup>9</sup> ≤ a[i] ≤ 10<sup>9</sup>
1 ≤ k ≤ 10<sup>5</sup>
-10<sup>9</sup> ≤ l ≤ r ≤ 10<sup>9</sup>The output must consist of k integers - responses for the queries.Sample Input
5
10 1 10 3 4
4
1 10
2 9
3 4
2 2
Sample Output
5 2 2 0, I have written this Solution Code: #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k) returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
#define int long long
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int64_t random_long(int l = LLONG_MIN, int r = LLONG_MAX) {
uniform_int_distribution<int64_t> generator(l, r);
return generator(rng);
}
vector<int> a;
int searchLeft(int x, int n) {
int l = 0, r = n - 1;
while (l <= r) {
int mid = l + (r - l) / 2;
if (a[mid] <= x) {
l = mid + 1;
} else {
r = mid - 1;
}
}
return l;
}
int searchRight(int x, int n) {
int l = 0, r = n - 1;
while (l <= r) {
int mid = l + (r - l) / 2;
if (a[mid] < x) {
l = mid + 1;
} else {
r = mid - 1;
}
}
return l + 1;
}
int32_t main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, k;
cin >> n;
a.assign(n, 0);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a.begin(), a.end());
cin >> k;
while (k--) {
int x, y;
cin >> x >> y;
x -= 1, y += 1;
int left = searchLeft(x, n);
int right = searchRight(y, n);
cout << right - left - 1 << " ";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given resistance value of N resistors. Find the net resistance of the system when all of these N resistors are connected in parallel.
If there are three resistors A1, A2, A3, when they are connected in parallel, the net resistance will be 1/((1/A1) + (1/A2) + (1/A3))
Since this number can also have a fraction part, you only have to print the floor of the result obtained.
For example, if value of 1/((1/A1) + (1/A2) + (1/A3)) if 7.54567, you only have to print 7.First line contains a single integer N denoting the number of resistors.
Next line contains N space separated integers containing the value of different resistors.
Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^9Print the integral part or floor of the value obtained from the formula 1/((1/A1) + (1/A2) + ..... + (1/AN)).Input
2
10 30
Output
7
1/((1/10) + (1/30)) = 30/4 = 7.5 and floor of 7.5 is 7, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int N = Integer.parseInt(br.readLine());
double arr[] = new double[N];
String str[] = br.readLine().trim().split(" ");
for(int i=0;i<N;i++)
arr[i]=Integer.parseInt(str[i]);
double resistance=0;
int equResistance=0;
for(int i=0;i<N;i++)
arr[i]=Integer.parseInt(str[i]);
for(int i=0;i<N;i++)
{
resistance=resistance+(1/arr[i]);
}
equResistance = (int)Math.floor((1/resistance));
System.out.println(equResistance);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given resistance value of N resistors. Find the net resistance of the system when all of these N resistors are connected in parallel.
If there are three resistors A1, A2, A3, when they are connected in parallel, the net resistance will be 1/((1/A1) + (1/A2) + (1/A3))
Since this number can also have a fraction part, you only have to print the floor of the result obtained.
For example, if value of 1/((1/A1) + (1/A2) + (1/A3)) if 7.54567, you only have to print 7.First line contains a single integer N denoting the number of resistors.
Next line contains N space separated integers containing the value of different resistors.
Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^9Print the integral part or floor of the value obtained from the formula 1/((1/A1) + (1/A2) + ..... + (1/AN)).Input
2
10 30
Output
7
1/((1/10) + (1/30)) = 30/4 = 7.5 and floor of 7.5 is 7, I have written this Solution Code: r = input("")
r = int(r)
n = input("").split()
resistance=0.0
for i in range(0,r):
resistor = float(n[i])
resistance = resistance + (1/resistor)
print(int(1/resistance)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given resistance value of N resistors. Find the net resistance of the system when all of these N resistors are connected in parallel.
If there are three resistors A1, A2, A3, when they are connected in parallel, the net resistance will be 1/((1/A1) + (1/A2) + (1/A3))
Since this number can also have a fraction part, you only have to print the floor of the result obtained.
For example, if value of 1/((1/A1) + (1/A2) + (1/A3)) if 7.54567, you only have to print 7.First line contains a single integer N denoting the number of resistors.
Next line contains N space separated integers containing the value of different resistors.
Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^9Print the integral part or floor of the value obtained from the formula 1/((1/A1) + (1/A2) + ..... + (1/AN)).Input
2
10 30
Output
7
1/((1/10) + (1/30)) = 30/4 = 7.5 and floor of 7.5 is 7, 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 n; cin >> n;
double s = 0;
for(int i = 1; i <= n; i++){
double p; cin >> p;
s = s + (1/p);
}
s = 1/s;
cout << floor(s);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S you have to remove all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, remove them as well. Repeat this untill no adjacent letter in the string is same.
Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.The input data consists of a single string S.
Constraints:
1 <= |S| <= 100000
S contains lowercase english letters only.Print the given string after it is processed. It is guaranteed that the result will contain at least one character.Sample Input
hhoowaaaareyyoouu
Sample Output
wre
Explanation:
First we remove "hh" then "oo" then "aa" then "yy" then "oo" then "uu" and we are left with "wre".
Now we cannot remove anything.
Sample Input:-
abcde
Sample Output:-
abcde
Sample Input:-
abcddcb
Sample Output:-
a, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader in =new BufferedReader(new InputStreamReader(System.in));
StringBuilder s = new StringBuilder();
String text=null;
while ((text = in.readLine ()) != null)
{
s.append(text);
}
int len=s.length();
for(int i=0;i<len-1;i++){
if(s.charAt(i)==s.charAt(i+1)){
int flag=0;
s.delete(i,i+2);
int left=i-1;
len=len-2;
i=i-2;
if(i<0){
i=-1;
}
}
}
System.out.println(s);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S you have to remove all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, remove them as well. Repeat this untill no adjacent letter in the string is same.
Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.The input data consists of a single string S.
Constraints:
1 <= |S| <= 100000
S contains lowercase english letters only.Print the given string after it is processed. It is guaranteed that the result will contain at least one character.Sample Input
hhoowaaaareyyoouu
Sample Output
wre
Explanation:
First we remove "hh" then "oo" then "aa" then "yy" then "oo" then "uu" and we are left with "wre".
Now we cannot remove anything.
Sample Input:-
abcde
Sample Output:-
abcde
Sample Input:-
abcddcb
Sample Output:-
a, I have written this Solution Code: s=input()
l=["aa","bb","cc","dd","ee","ff","gg","hh","ii","jj","kk","ll","mm","nn","oo","pp","qq","rr","ss","tt","uu","vv","ww","xx","yy","zz"]
while True:
do=False
for i in range(len(l)):
if l[i] in s:
do=True
while l[i] in s:
s=s.replace(l[i],"")
if do==False:
break
print(s), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S you have to remove all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, remove them as well. Repeat this untill no adjacent letter in the string is same.
Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.The input data consists of a single string S.
Constraints:
1 <= |S| <= 100000
S contains lowercase english letters only.Print the given string after it is processed. It is guaranteed that the result will contain at least one character.Sample Input
hhoowaaaareyyoouu
Sample Output
wre
Explanation:
First we remove "hh" then "oo" then "aa" then "yy" then "oo" then "uu" and we are left with "wre".
Now we cannot remove anything.
Sample Input:-
abcde
Sample Output:-
abcde
Sample Input:-
abcddcb
Sample Output:-
a, 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();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main(){
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
string s;
cin>>s;
int len=s.length();
char stk[410000];
int k = 0;
for (int i = 0; i < len; i++)
{
stk[k++] = s[i];
while (k > 1 && stk[k - 1] == stk[k - 2])
k -= 2;
}
for (int i = 0; i < k; i++)
cout << stk[i];
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Morgan was opening some boxes when she found one bearing a message “For Morgan, Dad loves you 3000". She wants to open the box but she needs to solve the question to open it. She asks Peter to do that. Help Peter and Morgan solve the question so that they can see what's inside the box.
Given two arrays Arr1 and Arr2 of size N1 and N2. Your task is to find the sum of all elements that are common to both arrays. If there are no common elements the output would be 0.
Note: The arrays may contain duplicate elements. However, you need to sum only unique elements that are common to both arrays.The first line of input contains N1 and N2 separated by a space. The second line contains N1 space separated elements of Arr1. The third line contains N2 space separated elements of Arr2.
Constraints:
1 <= N1, N2 <= 10<sup>6</sup>
1 <= Arr1[i], Arr2[i] <= 1000000000Print the sum of common elements.Sample Input:
5 6
1 2 3 4 5
2 3 4 5 6 7
Sample Output:
14
Explanation:- Common elements = 2, 3, 4 , 5
sum= 2 + 3 + 4 + 5 = 14
Sample Input:-
3 3
1 2 3
4 5 6
Sample Output:-
0, 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 void main (String[] args) {
FastReader sc = new FastReader();
int n1 = sc.nextInt();
int n2 = sc.nextInt();
HashSet<Integer> hs = new HashSet<>();
long sum = 0l;
for(int i=0;i<n1;i++){
int curr = sc.nextInt();
hs.add(curr);
}
for(int i=0;i<n2;i++){
int sl = sc.nextInt();
if(hs.contains(sl)){
sum+=sl;
hs.remove(sl);
}
}
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Morgan was opening some boxes when she found one bearing a message “For Morgan, Dad loves you 3000". She wants to open the box but she needs to solve the question to open it. She asks Peter to do that. Help Peter and Morgan solve the question so that they can see what's inside the box.
Given two arrays Arr1 and Arr2 of size N1 and N2. Your task is to find the sum of all elements that are common to both arrays. If there are no common elements the output would be 0.
Note: The arrays may contain duplicate elements. However, you need to sum only unique elements that are common to both arrays.The first line of input contains N1 and N2 separated by a space. The second line contains N1 space separated elements of Arr1. The third line contains N2 space separated elements of Arr2.
Constraints:
1 <= N1, N2 <= 10<sup>6</sup>
1 <= Arr1[i], Arr2[i] <= 1000000000Print the sum of common elements.Sample Input:
5 6
1 2 3 4 5
2 3 4 5 6 7
Sample Output:
14
Explanation:- Common elements = 2, 3, 4 , 5
sum= 2 + 3 + 4 + 5 = 14
Sample Input:-
3 3
1 2 3
4 5 6
Sample Output:-
0, I have written this Solution Code: N1,N2=(map(int,input().split()))
arr1=set(map(int,input().split()))
arr2=set(map(int,input().split()))
arr1=list(arr1)
arr2=list(arr2)
arr1.sort()
arr2.sort()
i=0
j=0
sum1=0
while i<len(arr1) and j<len(arr2):
if arr1[i]==arr2[j]:
sum1+=arr1[i]
i+=1
j+=1
elif arr1[i]>arr2[j]:
j+=1
else:
i+=1
print(sum1), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Morgan was opening some boxes when she found one bearing a message “For Morgan, Dad loves you 3000". She wants to open the box but she needs to solve the question to open it. She asks Peter to do that. Help Peter and Morgan solve the question so that they can see what's inside the box.
Given two arrays Arr1 and Arr2 of size N1 and N2. Your task is to find the sum of all elements that are common to both arrays. If there are no common elements the output would be 0.
Note: The arrays may contain duplicate elements. However, you need to sum only unique elements that are common to both arrays.The first line of input contains N1 and N2 separated by a space. The second line contains N1 space separated elements of Arr1. The third line contains N2 space separated elements of Arr2.
Constraints:
1 <= N1, N2 <= 10<sup>6</sup>
1 <= Arr1[i], Arr2[i] <= 1000000000Print the sum of common elements.Sample Input:
5 6
1 2 3 4 5
2 3 4 5 6 7
Sample Output:
14
Explanation:- Common elements = 2, 3, 4 , 5
sum= 2 + 3 + 4 + 5 = 14
Sample Input:-
3 3
1 2 3
4 5 6
Sample Output:-
0, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#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 sz 200000
#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);
signed main()
{
int n,m;
cin>>n>>m;
set<int> ss,yy;
for(int i=0;i<n;i++)
{
int a;
cin>>a;
ss.insert(a);
}
for(int i=0;i<m;i++)
{
int a;
cin>>a;
if(ss.find(a)!=ss.end()) yy.insert(a);
}
int sum=0;
for(auto it:yy)
{
sum+=it;
}
cout<<sum<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N. For all pairs (i, j) (1 <= i < j <= N), find the maximum value of abs(A<sub>i</sub> - A<sub>j</sub>) in the array.The first line of the input contains a single integer N.
The second line of the input contains N space seperated integers.
Constraints:
2 <= N <= 10<sup>5</sup>
1 <= A<sub>i</sub> <= 10<sup>9</sup>Print the maximum value of abs(A<sub>i</sub> - A<sub>j</sub>) in the array.Sample Input:
5
7 9 4 1 8
Sample Output:
8, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n;
cin >> n;
vector<int> a(n);
for(auto &i : a) cin >> i;
sort(a.begin(), a.end());
cout << a[n - 1] - a[0];
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alexa will turn on the air conditioner if, and only if, the temperature of the room is 30 degrees Celsius or above. The current temperature of the room is X degrees Celsius. Will she turn on the air conditioner?The input consists of a single integer X.
<b>Constraints</b>
−40≤X≤40
X is an integer.Print Yes if you will turn on the air conditioner; print No otherwise.<b>Sample Input 1</b>
25
<b>Sample Output 1</b>
No
<b>Sample Input 2</b>
30
<b>Sample Output 2</b>
Yes, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
int x;
cin >> x;
if (x >= 30) 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: 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: 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 an array Arr of N integers. You have to process Q queries on the array. Each query contains L and R, for each query you have to report the number of indices i in the array such that L <= Arr[i] <= R.First line of input contains two integers N and Q representing number of elements in the array and number of queries.
Second line of input contains N integers representing the array Arr.
Next Q lines of input contains L and R for that query.
Constraints
1 <= N,Q <= 100000
1 <= Arr[i] <= 100000
1 <= L <= R <= 100000For each query you have to print the number of index i in the array such that L <= Arr[i] <= R in a seperate line.Sample input
10 10
3 6 7 4 9 1 7 10 9 4
1 9
4 6
4 7
4 6
6 8
5 8
2 6
2 3
9 9
2 5
Sample output
9
3
5
3
3
3
4
1
2
3
Explanation :
For range 1 to 9 : all indexes except index 8 (with value 10) have values in the range 1 to 9
For range 4 to 6 : indexes 2(with value 6), 4(with value 4) and 10(with value 4) have value in the range 4 to 6, 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[] strArr = br.readLine().split(" ");
int N = Integer.parseInt(strArr[0]);
int Q = Integer.parseInt(strArr[1]);
int[] arr = new int[100001];
strArr = br.readLine().split(" ");
int max = 0;
for(int i=0; i<N; i++){
int curr = Integer.parseInt(strArr[i]);
arr[curr]++;
if(curr>max){
max = curr;
}
}
long[] prefixSum = new long[100001];
long sum = 0;
for(int i=0; i<=100000; i++){
sum = sum + arr[i];
prefixSum[i] = sum;
}
for(int i=0; i<Q; i++){
strArr = br.readLine().split(" ");
int L = Integer.parseInt(strArr[0]);
int R = Integer.parseInt(strArr[1]);
long count = prefixSum[R]-prefixSum[L-1];
System.out.println(count);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array Arr of N integers. You have to process Q queries on the array. Each query contains L and R, for each query you have to report the number of indices i in the array such that L <= Arr[i] <= R.First line of input contains two integers N and Q representing number of elements in the array and number of queries.
Second line of input contains N integers representing the array Arr.
Next Q lines of input contains L and R for that query.
Constraints
1 <= N,Q <= 100000
1 <= Arr[i] <= 100000
1 <= L <= R <= 100000For each query you have to print the number of index i in the array such that L <= Arr[i] <= R in a seperate line.Sample input
10 10
3 6 7 4 9 1 7 10 9 4
1 9
4 6
4 7
4 6
6 8
5 8
2 6
2 3
9 9
2 5
Sample output
9
3
5
3
3
3
4
1
2
3
Explanation :
For range 1 to 9 : all indexes except index 8 (with value 10) have values in the range 1 to 9
For range 4 to 6 : indexes 2(with value 6), 4(with value 4) and 10(with value 4) have value in the range 4 to 6, I have written this Solution Code: n,q=map(int,input().split())
a=list(map(int,input().split()))
b=[0]*(100001)
for i in range(n):
b[a[i]]+=1
s=0
for i in range(100001):
s+=b[i]
b[i]=s
for i in range(q):
v=[int(k) for k in input().split()]
print(b[v[1]]-b[v[0]-1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array Arr of N integers. You have to process Q queries on the array. Each query contains L and R, for each query you have to report the number of indices i in the array such that L <= Arr[i] <= R.First line of input contains two integers N and Q representing number of elements in the array and number of queries.
Second line of input contains N integers representing the array Arr.
Next Q lines of input contains L and R for that query.
Constraints
1 <= N,Q <= 100000
1 <= Arr[i] <= 100000
1 <= L <= R <= 100000For each query you have to print the number of index i in the array such that L <= Arr[i] <= R in a seperate line.Sample input
10 10
3 6 7 4 9 1 7 10 9 4
1 9
4 6
4 7
4 6
6 8
5 8
2 6
2 3
9 9
2 5
Sample output
9
3
5
3
3
3
4
1
2
3
Explanation :
For range 1 to 9 : all indexes except index 8 (with value 10) have values in the range 1 to 9
For range 4 to 6 : indexes 2(with value 6), 4(with value 4) and 10(with value 4) have value in the range 4 to 6, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n,q;
cin>>n>>q;
int c[100001]={};
int a[n+1];
for(int i=1;i<=n;++i)
{
cin>>a[i];
c[a[i]]++;
}
vector<int> ans;
for(int i=1;i<=100000;++i)
c[i]=c[i]+c[i-1];
for(int i=1;i<=q;++i)
{
int l,r;
cin>>l>>r;
ans.push_back(c[r]-c[l-1]);
}
for(auto r:ans)
cout<<r<<"\n";
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number is called Silly if it can be represented as the sum of the square of consecutive natural numbers starting from 1.
For a given number N, find the closest silly number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SillyNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the closest Silly number.
Note:- If more than one answer exists return the minimum one.Sample Input:-
18
Sample Output:-
14
Explanation:-
1*1 + 2*2 + 3*3 = 14
Sample Input:-
2
Sample Output:-
1, I have written this Solution Code: static int SillyNumber(int N){
int sum=0;
int x=1;
while(sum<N){
sum+=x*x;
x++;
}
x--;
if(sum-N < N-(sum-x*x)){
return sum;
}
else{
return sum-x*x;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number is called Silly if it can be represented as the sum of the square of consecutive natural numbers starting from 1.
For a given number N, find the closest silly number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SillyNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the closest Silly number.
Note:- If more than one answer exists return the minimum one.Sample Input:-
18
Sample Output:-
14
Explanation:-
1*1 + 2*2 + 3*3 = 14
Sample Input:-
2
Sample Output:-
1, I have written this Solution Code: int SillyNumber(int N){
int sum=0;
int x=1;
while(sum<N){
sum+=x*x;
x++;
}
x--;
if(sum-N < N-(sum-x*x)){
return sum;
}
else{
return sum-x*x;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number is called Silly if it can be represented as the sum of the square of consecutive natural numbers starting from 1.
For a given number N, find the closest silly number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SillyNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the closest Silly number.
Note:- If more than one answer exists return the minimum one.Sample Input:-
18
Sample Output:-
14
Explanation:-
1*1 + 2*2 + 3*3 = 14
Sample Input:-
2
Sample Output:-
1, I have written this Solution Code: int SillyNumber(int N){
int sum=0;
int x=1;
while(sum<N){
sum+=x*x;
x++;
}
x--;
if(sum-N < N-(sum-x*x)){
return sum;
}
else{
return sum-x*x;
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number is called Silly if it can be represented as the sum of the square of consecutive natural numbers starting from 1.
For a given number N, find the closest silly number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SillyNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the closest Silly number.
Note:- If more than one answer exists return the minimum one.Sample Input:-
18
Sample Output:-
14
Explanation:-
1*1 + 2*2 + 3*3 = 14
Sample Input:-
2
Sample Output:-
1, I have written this Solution Code: def SillyNumber(N):
sum=0
x=1
while sum<N:
sum=sum+x*x
x=x+1
x=x-1
if (sum-N) < (N-(sum-x*x)):
return sum;
else:
return sum - x*x
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 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: Given two integers a and b your task is to print the summation of these two numbersInput contains a single line containing two space separated integers a and b.
Constraints:-
1 <= a, b <= 10<sup>20000</sup>Print the sum of a and b.Sample Input:-
3 8
Sample Output:-
11
Sample Input:-
15 1
Sample Output:-
16, I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
import java.math.BigInteger;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
BigInteger sum;
String ip1 = sc.next();
String ip2 = sc.next();
BigInteger a = new BigInteger(ip1);
BigInteger b = new BigInteger(ip2);
sum = a.add(b);
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b your task is to print the summation of these two numbersInput contains a single line containing two space separated integers a and b.
Constraints:-
1 <= a, b <= 10<sup>20000</sup>Print the sum of a and b.Sample Input:-
3 8
Sample Output:-
11
Sample Input:-
15 1
Sample Output:-
16, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-02-03 02:46:30
**/
#include <bits/stdc++.h>
#define NX 105
#define MX 3350
using namespace std;
const int mod = 998244353;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
typedef long long INT;
const int pb = 10;
const int base_digits = 9;
const int base = 1000000000;
const int DIV = 100000;
struct bigint {
vector<int> a;
int sign;
bigint() : sign(1) {}
bigint(INT v) { *this = v; }
bigint(const string &s) { read(s); }
void operator=(const bigint &v) { sign = v.sign, a = v.a; }
void operator=(INT v) {
sign = 1;
if (v < 0) sign = -1, v = -v;
for (; v > 0; v = v / base) a.push_back(v % base);
}
bigint operator+(const bigint &v) const {
if (sign == v.sign) {
bigint res = v;
for (int i = 0, carry = 0; i < (int)max(a.size(), v.a.size()) ||
carry;
i++) {
if (i == (int)res.a.size()) res.a.push_back(0);
res.a[i] += carry + (i < (int)a.size() ? a[i] : 0);
carry = res.a[i] >= base;
if (carry) res.a[i] -= base;
}
return res;
}
return *this - (-v);
}
bigint operator-(const bigint &v) const {
if (sign == v.sign) {
if (abs() >= v.abs()) {
bigint res = *this;
for (int i = 0, carry = 0; i < (int)v.a.size() || carry; i++) {
res.a[i] -= carry + (i < (int)v.a.size() ? v.a[i] : 0);
carry = res.a[i] < 0;
if (carry) res.a[i] += base;
}
res.trim();
return res;
}
return -(v - *this);
}
return *this + (-v);
}
void operator*=(int v) {
if (v < 0) sign = -sign, v = -v;
for (int i = 0, carry = 0; i < (int)a.size() || carry; i++) {
if (i == (int)a.size()) a.push_back(0);
INT cur = a[i] * (INT)v + carry;
carry = (int)(cur / base);
a[i] = (int)(cur % base);
}
trim();
}
bigint operator*(int v) const {
bigint res = *this;
res *= v;
return res;
}
friend pair<bigint, bigint> DIVmod(const bigint &a1, const bigint &b1) {
int norm = base / (b1.a.back() + 1);
bigint a = a1.abs() * norm;
bigint b = b1.abs() * norm;
bigint q, r;
q.a.resize(a.a.size());
for (int i = a.a.size() - 1; i >= 0; i--) {
r *= base;
r += a.a[i];
int s1 = r.a.size() <= b.a.size() ? 0 : r.a[b.a.size()];
int s2 = r.a.size() <= b.a.size() - 1 ? 0 : r.a[b.a.size() - 1];
int d = ((INT)base * s1 + s2) / b.a.back();
r -= b * d;
while (r < 0) r += b, --d;
q.a[i] = d;
}
q.sign = a1.sign * b1.sign;
r.sign = a1.sign;
q.trim();
r.trim();
return make_pair(q, r / norm);
}
bigint operator/(const bigint &v) const { return DIVmod(*this, v).first; }
bigint operator%(const bigint &v) const { return DIVmod(*this, v).second; }
void operator/=(int v) {
if (v < 0) sign = -sign, v = -v;
for (int i = (int)a.size() - 1, rem = 0; i >= 0; i--) {
INT cur = a[i] + rem * (INT)base;
a[i] = (int)(cur / v);
rem = (int)(cur % v);
}
trim();
}
bigint operator/(int v) const {
bigint res = *this;
res /= v;
return res;
}
int operator%(int v) const {
if (v < 0) v = -v;
int m = 0;
for (int i = a.size() - 1; i >= 0; --i)
m = (a[i] + m * (INT)base) % v;
return m * sign;
}
void operator+=(const bigint &v) { *this = *this + v; }
void operator-=(const bigint &v) { *this = *this - v; }
void operator*=(const bigint &v) { *this = *this * v; }
void operator/=(const bigint &v) { *this = *this / v; }
bool operator<(const bigint &v) const {
if (sign != v.sign) return sign < v.sign;
if (a.size() != v.a.size()) return a.size() * sign < v.a.size() * v.sign;
for (int i = a.size() - 1; i >= 0; i--)
if (a[i] != v.a[i]) return a[i] * sign < v.a[i] * sign;
return false;
}
bool operator>(const bigint &v) const { return v < *this; }
bool operator<=(const bigint &v) const { return !(v < *this); }
bool operator>=(const bigint &v) const { return !(*this < v); }
bool operator==(const bigint &v) const { return !(*this < v) && !(v < *this); }
bool operator!=(const bigint &v) const { return *this < v || v < *this; }
void trim() {
while (!a.empty() && !a.back()) a.pop_back();
if (a.empty()) sign = 1;
}
bool isZero() const { return a.empty() || (a.size() == 1 && !a[0]); }
bigint operator-() const {
bigint res = *this;
res.sign = -sign;
return res;
}
bigint abs() const {
bigint res = *this;
res.sign *= res.sign;
return res;
}
INT longValue() const {
INT res = 0;
for (int i = a.size() - 1; i >= 0; i--) res = res * base + a[i];
return res * sign;
}
friend bigint gcd(const bigint &a, const bigint &b) { return b.isZero() ? a : gcd(b, a % b); }
friend bigint lcm(const bigint &a, const bigint &b) { return a / gcd(a, b) * b; }
void read(const string &s) {
sign = 1;
a.clear();
int pos = 0;
while (pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
if (s[pos] == '-') sign = -sign;
pos++;
}
for (int i = s.size() - 1; i >= pos; i -= base_digits) {
int x = 0;
for (int j = max(pos, i - base_digits + 1); j <= i; j++) x = x *
pb +
s[j] - '0';
a.push_back(x);
}
trim();
}
friend istream &operator>>(istream &stream, bigint &v) {
string s;
stream >> s;
v.read(s);
return stream;
}
friend ostream &operator<<(ostream &stream, const bigint &v) {
if (v.sign == -1) stream << '-';
stream << (v.a.empty() ? 0 : v.a.back());
for (int i = (int)v.a.size() - 2; i >= 0; --i) stream << setw(base_digits) << setfill('0') << v.a[i];
return stream;
}
static vector<int> convert_base(const vector<int> &a, int old_digits, int new_digits) {
vector<INT> p(max(old_digits, new_digits) + 1);
p[0] = 1;
for (int i = 1; i < (int)p.size(); i++) p[i] = p[i - 1] * pb;
vector<int> res;
INT cur = 0;
int cur_digits = 0;
for (int i = 0; i < (int)a.size(); i++) {
cur += a[i] * p[cur_digits];
cur_digits += old_digits;
while (cur_digits >= new_digits) {
res.push_back(int(cur % p[new_digits]));
cur /= p[new_digits];
cur_digits -= new_digits;
}
}
res.push_back((int)cur);
while (!res.empty() && !res.back()) res.pop_back();
return res;
}
typedef vector<INT> vll;
static vll karatsubaMultiply(const vll &a, const vll &b) {
int n = a.size();
vll res(n + n);
if (n <= 32) {
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
res[i + j] += a[i] * b[j];
return res;
}
int k = n >> 1;
vll a1(a.begin(), a.begin() + k);
vll a2(a.begin() + k, a.end());
vll b1(b.begin(), b.begin() + k);
vll b2(b.begin() + k, b.end());
vll a1b1 = karatsubaMultiply(a1, b1);
vll a2b2 = karatsubaMultiply(a2, b2);
for (int i = 0; i < k; i++) a2[i] += a1[i];
for (int i = 0; i < k; i++) b2[i] += b1[i];
vll r = karatsubaMultiply(a2, b2);
for (int i = 0; i < (int)a1b1.size(); i++) r[i] -= a1b1[i];
for (int i = 0; i < (int)a2b2.size(); i++) r[i] -= a2b2[i];
for (int i = 0; i < (int)r.size(); i++) res[i + k] += r[i];
for (int i = 0; i < (int)a1b1.size(); i++) res[i] += a1b1[i];
for (int i = 0; i < (int)a2b2.size(); i++) res[i + n] += a2b2[i];
return res;
}
bigint operator*(const bigint &v) const {
vector<int> a5 = convert_base(this->a, base_digits, 5);
vector<int> b5 = convert_base(v.a, base_digits, 5);
vll a(a5.begin(), a5.end());
vll b(b5.begin(), b5.end());
while (a.size() < b.size()) a.push_back(0);
while (b.size() < a.size()) b.push_back(0);
while (a.size() & (a.size() - 1)) a.push_back(0), b.push_back(0);
vll c = karatsubaMultiply(a, b);
bigint res;
res.sign = sign * v.sign;
for (int i = 0, carry = 0; i < (int)c.size(); i++) {
INT cur = c[i] + carry;
res.a.push_back((int)(cur % DIV));
carry = (int)(cur / DIV);
}
res.a = convert_base(res.a, 5, base_digits);
res.trim();
return res;
}
inline bool isOdd() { return a[0] & 1; }
};
int main() {
bigint n, m;
cin >> n >> m;
cout << n + m << "\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b your task is to print the summation of these two numbersInput contains a single line containing two space separated integers a and b.
Constraints:-
1 <= a, b <= 10<sup>20000</sup>Print the sum of a and b.Sample Input:-
3 8
Sample Output:-
11
Sample Input:-
15 1
Sample Output:-
16, I have written this Solution Code: n,m = map(int,input().split())
print(n+m) , In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Anya owns N triplets of integers (A<sub>i</sub>, B<sub>i</sub>, C<sub>i</sub>), for each i from 1 to N. She asks you to find a sequence of integers which satisfies the following conditions:
1. The sequence contains at least C<sub>i</sub> distinct integers from the closed interval [A<sub>i</sub>, B<sub>i</sub>], for each i from 1 to N.
2. Out of all sequences satisfying the first condition, choose a sequence with the minimum possible number of elements.
For simplicity, she asks you to just print the length of such a sequence.The first line of the input contains a single integer N denoting the number of triplets.
Then N lines follow, where the i<sup>th</sup> line contains three integers A<sub>i</sub>, B<sub>i</sub>, C<sub>i</sub> for each i from 1 to N.
<b> Constraints: </b>
1 ≤ N ≤ 2000
1 ≤ A<sub>i</sub> ≤ B<sub>i</sub> ≤ 2000
0 ≤ C<sub>i</sub> ≤ B<sub>i</sub> - A<sub>i</sub> + 1Print a single integer — the minimum possible sequence length.Sample Input 1:
1
1 3 3
Sample Output 1:
3
Sample Explanation 1:
Since there are only 3 elements in the closed interval [1,3], and we need to take 3 of them, clearly the smallest possible length is 3.
Sample Input 2:
2
1 3 1
3 5 1
Sample Output 2:
1
Sample Explanation 2:
We can take the sequence consisting of a single element {3}., I have written this Solution Code: l = [0]*2001
k = []
n = int(input())
for i in range(n):
a,b,c = map(int,input().split())
k.append([b,(a,c)])
k.sort()
for b,aa in k:
a = aa[0]
c = aa[1]
cnt = 0
for i in range(b,a-1,-1):
if l[i]:
cnt+=1
if cnt>=c:
continue
else:
for i in range(b,a-1,-1):
if not l[i]:
l[i]=1
cnt+=1
if cnt==c:
break
print(sum(l)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Anya owns N triplets of integers (A<sub>i</sub>, B<sub>i</sub>, C<sub>i</sub>), for each i from 1 to N. She asks you to find a sequence of integers which satisfies the following conditions:
1. The sequence contains at least C<sub>i</sub> distinct integers from the closed interval [A<sub>i</sub>, B<sub>i</sub>], for each i from 1 to N.
2. Out of all sequences satisfying the first condition, choose a sequence with the minimum possible number of elements.
For simplicity, she asks you to just print the length of such a sequence.The first line of the input contains a single integer N denoting the number of triplets.
Then N lines follow, where the i<sup>th</sup> line contains three integers A<sub>i</sub>, B<sub>i</sub>, C<sub>i</sub> for each i from 1 to N.
<b> Constraints: </b>
1 ≤ N ≤ 2000
1 ≤ A<sub>i</sub> ≤ B<sub>i</sub> ≤ 2000
0 ≤ C<sub>i</sub> ≤ B<sub>i</sub> - A<sub>i</sub> + 1Print a single integer — the minimum possible sequence length.Sample Input 1:
1
1 3 3
Sample Output 1:
3
Sample Explanation 1:
Since there are only 3 elements in the closed interval [1,3], and we need to take 3 of them, clearly the smallest possible length is 3.
Sample Input 2:
2
1 3 1
3 5 1
Sample Output 2:
1
Sample Explanation 2:
We can take the sequence consisting of a single element {3}., I have written this Solution Code: //Author: Xzirium
//Time and Date: 00:28:35 28 December 2021
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t clk;
clk = clock();
//-----------------------------------------------------------------------------------------------------------//
READV(N);
vector<pair<pll,ll>> Z;
FORI(i,0,N)
{
READV(a);
READV(b);
READV(c);
Z.pb({{b,a},c});
}
sort(Z.begin(),Z.end());
ordered_set1<ll> unused;
FORI(i,1,2001)
{
unused.insert(i);
}
ll ans=0;
FORI(i,0,N)
{
ll a=Z[i].fi.se;
ll b=Z[i].fi.fi;
ll c=Z[i].se;
ll curr=b-a+1-(unused.order_of_key(a-1)-unused.order_of_key(b));
while(curr<c)
{
ans++;
curr++;
auto it=unused.lower_bound(b);
unused.erase(it);
}
}
cout<<ans<<endl;
//-----------------------------------------------------------------------------------------------------------//
clk = clock() - clk;
cerr << fixed << setprecision(6) << "Time: " << ((double)clk)/CLOCKS_PER_SEC << endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers containing only 0 or 1. You can do the following operations on the array:
<ul><li>swap elements at two indices</li><li>choose one index and change its value from 0 to 1 or vice-versa.</li></ul>
You have to do the minimum number of the above operations such that the final array is non-decreasing.
<b>Note</b> Consider 1 based Array-indexingThe first line of input contains a single integer N.
The second line of input contains N space-separated integers denoting the array.
Constraints:
1 ≤ N ≤ 100000
elements of the array are 0 or 1.Minimum number of moves required such that the final array is non- decreasing.Sample Input 1
5
1 1 0 0 1
Sample Output 1
2
Explanation:
Swap indices (1, 3)
Swap indices (2, 4)
Sample Input 2
5
0 0 1 1 1
Sample Output 2
0, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine().trim());
String array[] = br.readLine().trim().split(" ");
boolean decreasingOrder = false;
int[] arr = new int[n];
int totalZeroCount = 0,
totalOneCount = 0;
for(int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(array[i]);
if(i != 0 && arr[i] < arr[i - 1])
decreasingOrder = true;
if(arr[i] % 2 == 0)
++totalZeroCount;
else
++totalOneCount;
}
if(!decreasingOrder) {
System.out.println("0");
} else {
int oneCount = 0;
for(int i = 0; i < totalZeroCount; i++) {
if(arr[i] == 1)
++oneCount;
}
System.out.println(oneCount);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers containing only 0 or 1. You can do the following operations on the array:
<ul><li>swap elements at two indices</li><li>choose one index and change its value from 0 to 1 or vice-versa.</li></ul>
You have to do the minimum number of the above operations such that the final array is non-decreasing.
<b>Note</b> Consider 1 based Array-indexingThe first line of input contains a single integer N.
The second line of input contains N space-separated integers denoting the array.
Constraints:
1 ≤ N ≤ 100000
elements of the array are 0 or 1.Minimum number of moves required such that the final array is non- decreasing.Sample Input 1
5
1 1 0 0 1
Sample Output 1
2
Explanation:
Swap indices (1, 3)
Swap indices (2, 4)
Sample Input 2
5
0 0 1 1 1
Sample Output 2
0, 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();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int a[n];
for(int i=0;i<n;++i){
cin>>a[i];
}
int cnt = 0;
for (int i = 0; i < n; i++) {
if (a[i]==0) cnt++;
}
int ans = 0;
for (int i = 0; i < cnt; i++) if (a[i] == 1) ans++;
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers containing only 0 or 1. You can do the following operations on the array:
<ul><li>swap elements at two indices</li><li>choose one index and change its value from 0 to 1 or vice-versa.</li></ul>
You have to do the minimum number of the above operations such that the final array is non-decreasing.
<b>Note</b> Consider 1 based Array-indexingThe first line of input contains a single integer N.
The second line of input contains N space-separated integers denoting the array.
Constraints:
1 ≤ N ≤ 100000
elements of the array are 0 or 1.Minimum number of moves required such that the final array is non- decreasing.Sample Input 1
5
1 1 0 0 1
Sample Output 1
2
Explanation:
Swap indices (1, 3)
Swap indices (2, 4)
Sample Input 2
5
0 0 1 1 1
Sample Output 2
0, I have written this Solution Code: n=int(input())
l=list(map(int,input().split()))
x=l.count(0)
c=0
for i in range(0,x):
if(l[i]==1):
c+=1
print(c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying to open a lock for that she needs a passcode. Sara knows that the passcode contains N digits in it and the digits at even indices(0- indexing) are even and the digits at odd indices are {2, 3, 5, 7}.
Now Sara wants to know how many different passcodes are possible so that she can open her lock.The input contains a single integer N.
Constraints:-
1 <= N <= 10^15Print the number of possible passcodes.
Note:-
Since the answer can be quite large so print your ans modulo 10^9+7.Sample Input:-
1
Sample Output:-
5
Explanation:-
Possible answer:- 0,2,4,6,8
Sample Input:-
4
Sample Output:-
400, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.math.*;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
long n = sc.nextLong();
BigInteger five = new BigInteger("5");
BigInteger four = new BigInteger("4");
BigInteger m = new BigInteger("1000000007");
long fourPower = n/2;
long fivePow = n - fourPower;
BigInteger res = five.modPow(BigInteger.valueOf(fivePow),m);
BigInteger res2 = four.modPow(BigInteger.valueOf(fourPower),m);
long ans = (res.longValue()* res2.longValue())%mod;
System.out.print(ans);
}
static long mod = 1000000007;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying to open a lock for that she needs a passcode. Sara knows that the passcode contains N digits in it and the digits at even indices(0- indexing) are even and the digits at odd indices are {2, 3, 5, 7}.
Now Sara wants to know how many different passcodes are possible so that she can open her lock.The input contains a single integer N.
Constraints:-
1 <= N <= 10^15Print the number of possible passcodes.
Note:-
Since the answer can be quite large so print your ans modulo 10^9+7.Sample Input:-
1
Sample Output:-
5
Explanation:-
Possible answer:- 0,2,4,6,8
Sample Input:-
4
Sample Output:-
400, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
long long fast_pow(int base, long long n,int M=1000000007)
{
if(n==0)
return 1;
if(n==1)
return base;
long long halfn=fast_pow(base,n/2);
if(n%2==0)
return ( halfn * halfn ) % M;
else
return ( ( ( halfn * halfn ) % M ) * base ) % M;
}
int solve(long long n) {
const int M = 1000000007;
return (int)(fast_pow(5,(n+1)/2)*fast_pow(4,(n/2))%M);
}
int main(){
long long n;
cin>>n;
cout<<solve(n);
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr of N integers. Find the number of subarrays of this array that are powerful.
A subarray [L, R] (1 <= L <= R <= N) is said to be powerful if the product A<sub>L</sub> * A<sub>L+1</sub> * ... * A<sub>R-1</sub> * A<sub>R</sub> is odd.The first line of input contains a single integer N
The second line of input contains N integers representing the elements of the array Arr
<b>Constraints </b>
1 <= N <= 100000
1 <= Arr[i] <= 100000Output the number of powerful subarrays of array Arr.Sample Input 1
5
2 4 4 5 3
Sample output 1
3
Sample Input 2
3
1 5 1
Sample Output 2
6
<b>Explanation:</b>
(3), (5), (3, 5) are the required subarrays.
(1), (2), (1), (1, 5), (5, 1]) (1, 5, 1) are the required subarrays., 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 cnt=0;
long ans=0;
for(int i=0;i<n;i++){
if(a[i]&1){cnt++;}
else{
ans+=(cnt*(cnt+1))/2;
cnt=0;
}
}
ans+=(cnt*(cnt+1))/2;
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr of N integers. Find the number of subarrays of this array that are powerful.
A subarray [L, R] (1 <= L <= R <= N) is said to be powerful if the product A<sub>L</sub> * A<sub>L+1</sub> * ... * A<sub>R-1</sub> * A<sub>R</sub> is odd.The first line of input contains a single integer N
The second line of input contains N integers representing the elements of the array Arr
<b>Constraints </b>
1 <= N <= 100000
1 <= Arr[i] <= 100000Output the number of powerful subarrays of array Arr.Sample Input 1
5
2 4 4 5 3
Sample output 1
3
Sample Input 2
3
1 5 1
Sample Output 2
6
<b>Explanation:</b>
(3), (5), (3, 5) are the required subarrays.
(1), (2), (1), (1, 5), (5, 1]) (1, 5, 1) are the required subarrays., I have written this Solution Code: n = int(input())
arr = list(map(int,input().split()))
c=0
result=0
for i in arr:
if i % 2 != 0:
c += 1
else:
result += (c*(c+1)) / 2
c=0
result += (c*(c+1)) / 2
print(int(result)), 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. Find the number of subarrays of this array that are powerful.
A subarray [L, R] (1 <= L <= R <= N) is said to be powerful if the product A<sub>L</sub> * A<sub>L+1</sub> * ... * A<sub>R-1</sub> * A<sub>R</sub> is odd.The first line of input contains a single integer N
The second line of input contains N integers representing the elements of the array Arr
<b>Constraints </b>
1 <= N <= 100000
1 <= Arr[i] <= 100000Output the number of powerful subarrays of array Arr.Sample Input 1
5
2 4 4 5 3
Sample output 1
3
Sample Input 2
3
1 5 1
Sample Output 2
6
<b>Explanation:</b>
(3), (5), (3, 5) are the required subarrays.
(1), (2), (1), (1, 5), (5, 1]) (1, 5, 1) are the required subarrays., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
InputStreamReader inputStreamReader=new InputStreamReader(System.in);
BufferedReader reader=new BufferedReader(inputStreamReader);
int n =Integer.parseInt(reader.readLine());
String str=reader.readLine();
String[] strarr=str.split(" ");
int[] arr=new int[n];
for(int i=0;i<n;i++){
arr[i]=Integer.parseInt(strarr[i]);
}
long noOfsubArrays=0;
int start=0;
boolean sFlag=false;
for(int i=0;i<n;i++){
if(arr[i]%2!=0){
if(!sFlag){
start=i;
noOfsubArrays++;
sFlag=true;
continue;
}
noOfsubArrays+=2;
int temp=start;
temp++;
while(temp<i){
temp++;
noOfsubArrays++;
}
}else if(arr[i]%2==0 && sFlag){
sFlag=false;
}
}
System.out.println(noOfsubArrays);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement the function <code>ceil</code>, which should take a number which can be a float(decimal)
and return its result as an integer with ceil function applied to it (Use JS In built functions)Function will take a float as inputFunction will return a numberconsole.log(ceil(1.99)) // prints 2
console.log(ceil(2.1)) // prints 3
console.log(ceil(-1.1)) // prints -1, I have written this Solution Code: function ceil(num){
// write code here
// return the output , do not use console.log here
return Math.ceil(num)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement the function <code>ceil</code>, which should take a number which can be a float(decimal)
and return its result as an integer with ceil function applied to it (Use JS In built functions)Function will take a float as inputFunction will return a numberconsole.log(ceil(1.99)) // prints 2
console.log(ceil(2.1)) // prints 3
console.log(ceil(-1.1)) // prints -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);
float a =sc.nextFloat();
int b=(int)(Math.ceil(a));
System.out.println(b);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary tree containing with N nodes and an integer X. Your task is to complete the function countSubtreesWithSumX() that returns the count of the number of subtress having total node’s data sum equal to a value X.
Example: A tree given below<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>countSubtreesWithSumX()</b> that takes "root" node and the integer x as parameter.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^3
1 <= node values <= 10^4
<b>Sum of "N" over all testcases does not exceed 10^5</b>Return the number of subtrees with sum X. The driver code will take care of printing it.Sample Input:
1
3 5
1 2 3
Sum=5
Tree:-
1
/ \
2 3
Sample Output:
0
Explanation:
No subtree has a sum equal to 5.
Sample Input:-
1
5 5
2 1 3 4 5
Sum=5
Tree:-
2
/ \
1 3
/ \
4 5
Sample Output:-
1, I have written this Solution Code: static int c = 0;
static int countSubtreesWithSumXUtil(Node root,int x)
{
// if tree is empty
if (root==null)return 0;
// sum of nodes in the left subtree
int ls = countSubtreesWithSumXUtil(root.left,x);
// sum of nodes in the right subtree
int rs = countSubtreesWithSumXUtil(root.right, x);
int sum = ls + rs + root.data;
// if tree's nodes sum == x
if (sum == x)c++;
return sum;
}
static int countSubtreesWithSumX(Node root, int x)
{
c = 0;
// if tree is empty
if (root==null)return 0;
// sum of nodes in the left subtree
int ls = countSubtreesWithSumXUtil(root.left, x);
// sum of nodes in the right subtree
int rs = countSubtreesWithSumXUtil(root.right, x);
// check if above sum is equal to x
if ((ls + rs + root.data) == x)c++;
return c;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara wants to be the best programmer in the world but today she is stuck on an easy problem. Help her to solve it.
<b>Problem description:- </b>
Choose an integer N and subtract the sum of digits of the number N from it i. e. if N is 245 then subtract 2+4+5 from it making it 245 - 11 = 234.
Keep on doing this operation until the number becomes 0 (for eg. 25 requires 3 operations to reduce to 0
25 - > 18 - > 9 - > 0).
Given a number N, your task is to print the number of operations required to make the number 0.The first line of input contains a single integer containing the number of test cases T. Next T lines contain a single integer N.
Constraints:-
1 <= T <= 10000
1 <= N <= 1000000For each test case print the number of operations required to make the number 0.Sample Input:-
4
25
8
17
842
Sample Output:-
3
1
2
72
Explanation:-
25 - > 18 - > 9 - > 0
8 - > 0
17 - > 9 - > 0, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int[] dp = new int[1000002];
public static void main (String[] args) throws IOException {
int x = 1000000;
int sum = 0;
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
for(int i = 1; i <= x; ++i) {
int j = i;
sum = total(j);
dp[i] = dp[i-sum]+1;
}
int t = Integer.parseInt(br.readLine());
while(t-- > 0) {
int count = 0;
int n = Integer.parseInt(br.readLine());
System.out.println(dp[n]);
}
}
static int total(int n) {
int temp = 0;
while(n > 0) {
temp += n%10;
n = n/10;
}
return temp;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara wants to be the best programmer in the world but today she is stuck on an easy problem. Help her to solve it.
<b>Problem description:- </b>
Choose an integer N and subtract the sum of digits of the number N from it i. e. if N is 245 then subtract 2+4+5 from it making it 245 - 11 = 234.
Keep on doing this operation until the number becomes 0 (for eg. 25 requires 3 operations to reduce to 0
25 - > 18 - > 9 - > 0).
Given a number N, your task is to print the number of operations required to make the number 0.The first line of input contains a single integer containing the number of test cases T. Next T lines contain a single integer N.
Constraints:-
1 <= T <= 10000
1 <= N <= 1000000For each test case print the number of operations required to make the number 0.Sample Input:-
4
25
8
17
842
Sample Output:-
3
1
2
72
Explanation:-
25 - > 18 - > 9 - > 0
8 - > 0
17 - > 9 - > 0, I have written this Solution Code: import sys
sys.setrecursionlimit(40000)
dic = {}
def sumi(n):
x = 0
while(n>0):
x += n%10
n //= 10
return x
def Num_of_times(n):
if (n==0):
return 0
if dic.get(n):
count = dic[n]
return count
else:
x = sumi(n)
count = 1+Num_of_times(n-x)
dic[n] = count
return count
arr = []
maxi = 0
for i in range(int(input())):
x = int(input())
maxi = max(maxi,x)
arr.append(x)
maxi_ans = Num_of_times(maxi)
for i in arr:
if i == maxi:
print(maxi_ans)
else:
print(Num_of_times(i)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara wants to be the best programmer in the world but today she is stuck on an easy problem. Help her to solve it.
<b>Problem description:- </b>
Choose an integer N and subtract the sum of digits of the number N from it i. e. if N is 245 then subtract 2+4+5 from it making it 245 - 11 = 234.
Keep on doing this operation until the number becomes 0 (for eg. 25 requires 3 operations to reduce to 0
25 - > 18 - > 9 - > 0).
Given a number N, your task is to print the number of operations required to make the number 0.The first line of input contains a single integer containing the number of test cases T. Next T lines contain a single integer N.
Constraints:-
1 <= T <= 10000
1 <= N <= 1000000For each test case print the number of operations required to make the number 0.Sample Input:-
4
25
8
17
842
Sample Output:-
3
1
2
72
Explanation:-
25 - > 18 - > 9 - > 0
8 - > 0
17 - > 9 - > 0, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define LL long long
int dp[1000001];
void solve() {
int n; cin >> n;
cout << dp[n] << '\n';
}
int main() {
ios::sync_with_stdio(0), cin.tie(0);
for(int i = 1; i <= 1000000; i++) {
int s = 0, j = i;
while(j > 0) {
s += j % 10;
j /= 10;
}
dp[i] = dp[i - s] + 1;
}
int tt; cin >> tt;
while(tt--) solve();
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary array (a binary array is an array consisting only of zeroes and ones) of size N. You can do following operation on the array any number of times. In an operation you can select any subarray of size 3 of the array such that all three elements of that subarray are not same and replace that subarray with a single element which is the element with frequency 2 in that subarray. For example if we are given array [0, 1, 0, 1, 1, 0] if we select subarray (0-2) then we will replace [0, 1, 0] with [0] so the array finally becomes [0, 1, 1, 0]. You have to find if it is possible to apply this operation some number of times to finally reduce the array to size 1.First line of input contains a single integer N, size of the array.
Second line contains N integers denoting the binary array.
Constraints:
3 <= N <= 100000
0 <= Arr[i] <= 1Print 1 if it is possible to finally reduce the array to size 1 else print 0.Sample Input 1
5
0 1 0 1 0
Sample Output 1
1
Explanation:
First we select subarray (0-2) array gets changed to 0 1 0
Second we select subarray (0-2) array gets changed to 0
As the size became 1 therefore it is possible.
Sample Input 2
5
1 1 1 1 0
Sample Output 2
0
Explanation -
First we select subarray (2-4) array gets changed to 1 1 1 1
Now we cannot do the operation anymore., I have written this Solution Code: a=int(input())
b=input().split(" ")
p=0
if b[0]=="1" and b[1]=="1" and b[2]=="1" and b[3]=="1" and b[4]=="1" and b[5]=="1":
print(1)
p=1
exit(0)
bb=""
for i in b:
bb=bb+i
b=bb
while len(b)!=1 and p==0:
if "101" in b:
b=b.replace("101","1")
elif "110" in b:
b=b.replace("110","1")
elif "011" in b:
b=b.replace("011","1")
elif "010" in b:
b=b.replace("010","0")
elif "001" in b:
b=b.replace("001","0")
elif "100" in b:
b=b.replace("100","0")
else:
break
if len(b)==1 and p==0:
print("1")
elif p==0 and len(b)!=1:
print("0"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary array (a binary array is an array consisting only of zeroes and ones) of size N. You can do following operation on the array any number of times. In an operation you can select any subarray of size 3 of the array such that all three elements of that subarray are not same and replace that subarray with a single element which is the element with frequency 2 in that subarray. For example if we are given array [0, 1, 0, 1, 1, 0] if we select subarray (0-2) then we will replace [0, 1, 0] with [0] so the array finally becomes [0, 1, 1, 0]. You have to find if it is possible to apply this operation some number of times to finally reduce the array to size 1.First line of input contains a single integer N, size of the array.
Second line contains N integers denoting the binary array.
Constraints:
3 <= N <= 100000
0 <= Arr[i] <= 1Print 1 if it is possible to finally reduce the array to size 1 else print 0.Sample Input 1
5
0 1 0 1 0
Sample Output 1
1
Explanation:
First we select subarray (0-2) array gets changed to 0 1 0
Second we select subarray (0-2) array gets changed to 0
As the size became 1 therefore it is possible.
Sample Input 2
5
1 1 1 1 0
Sample Output 2
0
Explanation -
First we select subarray (2-4) array gets changed to 1 1 1 1
Now we cannot do the operation anymore., I have written this Solution Code: #pragma GCC optimize ("O3")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int v=0;
for(int i=0;i<n;++i){
int d;
cin>>d;
if(d==0)
++v;
else
--v;
}
if(abs(v)==1){
cout<<1;
}
else{
cout<<0;
}
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary array (a binary array is an array consisting only of zeroes and ones) of size N. You can do following operation on the array any number of times. In an operation you can select any subarray of size 3 of the array such that all three elements of that subarray are not same and replace that subarray with a single element which is the element with frequency 2 in that subarray. For example if we are given array [0, 1, 0, 1, 1, 0] if we select subarray (0-2) then we will replace [0, 1, 0] with [0] so the array finally becomes [0, 1, 1, 0]. You have to find if it is possible to apply this operation some number of times to finally reduce the array to size 1.First line of input contains a single integer N, size of the array.
Second line contains N integers denoting the binary array.
Constraints:
3 <= N <= 100000
0 <= Arr[i] <= 1Print 1 if it is possible to finally reduce the array to size 1 else print 0.Sample Input 1
5
0 1 0 1 0
Sample Output 1
1
Explanation:
First we select subarray (0-2) array gets changed to 0 1 0
Second we select subarray (0-2) array gets changed to 0
As the size became 1 therefore it is possible.
Sample Input 2
5
1 1 1 1 0
Sample Output 2
0
Explanation -
First we select subarray (2-4) array gets changed to 1 1 1 1
Now we cannot do the operation anymore., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n= Integer.parseInt(br.readLine());
int zero = 0,one = 0;
String[] s = br.readLine().trim().split(" ");
for(int i=0;i<n;i++){
int x = Integer.parseInt(s[i]);
if(x%2==0)
zero++;
else one++;
}
if(zero-one==1 || zero-one==-1 )
System.out.print(1);
else
System.out.print(0);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of numbers. Use appropriate array methods to find all numbers that are greater than 5. Complete the function <code>getNumbersGreaterThan5</code> that accepts an array of integers <code>nums</code> and returns an array of numbers that are greater than 5.An array <code>nums</code> of numbersAn array of the numbers greater than 5 that are present in <code>nums</code>const inputArr = [1,2,3,9,10,7,5,4,3]
const ans = getNumbersGreaterThan5(inputArr)
console.log(ans) // prints [9, 10, 7], I have written this Solution Code: function getNumbersGreaterThan5(nums) {
return nums.filter((num) => num > 5);
}, In this Programming Language: JavaScript, 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 a boolean Matrix of size N*M, A cell of the matrix is called "Good" if it is completely surrounded by the cells containing '1' i.e. each adjacent cell (which shares a common edge) contains '1'. Your task is to find the number of such cells.
See the below example for a better understandingFirst line of input contains two space- separated integers N and M. Next N lines of input contain M space- separated integers depicting the values of the matrix.
Constraints:-
3 <= N, M <= 500
0 <= Matrix[][] <= 1Print the number of good cells.Sample Input:-
3 3
1 1 0
1 1 1
1 1 1
Sample Output:-
1
Explanation:-
Only cell at position 1, 1 is good
Sample Input:-
5 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
Sample Output:-
3
Explanation:-
(1, 2), (2, 1) and (3, 2) are good cells, I have written this Solution Code: // mat is the matrix/ 2d array
// n,m are dimensions
function goodCell(mat, n, m) {
// write code here
// do not console.log
// return the answer as a number
let cnt = 0;
for (let i = 1; i < n - 1; i++) {
for (let j = 1; j < m - 1; j++) {
if (mat[i - 1][j] == 1 && mat[i + 1][j] == 1 && mat[i][j - 1] == 1 && mat[i][j + 1] == 1) {
cnt++;
}
}
}
return cnt
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean Matrix of size N*M, A cell of the matrix is called "Good" if it is completely surrounded by the cells containing '1' i.e. each adjacent cell (which shares a common edge) contains '1'. Your task is to find the number of such cells.
See the below example for a better understandingFirst line of input contains two space- separated integers N and M. Next N lines of input contain M space- separated integers depicting the values of the matrix.
Constraints:-
3 <= N, M <= 500
0 <= Matrix[][] <= 1Print the number of good cells.Sample Input:-
3 3
1 1 0
1 1 1
1 1 1
Sample Output:-
1
Explanation:-
Only cell at position 1, 1 is good
Sample Input:-
5 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
Sample Output:-
3
Explanation:-
(1, 2), (2, 1) and (3, 2) are good cells, I have written this Solution Code: N, M= list(map(int,input().split()))
mat =[]
for i in range(N):
List =list(map(int,input().split()))[:M]
mat.append(List)
count =0
for i in range(1,N-1):
for j in range(1,M-1):
if (mat[i][j-1] == 1 and mat[i][j+1] == 1 and mat[i-1][j] == 1 and mat[i+1][j] == 1):
count +=1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean Matrix of size N*M, A cell of the matrix is called "Good" if it is completely surrounded by the cells containing '1' i.e. each adjacent cell (which shares a common edge) contains '1'. Your task is to find the number of such cells.
See the below example for a better understandingFirst line of input contains two space- separated integers N and M. Next N lines of input contain M space- separated integers depicting the values of the matrix.
Constraints:-
3 <= N, M <= 500
0 <= Matrix[][] <= 1Print the number of good cells.Sample Input:-
3 3
1 1 0
1 1 1
1 1 1
Sample Output:-
1
Explanation:-
Only cell at position 1, 1 is good
Sample Input:-
5 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
Sample Output:-
3
Explanation:-
(1, 2), (2, 1) and (3, 2) are good cells, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1000001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
signed main(){
int n,m;
cin>>n>>m;
int a[n][m];
FOR(i,n){
FOR(j,m){
cin>>a[i][j];}}
int sum=0,sum1=0;;
FOR1(i,1,n-1){
FOR1(j,1,m-1){
if(a[i-1][j]==1 && a[i+1][j]==1 && a[i][j-1]==1 && a[i][j+1]==1){
sum++;
}
}
}
out1(sum);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean Matrix of size N*M, A cell of the matrix is called "Good" if it is completely surrounded by the cells containing '1' i.e. each adjacent cell (which shares a common edge) contains '1'. Your task is to find the number of such cells.
See the below example for a better understandingFirst line of input contains two space- separated integers N and M. Next N lines of input contain M space- separated integers depicting the values of the matrix.
Constraints:-
3 <= N, M <= 500
0 <= Matrix[][] <= 1Print the number of good cells.Sample Input:-
3 3
1 1 0
1 1 1
1 1 1
Sample Output:-
1
Explanation:-
Only cell at position 1, 1 is good
Sample Input:-
5 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
Sample Output:-
3
Explanation:-
(1, 2), (2, 1) and (3, 2) are good cells, 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 m= sc.nextInt();
int a[][]= new int[n][m];
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
a[i][j]=sc.nextInt();}}
int cnt=0;
for(int i=1;i<n-1;i++){
for(int j=1;j<m-1;j++){
if(a[i-1][j]==1 && a[i+1][j]==1 && a[i][j-1]==1 && a[i][j+1]==1){
cnt++;
}
}
}
System.out.print(cnt);
}
}
, 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: 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: Given a string S. Two position i and j of the string are friends if they have the same character. The distance between two friends at positions i and j is defined as |i- j|. Find the sum of distances of all the pairs of friends in the given strings.First line of input contains a single string S.
Constraints:
1 <= |S| <= 1000000
String contains lowercase english letters.Output a single integer which is the sum of distance of all the pair of friends in the given strings.Sample Input
ababa
Sample Output
10
Explanation: Friend pairs - (1, 3) (1, 5) (2, 4) (3, 5), I have written this Solution Code: import java.io.*;
import java.io.IOException;
import java.util.*;
class Main {
public static long mod = (long)Math.pow(10,9)+7 ;
public static double epsilon=0.00000000008854;
public static InputReader sc = new InputReader(System.in);
public static PrintWriter pw = new PrintWriter(System.out);
public static void main(String[] args) {
String s=sc.nextLine();
int n=s.length();
int hnum[]=new int[26];
int hpast[]=new int[26];
Arrays.fill(hpast,-1);
long hsum[]=new long[26];
long ans=0;
for(int i=0;i<n;i++){
int k=s.charAt(i)-'a';
if(hpast[k]!=-1)
hsum[k]=hsum[k]+(i-hpast[k])*hnum[k];
ans+=hsum[k];
hnum[k]++;
hpast[k]=i;
}
pw.println(ans);
pw.flush();
pw.close();
}
public static Comparator<Long[]> column(int i){
return
new Comparator<Long[]>() {
@Override
public int compare(Long[] o1, Long[] o2) {
return o1[i].compareTo(o2[i]);
}
};
}
public static Comparator<Integer[]> col(int i){
return
new Comparator<Integer[]>() {
@Override
public int compare(Integer[] o1, Integer[] o2) {
return o1[i].compareTo(o2[i]);
}
};
}
public static String reverseString(String s){
StringBuilder input1 = new StringBuilder();
input1.append(s);
input1 = input1.reverse();
return input1.toString();
}
public static int[] scanArray(int n){
int a[]=new int [n];
for(int i=0;i<n;i++)
a[i]=sc.nextInt();
return a;
}
public static long[] scanLongArray(int n){
long a[]=new long [n];
for(int i=0;i<n;i++)
a[i]=sc.nextLong();
return a;
}
public static String [] scanStrings(int n){
String a[]=new String [n];
for(int i=0;i<n;i++)
a[i]=sc.nextLine();
return a;
}
public static class InputReader {
private final InputStream stream;
private final byte[] buf = new byte[8192];
private int curChar, snumChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int snext() {
if (snumChars == -1)
throw new InputMismatchException();
if (curChar >= snumChars) {
curChar = 0;
try {
snumChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (snumChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public int[] nextIntArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
return a;
}
public String readString() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isSpaceChar(c));
return res.toString();
}
public String nextLine() {
int c = snext();
while (isSpaceChar(c))
c = snext();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isEndOfLine(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S. Two position i and j of the string are friends if they have the same character. The distance between two friends at positions i and j is defined as |i- j|. Find the sum of distances of all the pairs of friends in the given strings.First line of input contains a single string S.
Constraints:
1 <= |S| <= 1000000
String contains lowercase english letters.Output a single integer which is the sum of distance of all the pair of friends in the given strings.Sample Input
ababa
Sample Output
10
Explanation: Friend pairs - (1, 3) (1, 5) (2, 4) (3, 5), I have written this Solution Code: def findS(s):
visited= [ 0 for i in range(256)];
distance =[0 for i in range (256)];
for i in range(256):
visited[i]=0;
distance[i]=0;
sum=0;
for i in range(len(s)):
sum+=visited[ord(s[i])] * i - distance[ord(s[i])];
visited[ord(s[i])] +=1;
distance[ord(s[i])] +=i;
return sum;
if __name__ == '__main__':
s=input("");
print(findS(s));, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S. Two position i and j of the string are friends if they have the same character. The distance between two friends at positions i and j is defined as |i- j|. Find the sum of distances of all the pairs of friends in the given strings.First line of input contains a single string S.
Constraints:
1 <= |S| <= 1000000
String contains lowercase english letters.Output a single integer which is the sum of distance of all the pair of friends in the given strings.Sample Input
ababa
Sample Output
10
Explanation: Friend pairs - (1, 3) (1, 5) (2, 4) (3, 5), I have written this Solution Code: #pragma GCC optimize ("O3")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
string s;
cin>>s;
int c[26]={};
int f[26]={};
int ans=0;
int n=s.length();
for(int i=0;i<n;++i){
ans+=f[s[i]-'a']*i-c[s[i]-'a'];
f[s[i]-'a']++;
c[s[i]-'a']+=i;
}
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is fond of ice cream initially she had N ice creams with her. If Sara eats exactly half of the ice cream she has in a day and the remaining icecreams get tripled each night. How many ice creams does Sara have at the end of D-days?
<b>Note:- </b>
Take the floor value while dividing by 2.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Icecreams()</b> that takes integers N and D as parameters.
<b>Constraints:-</b>
1 <= N <= 100
1 <= D <= 20Print a single integer denoting the number of ice creams at the end of D days.Sample Input 1:-
5 1
Sample Output 1:-
9
</b>Explanation:-</b>
Sara will eat 2 ice creams and the remaining 3 will be tripled i. e 9.
Sample Input 2:-
5 3
Sample Output 2:-
24
<b>Explanation:-</b>
Day 1:- N=5, eaten = 2, rem = 3 => remaining = 3*3 = 9
Day 2:- N=9, eaten = 4, rem = 5 => remaining = 5*3 = 15
Day 3:- N=15, eaten = 7, rem = 8 => remaining = 8*3 = 24, I have written this Solution Code: static void Icecreams (int N, int D){
int x=N;
while(D-->0){
x-=x/2;
x*=3;
}
System.out.println(x);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is fond of ice cream initially she had N ice creams with her. If Sara eats exactly half of the ice cream she has in a day and the remaining icecreams get tripled each night. How many ice creams does Sara have at the end of D-days?
<b>Note:- </b>
Take the floor value while dividing by 2.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Icecreams()</b> that takes integers N and D as parameters.
<b>Constraints:-</b>
1 <= N <= 100
1 <= D <= 20Print a single integer denoting the number of ice creams at the end of D days.Sample Input 1:-
5 1
Sample Output 1:-
9
</b>Explanation:-</b>
Sara will eat 2 ice creams and the remaining 3 will be tripled i. e 9.
Sample Input 2:-
5 3
Sample Output 2:-
24
<b>Explanation:-</b>
Day 1:- N=5, eaten = 2, rem = 3 => remaining = 3*3 = 9
Day 2:- N=9, eaten = 4, rem = 5 => remaining = 5*3 = 15
Day 3:- N=15, eaten = 7, rem = 8 => remaining = 8*3 = 24, I have written this Solution Code: void Icecreams (int N, int D){
int x=N;
while(D--){
x-=x/2;
x*=3;
}
cout << x;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is fond of ice cream initially she had N ice creams with her. If Sara eats exactly half of the ice cream she has in a day and the remaining icecreams get tripled each night. How many ice creams does Sara have at the end of D-days?
<b>Note:- </b>
Take the floor value while dividing by 2.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Icecreams()</b> that takes integers N and D as parameters.
<b>Constraints:-</b>
1 <= N <= 100
1 <= D <= 20Print a single integer denoting the number of ice creams at the end of D days.Sample Input 1:-
5 1
Sample Output 1:-
9
</b>Explanation:-</b>
Sara will eat 2 ice creams and the remaining 3 will be tripled i. e 9.
Sample Input 2:-
5 3
Sample Output 2:-
24
<b>Explanation:-</b>
Day 1:- N=5, eaten = 2, rem = 3 => remaining = 3*3 = 9
Day 2:- N=9, eaten = 4, rem = 5 => remaining = 5*3 = 15
Day 3:- N=15, eaten = 7, rem = 8 => remaining = 8*3 = 24, I have written this Solution Code: void Icecreams (int N, int D){
int x=N;
while(D--){
x-=x/2;
x*=3;
}
printf("%d", x);
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is fond of ice cream initially she had N ice creams with her. If Sara eats exactly half of the ice cream she has in a day and the remaining icecreams get tripled each night. How many ice creams does Sara have at the end of D-days?
<b>Note:- </b>
Take the floor value while dividing by 2.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Icecreams()</b> that takes integers N and D as parameters.
<b>Constraints:-</b>
1 <= N <= 100
1 <= D <= 20Print a single integer denoting the number of ice creams at the end of D days.Sample Input 1:-
5 1
Sample Output 1:-
9
</b>Explanation:-</b>
Sara will eat 2 ice creams and the remaining 3 will be tripled i. e 9.
Sample Input 2:-
5 3
Sample Output 2:-
24
<b>Explanation:-</b>
Day 1:- N=5, eaten = 2, rem = 3 => remaining = 3*3 = 9
Day 2:- N=9, eaten = 4, rem = 5 => remaining = 5*3 = 15
Day 3:- N=15, eaten = 7, rem = 8 => remaining = 8*3 = 24, I have written this Solution Code: def Icecreams(N,D):
ans = N
while D > 0:
ans = ans - ans//2
ans = ans*3
D = D-1
return ans
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
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