# C++ code to count number of operations to make two arrays same

Suppose we have two arrays A and B with n number of elements. Consider an operation: Select two indices i and j, then decrease ith element by 1 and increase jth element by 1. Each element of the array must be non-negative after performing an operation. We want to make A and B same. We have to find the sequence of operations to make A and B same. If not possible, return -1.

So, if the input is like A = [1, 2, 3, 4]; B = [3, 1, 2, 4], then the output will be [(1, 0), (2, 0)], because for i = 1 and j = 0 the array will be [2, 1, 3, 4], then for i = 2 and j = 0, it will be [3, 1, 2, 4]

## Steps

To solve this, we will follow these steps −

a := 0, b := 0, c := 0
n := size of A
Define an array C of size n and fill with 0
for initialize i := 0, when i < n, update (increase i by 1), do:
a := a + A[i]
for initialize i := 0, when i < n, update (increase i by 1), do:
b := b + A[i]
if a is not equal to b, then:
return -1
Otherwise
for initialize i := 0, when i < n, update (increase i by 1),
do:
c := c + |A[i] - B[i]|
C[i] := A[i] - B[i]
c := c / 2
i := 0
j := 0
while c is non-zero, decrease c after each iteration, do:
while C[i] <= 0, do:
(increase i by 1)
while C[j] >= 0, do:
(increase j by 1)
print i and j
decrease C[i] and increase C[j] by 1

## Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
void solve(vector<int> A, vector<int> B){
int a = 0, b = 0, c = 0;
int n = A.size();
vector<int> C(n, 0);
for (int i = 0; i < n; i++)
a += A[i];
for (int i = 0; i < n; i++)
b += A[i];
if (a != b){
cout << -1;
return;
}
else{
for (int i = 0; i < n; i++){
c += abs(A[i] - B[i]);
C[i] = A[i] - B[i];
}
c = c / 2;
int i = 0, j = 0;
while (c--){
while (C[i] <= 0)
i++;
while (C[j] >= 0)
j++;
cout << "(" << i << ", " << j << "), ";
C[i]--, C[j]++;
}
}
}
int main(){
vector<int> A = { 1, 2, 3, 4 };
vector<int> B = { 3, 1, 2, 4 };
solve(A, B);
}

## Input

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

## Output

(1, 0), (2, 0),

Updated on: 15-Mar-2022

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