# 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.

## Steps

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

## Example

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;
}

## Input

{ 1, 2, 3 }, 5

## Output

5

Updated on: 11-Mar-2022

118 Views