# C++ Program to check we can remove all stones by selecting boxes

Suppose we have an array A with N elements. Consider there are N boxes and they are arranged in a circle. The ith box contains A[i] stones. We have to check whether we can remove all stones from the boxes by repeatedly performing the operation: Select a box say ith box. For each j in range 1 to N, remove exactly j stones from the (i+j)th box. Here (N+k)th box is termed as kth box. This operation cannot be performed if a box does not contain sufficient number of stones.

So, if the input is like A = [4, 5, 1, 2, 3], then the output will be True, because we can remove all stones by starting from second box.

To solve this, we will follow these steps −

n := size of A
Define an array a of size (n + 1)
Define an array b of size (n + 1)
sum := 0, p := n * (n + 1)
for initialize i := 1, when i <= n, update (increase i by 1), do:
a[i] := A[i - 1]
sum := sum + a[i]
if sum mod p is not equal to 0, then:
return false
k := sum / p
for initialize i := 1, when i <= n, update (increase i by 1), do:
b[i] := a[i] - a[(i mod n) + 1]
sum := 0
for initialize i := 1, when i <= n, update (increase i by 1), do:
a[i] := b[i]
sum := sum + a[i]
if sum is not equal to 0, then:
return false
for initialize i := 1, when i <= n, update (increase i by 1), do:
if (a[i] + k) mod n is not equal to 0 or a[i] + k < 0, then:
return false
return true

## Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
bool solve(vector<int> A) {
int n = A.size();
vector<int> a(n + 1);
vector<int> b(n + 1);
int sum = 0, p = n * (n + 1) / 2;
for (int i = 1; i <= n; i++) {
a[i] = A[i - 1];
sum += a[i];
}
if (sum % p != 0) {
return false;
}
int k = sum / p;
for (int i = 1; i <= n; i++) {
b[i] = a[i] - a[i % n + 1];
}
sum = 0;
for (int i = 1; i <= n; i++) {
a[i] = b[i];
sum += a[i];
}
if (sum != 0) {
return false;
}
for (int i = 1; i <= n; i++) {
if ((a[i] + k) % n != 0 || a[i] + k < 0) {
return false;
}
}
return true;
}
int main(){
vector<int> A = { 4, 5, 1, 2, 3 };
cout << solve(A) << endl;
}

## Input

{ 4, 5, 1, 2, 3 }


## Output

1