Given an NxN matrix. The task is to check if after reversing all the rows of the given Matrix, the matrix remains the same or not.
Examples:
Input : N = 3 1 2 1 2 2 2 3 4 3 Output : Yes If all the rows are reversed then matrix will become: 1 2 1 2 2 2 3 4 3 which is same. Input : N = 3 1 2 2 2 2 2 3 4 3 Output : No
Approach:
- A most important observation is for the matrix to be the same after row reversals, each single row must be palindromic.
- Now to check if a row is palindromic, maintain two pointers, one pointing to start and the other to the end of row. Start comparing the values present and do start++ and end–. Repeat the process until all elements are checked till the middle of the row. If at each step elements are the same, then the row is palindromic otherwise not.
- If any of the Row is not palindromic then the answer is No.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>#define ll long long intusing namespace std;// Function to check Palindromic Conditionvoid specialMatrix(int matrix[3][3], int N){ for (int i = 0; i < N; i++) { // Pointer to start of row int start = 0; // Pointer to end of row int end = N - 1; while (start <= end) { // Single Mismatch means row is not palindromic if (matrix[i][start] != matrix[i][end]) { cout << "No" << endl; return; } start++; end--; } } cout << "Yes" << endl; return;}// Driver Codeint main(){ int matrix[3][3] = { { 1, 2, 1 }, { 2, 2, 2 }, { 3, 4, 3 } }; int N = 3; specialMatrix(matrix, N); return 0;} |
Java
// Java implementation of the above approachclass GFG {// Function to check Palindromic Conditionstatic void specialMatrix(int matrix[][], int N){ for (int i = 0; i < N; i++) { // Pointer to start of row int start = 0; // Pointer to end of row int end = N - 1; while (start <= end) { // Single Mismatch means // row is not palindromic if (matrix[i][start] != matrix[i][end]) { System.out.println("No"); return; } start++; end--; } } System.out.println("Yes"); return;}// Driver Codepublic static void main(String[] args) { int matrix[][] = { { 1, 2, 1 }, { 2, 2, 2 }, { 3, 4, 3 } }; int N = 3; specialMatrix(matrix, N);}}/* This code contributed by PrinciRaj1992 */ |
Python3
# Python 3 implementation of the above approach# Function to check Palindromic Conditiondef specialMatrix(matrix, N): for i in range(N): # Pointer to start of row start = 0 # Pointer to end of row end = N - 1 while (start <= end): # Single Mismatch means row is not palindromic if (matrix[i][start] != matrix[i][end]): print("No") return start += 1 end -= 1 print("Yes") return# Driver Codeif __name__ == '__main__': matrix = [[1, 2, 1], [2, 2, 2], [3, 4, 3]] N = 3 specialMatrix(matrix, N) # This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the above approachusing System;class GFG {// Function to check Palindromic Conditionstatic void specialMatrix(int [,]matrix, int N){ for (int i = 0; i < N; i++) { // Pointer to start of row int start = 0; // Pointer to end of row int end = N - 1; while (start <= end) { // Single Mismatch means // row is not palindromic if (matrix[i, start] != matrix[i, end]) { Console.WriteLine("No"); return; } start++; end--; } } Console.WriteLine("Yes"); return;}// Driver Codepublic static void Main(String[] args) { int [,]matrix = { { 1, 2, 1 }, { 2, 2, 2 }, { 3, 4, 3 } }; int N = 3; specialMatrix(matrix, N);}}// This code has been contributed by 29AjayKumar |
PHP
<?php// PHP implementation of the above approach// Function to check Palindromic Conditionfunction specialMatrix($matrix, $N){ for ($i = 0; $i < $N; $i++) { // Pointer to start of row $start = 0; // Pointer to end of row $end = ($N - 1); while ($start <= $end) { // Single Mismatch means row is not palindromic if ($matrix[$i][$start] != $matrix[$i][$end]) { echo "No", "\n"; return; } $start++; $end--; } } echo "Yes", "\n"; return;}// Driver Code$matrix = array(array(1, 2, 1), array(2, 2, 2), array(3, 4, 3));$N = 3;specialMatrix($matrix, $N);// This code is contributed by ajit.?> |
Javascript
<script> // Javascript implementation of the above approach // Function to check Palindromic Condition function specialMatrix(matrix, N) { for (let i = 0; i < N; i++) { // Pointer to start of row let start = 0; // Pointer to end of row let end = N - 1; while (start <= end) { // Single Mismatch means // row is not palindromic if (matrix[i][start] != matrix[i][end]) { document.write("No"); return; } start++; end--; } } document.write("Yes"); return; } let matrix = [ [ 1, 2, 1 ], [ 2, 2, 2 ], [ 3, 4, 3 ] ]; let N = 3; specialMatrix(matrix, N); </script> |
Output:
Yes
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