# C++ code to find out number of battery combos

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Suppose, we have n batteries that can be used a maximum of 5 times. We have some devices that need three batteries and each usage of the device increases the usage count of the batteries by 1. If we have to use the devices k times, we have to find out how many battery combinations we can make to power the devices. A battery cannot be used in two devices simultaneously and a battery that has been used 5 times cannot be included. The usage count of the batteries is given in the array batt.

So, if the input is like n = 6, k = 2, batt = {2, 4, 4, 2, 1, 3}, then the output will be 1.

There can be only one battery combination made to power devices for k times.

## Steps

To solve this, we will follow these steps −

ans := 0
for initialize i := 0, when i < n, update (increase i by 1), do:
if batt[i] + k <= 5, then:
(increase ans by 1)
return ans / 3

## Example

Let us see the following implementation to get better understanding

#include <bits/stdc++.h>
using namespace std;
#define N 100
int solve(int n, int k, int batt[]) {
int ans = 0;
for(int i = 0; i < n; i++){
if(batt[i] + k <= 5)
ans++;
}
return ans / 3;
}
int main() {
int n = 6, k = 2, batt[] = {2, 4, 4, 2, 1, 3};
cout<< solve(n, k, batt);
return 0;
}

## Input

6, 2, {2, 4, 4, 2, 1, 3}

## Output

1