# Check if the Matrix satisfies the given conditions

Given a matrix mat[][], the task is to check whether the matrix is valid or not. A valid matrix is the matrix that satisfies the given conditions:

1. For every row, it must contain a single distinct character.
2. No two consecutive rows have a character in common.

Examples:

Input: mat[][] = {
{0, 0, 0},
{1, 1, 1},
{0, 0, 2}}
Output: No
The last row doesn’t consist of a single distinct character.

Input: mat[][] = {
{8, 8, 8, 8, 8},
{4, 4, 4, 4, 4},
{6, 6, 6, 6, 6},
{5, 5, 5, 5, 5},
{8, 8, 8, 8, 8}}
Output: Yes

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

Approach: First of all check the first row if it contains same characters or not. If it contains same characters then iterate the second row and compare the characters of the current row with the first element of the previous row, if the two elements are equal or the characters of the current row are different, return false. Repeat the above process for all the consecutive row.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Storing the size of the matrix ` `const` `int` `M = 5, N = 5; ` ` `  `// Function that returns true if ` `// the matrix is valid ` `bool` `isPerfect(``char` `board[M][N + 1]) ` `{ ` `    ``char` `m; ` ` `  `    ``for` `(``int` `i = 0; i < M; i++) { ` ` `  `        ``// Get the current row ` `        ``string s = board[i]; ` ` `  `        ``// Comparing first element ` `        ``// of the row with the element ` `        ``// of previous row ` `        ``if` `(i > 0 && s[0] == m) ` `            ``return` `false``; ` ` `  `        ``// Checking if all the characters of the ` `        ``// current row are same or not ` `        ``for` `(``int` `j = 0; j < N; j++) { ` ` `  `            ``// Storing the first character ` `            ``if` `(j == 0) ` `                ``m = s[0]; ` ` `  `            ``// Comparing all the elements ` `            ``// with the first element ` `            ``else` `{ ` `                ``if` `(m != s[j]) ` `                    ``return` `false``; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``return` `true``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``char` `board[M][N + 1] = { ``"88888"``, ` `                             ``"44444"``, ` `                             ``"66666"``, ` `                             ``"55555"``, ` `                             ``"88888"` `}; ` ` `  `    ``if` `(isPerfect(board)) ` `        ``cout << ``"Yes"``; ` `    ``else` `        ``cout << ``"No"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG  ` `{ ` `     `  `    ``// Storing the size of the matrix  ` `    ``static` `final` `int` `M = ``5``, N = ``5``;  ` `     `  `    ``// Function that returns true if  ` `    ``// the matrix is valid  ` `    ``static` `boolean` `isPerfect(String board[])  ` `    ``{  ` `        ``char` `m = ``0``; ` `     `  `        ``for` `(``int` `i = ``0``; i < M; i++) ` `        ``{  ` `     `  `            ``// Get the current row  ` `            ``String s = board[i];  ` `     `  `            ``// Comparing first element  ` `            ``// of the row with the element  ` `            ``// of previous row  ` `            ``if` `(i > ``0` `&& s.charAt(``0``) == m)  ` `                ``return` `false``;  ` `     `  `            ``// Checking if all the characters of the  ` `            ``// current row are same or not  ` `            ``for` `(``int` `j = ``0``; j < N; j++)  ` `            ``{  ` `     `  `                ``// Storing the first character  ` `                ``if` `(j == ``0``)  ` `                    ``m = s.charAt(``0``);  ` `     `  `                ``// Comparing all the elements  ` `                ``// with the first element  ` `                ``else`  `                ``{  ` `                    ``if` `(m != s.charAt(j))  ` `                        ``return` `false``;  ` `                ``}  ` `            ``}  ` `        ``}  ` `        ``return` `true``;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{  ` `        ``String board[] = { ``"88888"``, ``"44444"``,  ` `                           ``"66666"``, ``"55555"``,  ` `                           ``"88888"` `};  ` `     `  `        ``if` `(isPerfect(board))  ` `            ``System.out.println(``"Yes"``);  ` `        ``else` `            ``System.out.println(``"No"``);  ` `    ``}  ` `} ` `     `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Storing the size of the matrix ` `M ``=` `5` `N ``=` `5` ` `  `# Function that returns true if ` `# the matrix is valid ` `def` `isPerfect(board): ` `    ``m ``=` `'' ` ` `  `    ``for` `i ``in` `range``(M): ` ` `  `        ``# Get the current row ` `        ``s ``=` `board[i] ` ` `  `        ``# Comparing first element ` `        ``# of the row with the element ` `        ``# of previous row ` `        ``if` `(i > ``0` `and` `s[``0``] ``=``=` `m): ` `            ``return` `False` ` `  `        ``# Checking if all the characters of the ` `        ``# current row are same or not ` `        ``for` `j ``in` `range``(N): ` ` `  `            ``# Storing the first character ` `            ``if` `(j ``=``=` `0``): ` `                ``m ``=` `s[``0``] ` ` `  `            ``# Comparing all the elements ` `            ``# with the first element ` `            ``else``: ` `                ``if` `(m !``=` `s[j]): ` `                    ``return` `False` ` `  `    ``return` `True` ` `  `# Driver code ` `board ``=` `[``"88888"``, ` `         ``"44444"``, ` `         ``"66666"``, ` `         ``"55555"``, ` `         ``"88888"``] ` ` `  `if` `(isPerfect(board)): ` `    ``print``(``"Yes"``) ` `else``: ` `    ``print``(``"No"``) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` `     `  `class` `GFG  ` `{ ` `     `  `    ``// Storing the size of the matrix  ` `    ``static` `readonly` `int` `M = 5, N = 5;  ` `     `  `    ``// Function that returns true if  ` `    ``// the matrix is valid  ` `    ``static` `bool` `isPerfect(String []board)  ` `    ``{  ` `        ``char` `m = ``'0'``; ` `     `  `        ``for` `(``int` `i = 0; i < M; i++) ` `        ``{  ` `     `  `            ``// Get the current row  ` `            ``String s = board[i];  ` `     `  `            ``// Comparing first element  ` `            ``// of the row with the element  ` `            ``// of previous row  ` `            ``if` `(i > 0 && s[0] == m)  ` `                ``return` `false``;  ` `     `  `            ``// Checking if all the characters of the  ` `            ``// current row are same or not  ` `            ``for` `(``int` `j = 0; j < N; j++)  ` `            ``{  ` `     `  `                ``// Storing the first character  ` `                ``if` `(j == 0)  ` `                    ``m = s[0];  ` `     `  `                ``// Comparing all the elements  ` `                ``// with the first element  ` `                ``else` `                ``{  ` `                    ``if` `(m != s[j])  ` `                        ``return` `false``;  ` `                ``}  ` `            ``}  ` `        ``}  ` `        ``return` `true``;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main (String[] args)  ` `    ``{  ` `        ``String []board = { ``"88888"``, ``"44444"``,  ` `                           ``"66666"``, ``"55555"``,  ` `                           ``"88888"` `};  ` `     `  `        ``if` `(isPerfect(board))  ` `            ``Console.WriteLine(``"Yes"``);  ` `        ``else` `            ``Console.WriteLine(``"No"``);  ` `    ``}  ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output:

```Yes
```

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