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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
