Given two matrices A[][] and B[][] of size M × N and an integer X, the task is to check if it is possible to convert matrix A[][] to matrix B[][] by adding any value to X consecutive cells in the same row or same column any number of times (possibly zero).
Examples:
Input: A[][] = { {0, 0}, {0, 0}}, B[][] = {{1, 2}, {0, 1}}, X = 2
Output: Yes
Explanation:
Operation 1: Adding 1 to A[0][0] and A[0][1] modifies A[][] to {{1, 1}, {0, 0}}.
Operation 2: Adding 1 to A[0][1] and A[1][1] modifies A[][] to {{1, 2}, {0, 1}}.
After performing this two operations, matrix A[][] and B[][] are equal.Input: A= {{0, 0, 0}, {0, 0, 0}}, B = {{1, 2, 3}, {4, 5, 6}}, X = 4
Output: False
Approach: The problem can be solved greedily by performing all the horizontal operations followed by the vertical operations.
Follow the steps below to solve the problem:
- Traverse the matrix to perform horizontal operations, using variables i and j over the ranges [0, M – 1] and [0, N – X], and perform the following operations:
- If A[i][j] is not equal to B[i][j], increment next X elements in the same row by A[i][j] – B[i][j].
- Now, traverse the matrix to perform vertical operations, using variables i and j over the ranges [0, M – X] and [0, N – 1] and perform the following operations:
- Check if A[i][j] is equal to B[i][j] or not.
- If found to be false, increment next X elements in the same column by A[i][j] – B[i][j].
- Print “True” if matrices A[][] and B[][] are equal. Otherwise, print “False”.
Below is the implementation of the above approach:
C++
// C++ Program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check whether Matrix A[][]// can be transformed to Matrix B[][] or notbool Check(int A[][2], int B[][2], int M, int N, int X){ // Traverse the matrix to perform // horizontal operations for (int i = 0; i < M; i++) { for (int j = 0; j <= N - X; j++) { if (A[i][j] != B[i][j]) { // Calculate difference int diff = B[i][j] - A[i][j]; for (int k = 0; k < X; k++) { // Update next X elements A[i][j + k] = A[i][j + k] + diff; } } } } // Traverse the matrix to perform // vertical operations for (int i = 0; i <= M - X; i++) { for (int j = 0; j < N; j++) { if (A[i][j] != B[i][j]) { // Calculate difference int diff = B[i][j] - A[i][j]; for (int k = 0; k < X; k++) { // Update next K elements A[i + k][j] = A[i + k][j] + diff; } } } } for (int i = 0; i < M; i++) { for (int j = 0; j < N; j++) { // A[i][j] is not equal to B[i][j] if (A[i][j] != B[i][j]) { // Conversion is not possible return 0; } } } // Conversion is possible return 1;}// Driver Codeint main(){ // Input int M = 2, N = 2, X = 2; int A[2][2] = { { 0, 0 }, { 0, 0 } }; int B[2][2] = { { 1, 2 }, { 0, 1 } }; if (Check(A, B, M, N, X)) { cout << "Yes" << endl; } else { cout << "No" << endl; } return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.util.*;class GFG{// Function to check whether Matrix A[][]// can be transformed to Matrix B[][] or notstatic int Check(int A[][], int B[][], int M, int N, int X){ // Traverse the matrix to perform // horizontal operations for(int i = 0; i < M; i++) { for(int j = 0; j <= N - X; j++) { if (A[i][j] != B[i][j]) { // Calculate difference int diff = B[i][j] - A[i][j]; for(int k = 0; k < X; k++) { // Update next X elements A[i][j + k] = A[i][j + k] + diff; } } } } // Traverse the matrix to perform // vertical operations for(int i = 0; i <= M - X; i++) { for(int j = 0; j < N; j++) { if (A[i][j] != B[i][j]) { // Calculate difference int diff = B[i][j] - A[i][j]; for(int k = 0; k < X; k++) { // Update next K elements A[i + k][j] = A[i + k][j] + diff; } } } } for(int i = 0; i < M; i++) { for(int j = 0; j < N; j++) { // A[i][j] is not equal to B[i][j] if (A[i][j] != B[i][j]) { // Conversion is not possible return 0; } } } // Conversion is possible return 1;}// Driver Codepublic static void main(String[] args){ // Input int M = 2, N = 2, X = 2; int A[][] = { { 0, 0 }, { 0, 0 } }; int B[][] = { { 1, 2 }, { 0, 1 } }; if (Check(A, B, M, N, X) != 0) { System.out.println("Yes"); } else { System.out.println("No"); }}}// This code is contributed by sanjoy_62 |
Python3
# Python3 program for the above approach# Function to check whether Matrix A[][]# can be transformed to Matrix B[][] or notdef Check(A, B, M, N, X): # Traverse the matrix to perform # horizontal operations for i in range(M): for j in range(N - X + 1): if (A[i][j] != B[i][j]): # Calculate difference diff = B[i][j] - A[i][j] for k in range(X): # Update next X elements A[i][j + k] = A[i][j + k] + diff # Traverse the matrix to perform # vertical operations for i in range(M - X + 1): for j in range(N): if (A[i][j] != B[i][j]): # Calculate difference diff = B[i][j] - A[i][j] for k in range(X): # Update next K elements A[i + k][j] = A[i + k][j] + diff for i in range(M): for j in range(N): # A[i][j] is not equal to B[i][j] if (A[i][j] != B[i][j]): # Conversion is not possible return 0 # Conversion is possible return 1# Driver Codeif __name__ == '__main__': # Input M, N, X = 2, 2, 2 A = [ [ 0, 0 ], [ 0, 0 ] ] B = [ [ 1, 2 ], [ 0, 1 ] ] if (Check(A, B, M, N, X)): print("Yes") else: print("No")# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{ // Function to check whether Matrix A[][]// can be transformed to Matrix B[][] or notstatic int Check(int [,]A, int [,]B, int M, int N, int X){ // Traverse the matrix to perform // horizontal operations for(int i = 0; i < M; i++) { for(int j = 0; j <= N - X; j++) { if (A[i, j] != B[i, j]) { // Calculate difference int diff = B[i, j] - A[i, j]; for(int k = 0; k < X; k++) { // Update next X elements A[i, j + k] = A[i, j + k] + diff; } } } } // Traverse the matrix to perform // vertical operations for(int i = 0; i <= M - X; i++) { for(int j = 0; j < N; j++) { if (A[i, j] != B[i, j]) { // Calculate difference int diff = B[i,j] - A[i,j]; for(int k = 0; k < X; k++) { // Update next K elements A[i + k, j] = A[i + k, j] + diff; } } } } for(int i = 0; i < M; i++) { for(int j = 0; j < N; j++) { // A[i][j] is not equal to B[i][j] if (A[i, j] != B[i, j]) { // Conversion is not possible return 0; } } } // Conversion is possible return 1;}// Driver Codepublic static void Main(){ // Input int M = 2, N = 2, X = 2; int [,]A = { { 0, 0 }, { 0, 0 } }; int [,]B = { { 1, 2 }, { 0, 1 } }; if (Check(A, B, M, N, X) == 1) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); }}}// This code is contributed by SURENDRA_GANGWAR |
Javascript
<script>//Javascript program for the above approach// Function to check whether Matrix A[][]// can be transformed to Matrix B[][] or notfunction Check( A, B,M, N, X){ // Traverse the matrix to perform // horizontal operations for (var i = 0; i < M; i++) { for (var j = 0; j <= N - X; j++) { if (A[i][j] != B[i][j]) { // Calculate difference var diff = B[i][j] - A[i][j]; for (var k = 0; k < X; k++) { // Update next X elements A[i][j + k] = A[i][j + k] + diff; } } } } // Traverse the matrix to perform // vertical operations for (var i = 0; i <= M - X; i++) { for (var j = 0; j < N; j++) { if (A[i][j] != B[i][j]) { // Calculate difference var diff = B[i][j] - A[i][j]; for (var k = 0; k < X; k++) { // Update next K elements A[i + k][j] = A[i + k][j] + diff; } } } } for (var i = 0; i < M; i++) { for (var j = 0; j < N; j++) { // A[i][j] is not equal to B[i][j] if (A[i][j] != B[i][j]) { // Conversion is not possible return 0; } } } // Conversion is possible return 1;}var M = 2, N = 2, X = 2; var A = [ [ 0, 0 ], [ 0, 0 ]]; var B = [ [ 1, 2 ], [ 0, 1 ] ]; if (Check(A, B, M, N, X)) { document.write( "Yes" + "<br>"); } else { document.write( "No" + "<br>"); }//This code is contributed by SoumikMondal</script> |
Output:
Yes
Time Complexity: O(M * N * X)
Auxiliary Space: O(1)
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