# Check if a matrix can be converted to another by repeatedly adding any value to X consecutive elements in a row or column

Given two matrices A[][] and B[][] of size M Ã— 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 ` `using` `namespace` `std;`   `// Function to check whether Matrix A[][]` `// can be transformed to Matrix B[][] or not` `bool` `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 Code` `int` `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 approach` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG{`   `// Function to check whether Matrix A[][]` `// can be transformed to Matrix B[][] or not` `static` `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 Code` `public` `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 not` `def` `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 Code` `if` `__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 approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG{` ` `  `// Function to check whether Matrix A[][]` `// can be transformed to Matrix B[][] or not` `static` `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 Code` `public` `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

 ``

Output:

`Yes`

Time Complexity: O(M * N * X)
Auxiliary Space: O(1)

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