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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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