C++ code to count copy operations without exceeding k

Suppose we have an array A with n elements and another number k. There are n piles of candies. The ith pile has A[i] number of candies. We can perform the operation on two indices i and j (i != j), then add another A[i] number of candies to A[i] (A[i] will not be reduced). We can perform this operation any number of times, but unfortunately if some pile contains strictly more than k candies we cannot perform the operation anymore. We have to find the maximum number of times we can perform this operation.

So, if the input is like A = [1, 2, 3]; k = 5, then the output will be 5, because we can take i = 0 and for j = 1 we can perform three times and for j = 2 we can perform two times. So in total 5 times.


To solve this, we will follow these steps −

ans := 0
n := size of A
sort the array A
for initialize i := 1, when i < n, update (increase i by 1), do:
   ans := ans + (k - A[i])
return ans


Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
int solve(vector<int> A, int k){
   int ans = 0;
   int n = A.size();
   sort(A.begin(), A.end());
   for (int i = 1; i < n; i++){
      ans += (k - A[i]) / A[0];
   return ans;
int main(){
   vector<int> A = { 1, 2, 3 };
   int k = 5;
   cout << solve(A, k) << endl;


{ 1, 2, 3 }, 5



Updated on: 11-Mar-2022


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