Find number of transformation to make two Matrix Equal in C++

In this problem, we are given two matrices mat1[][] and mat2[][] of the same size. Our task is to find the number of transformations to make two Matrices Equal.

The transformation one matrices are −

• Select any matrix of the two matrices.

• Select a row or column from the matrices

• Add 1 to all elements of the selected row or column.

Let’s take an example to understand the problem,

Input

mat1[][] = {{1 2}
{2 1}}
mat1[][] = {{2 3}
{4 3}}

Output

3

Explanation

1 2 => 2 2 => 2 3 => 2 3
2 1 => 3 1 => 3 2 => 4 3

Solution Approach

A simple solution to the problem is by finding whether transformation is possible or not. For this, we need to check −

if( mat[i][j] - mat[i][0] - mat[0][j] + mat[0][0] != 0 )

Then no solution exists.

Now, if transformation is possible, we will count transformations for rows and columns.

Program to illustrate the working of our solution,

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
int countTransformationReq(int mat1[][MAX], int mat2[][MAX],
int m, int n) {
   for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++)
         mat1[i][j] -= mat2[i][j];
      for (int i = 1; i < n; i++)
         for (int j = 1; j < m; j++)
            if (mat1[i][j] - mat1[i][0] - mat1[0][j] +
               mat1[0][0] != 0)
   return -1;
   int trxCount = 0;
   for (int i = 0; i < n; i++)
      trxCount += abs(mat1[i][0]);
   for (int j = 0; j < m; j++)
      trxCount += abs(mat1[0][j] - mat1[0][0]);
   return (trxCount);
}
int main() {
   int mat1[MAX][MAX] = { {1, 2}, {2, 1}};
   int mat2[MAX][MAX] = { {2, 3}, {4, 3}};
   cout<<"The number of transformation to make the teo
   maxtrces equal are "<<countTransformationReq(mat1, mat2, 2,
   2) ;
   return 0;
}

Output

The number of transformation to make the teo maxtrces equal are 3

Efficient approach

A more effective solution to the problem is by using the handshake formula.

We will create a temp[] array to count the occurrence of 0 and 1 in the modulo of the array. And then return the count value.

Program to illustrate the working of our solution,

Example

 Live Demo

#include<iostream>
using namespace std;
int countEvenSumSubArray(int arr[], int n){
   int temp[2] = {1, 0};
   int count = 0, sum = 0;
   for (int i=0; i<=n-1; i++){
      sum = ( (sum + arr[i]) % 2 + 2) % 2;
      temp[sum]++;
   }
   count += (temp[0]*(temp[0]-1)/2);
   count += (temp[1]*(temp[1]-1)/2);
   return (count);
}
int main(){
   int arr[] = {2, 1, 4, 2};
   int n = sizeof (arr) / sizeof (arr[0]);
   cout<<"The count of Subarrays with even sum is "<<countEvenSumSubArray(arr, n);
   return (0);
}

Output

The count of Subarrays with even sum is 4