# Check if Matrix remains unchanged after row reversals

Given a NxN matrix. The task is to check if after reversing all of the rows of the given Matrix, the matrix remains 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

1. A most important observation is for the matrix to be same after row reversals, each single row must be palindromic.
2. Now to check if a row is palindromic, maintain two pointers, one pointing to start and other to 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 same, then row is palindromic otherwise not.
3. If any of the Row is not palindromic then answer is No.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` ` `  `#include ` `#define ll long long int ` `using` `namespace` `std; ` ` `  `// Function to check Palindromic Condition ` `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]) { ` `                ``cout << ``"No"` `<< endl; ` `                ``return``; ` `            ``} ` ` `  `            ``start++; ` `            ``end--; ` `        ``} ` `    ``} ` ` `  `    ``cout << ``"Yes"` `<< endl; ` `    ``return``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `matrix = { { 1, 2, 1 }, ` `                         ``{ 2, 2, 2 }, ` `                         ``{ 3, 4, 3 } }; ` `    ``int` `N = 3; ` `    ``specialMatrix(matrix, N); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach ` `class` `GFG  ` `{ ` ` `  `// Function to check Palindromic Condition ` `static` `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 Code ` `public` `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 Condition ` `def` `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 Code ` `if` `__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 approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to check Palindromic Condition ` `static` `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 Code ` `public` `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

 ` `

Output:

```Yes
```

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