// C++ program to find number of transformation // to make two Matrix Equal #include    using namespace std; int countOperations(vector<vector<int>> &a vector<vector<int>> &b) {  int n = a.size();  int m = a[0].size();   // Create difference matrix (a = a - b)  for (int i = 0; i < n; i++) {  for (int j = 0; j < m; j++) {  a[i][j] -= b[i][j];  }  }  // Check if transformation is possible using the property  // a[i][j] - a[i][0] - a[0][j] + a[0][0] should be 0  for (int i = 1; i < n; i++) {  for (int j = 1; j < m; j++) {  if (a[i][j] - a[i][0] - a[0][j] + a[0][0] != 0) {  return -1;  }  }  }  int result = 0;  // Add operations needed for first column  for (int i = 0; i < n; i++) {  result += abs(a[i][0]);  }  // Add operations needed for  // first row (excluding a[0][0])  for (int j = 0; j < m; j++) {  result += abs(a[0][j] - a[0][0]);  }  return result; } int main() {    vector<vector<int>> a = {{1 1} {1 1}};  vector<vector<int>> b = {{1 2} {3 4}};  cout << countOperations(a b);  return 0; } 

// Java program to find number of transformation // to make two Matrix Equal import java.util.*; class GfG {  static int countOperations(int[][] a int[][] b) {  int n = a.length;  int m = a[0].length;  // Create difference matrix (a = a - b)  for (int i = 0; i < n; i++) {  for (int j = 0; j < m; j++) {  a[i][j] -= b[i][j];  }  }  // Check if transformation is possible using the  // property a[i][j] - a[i][0] - a[0][j] + a[0][0]  // should be 0  for (int i = 1; i < n; i++) {  for (int j = 1; j < m; j++) {  if (a[i][j] - a[i][0] - a[0][j] + a[0][0]  != 0) {  return -1;  }  }  }  int result = 0;  // Add operations needed for first column  for (int i = 0; i < n; i++) {  result += Math.abs(a[i][0]);  }  // Add operations needed for  // first row (excluding a[0][0])  for (int j = 0; j < m; j++) {  result += Math.abs(a[0][j] - a[0][0]);  }  return result;  }  public static void main(String[] args) {  int[][] a = { { 1 1 } { 1 1 } };  int[][] b = { { 1 2 } { 3 4 } };  System.out.println(countOperations(a b));  } } 

# Python program to find number of transformation # to make two Matrix Equal def countOperations(a b): n = len(a) m = len(a[0]) # Create difference matrix (a = a - b) for i in range(n): for j in range(m): a[i][j] -= b[i][j] # Check if transformation is possible using the property # a[i][j] - a[i][0] - a[0][j] + a[0][0] should be 0 for i in range(1 n): for j in range(1 m): if a[i][j] - a[i][0] - a[0][j] + a[0][0] != 0: return -1 result = 0 # Add operations needed for first column for i in range(n): result += abs(a[i][0]) # Add operations needed for # first row (excluding a[0][0]) for j in range(m): result += abs(a[0][j] - a[0][0]) return result if __name__ == '__main__': a = [ [1 1] [1 1] ] b = [ [1 2] [3 4] ] print(countOperations(a b)) 

// C# program to find number of transformation // to make two Matrix Equal using System; class GfG {  static int countOperations(int[ ] a int[ ] b) {  int n = a.GetLength(0);  int m = a.GetLength(1);  // Create difference matrix (a = a - b)  for (int i = 0; i < n; i++) {  for (int j = 0; j < m; j++) {  a[i j] -= b[i j];  }  }  // Check if transformation is possible using the  // property a[i j] - a[i 0] - a[0 j] + a[0 0]  // should be 0  for (int i = 1; i < n; i++) {  for (int j = 1; j < m; j++) {  if (a[i j] - a[i 0] - a[0 j] + a[0 0]  != 0) {  return -1;  }  }  }  int result = 0;  // Add operations needed for first column  for (int i = 0; i < n; i++) {  result += Math.Abs(a[i 0]);  }  // Add operations needed for  // first row (excluding a[0 0])  for (int j = 0; j < m; j++) {  result += Math.Abs(a[0 j] - a[0 0]);  }  return result;  }  static void Main(string[] args) {    int[ ] a = { { 1 1 } { 1 1 } };  int[ ] b = { { 1 2 } { 3 4 } };  Console.WriteLine(countOperations(a b));  } } 

// JavaScript program to find number of transformation // to make two Matrix Equal function countOperations(a b) {  let n = a.length;  let m = a[0].length;  // Create difference matrix (a = a - b)  for (let i = 0; i < n; i++) {  for (let j = 0; j < m; j++) {  a[i][j] -= b[i][j];  }  }  // Check if transformation is possible using the  // property a[i][j] - a[i][0] - a[0][j] + a[0][0] should  // be 0  for (let i = 1; i < n; i++) {  for (let j = 1; j < m; j++) {  if (a[i][j] - a[i][0] - a[0][j] + a[0][0]  !== 0) {  return -1;  }  }  }  let result = 0;  // Add operations needed for first column  for (let i = 0; i < n; i++) {  result += Math.abs(a[i][0]);  }  // Add operations needed for  // first row (excluding a[0][0])  for (let j = 0; j < m; j++) {  result += Math.abs(a[0][j] - a[0][0]);  }  return result; } //Driver code let a = [ [ 1 1 ] [ 1 1 ] ]; let b = [ [ 1 2 ] [ 3 4 ] ]; console.log(countOperations(a b));