# Row-wise common elements in two diagonals of a square matrix

Given a square matrix, find out count of numbers that are same in same row and same in both primary and secondary diagonals.**Examples : **

Input : 1 2 1 4 5 2 0 5 1 Output : 2 Primary diagonal is 1 5 1 Secondary diagonal is 1 5 0 Two elements (1 and 5) match in two diagonals and same. Input : 1 0 0 0 1 0 0 0 1 Output : 1 Primary diagonal is 1 1 1 Secondary diagonal is 0 1 0 Only one element is same.

We can achieve this in O(n) time, O(1) space and only one traversal. We can find current element in i-th row of primary diagonal as mat[i][i] and i-th element of secondary diagonal as mat[i][n-i-1].

## C++

`// CPP program to find common elements in` `// two diagonals.` `#include <iostream>` `#define MAX 100` `using` `namespace` `std;` `// Returns count of row wise same` `// elements in two diagonals of` `// mat[n][n]` `int` `countCommon(` `int` `mat[][MAX], ` `int` `n)` `{` ` ` `int` `res = 0;` ` ` `for` `(` `int` `i=0;i<n;i++)` ` ` `if` `(mat[i][i] == mat[i][n-i-1])` ` ` `res++;` ` ` `return` `res;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `mat[][MAX] = {{1, 2, 3},` ` ` `{4, 5, 6},` ` ` `{7, 8, 9}};` ` ` `cout << countCommon(mat, 3);` ` ` `return` `0;` `}` |

## Java

`// Java program to find common` `// elements in two diagonals.` `import` `java.io.*;` `class` `GFG` `{` ` ` `int` `MAX = ` `100` `;` ` ` ` ` `// Returns count of row wise same elements` ` ` `// in two diagonals of mat[n][n]` ` ` `static` `int` `countCommon(` `int` `mat[][], ` `int` `n)` ` ` `{` ` ` `int` `res = ` `0` `;` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++)` ` ` `if` `(mat[i][i] == mat[i][n - i - ` `1` `])` ` ` `res++;` ` ` `return` `res;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String args[])` `throws` `IOException` ` ` `{` ` ` `int` `mat[][] = {{` `1` `, ` `2` `, ` `3` `},` ` ` `{` `4` `, ` `5` `, ` `6` `},` ` ` `{` `7` `, ` `8` `, ` `9` `}};` ` ` `System.out.println(countCommon(mat, ` `3` `));` ` ` `}` `}` `// This code is contributed by Anshika Goyal.` |

## Python3

`# Python3 program to find common` `# elements in two diagonals.` `Max` `=` `100` `# Returns count of row wise same` `# elements in two diagonals of` `# mat[n][n]` `def` `countCommon(mat, n):` ` ` `res ` `=` `0` ` ` ` ` `for` `i ` `in` `range` `(n):` ` ` ` ` `if` `mat[i][i] ` `=` `=` `mat[i][n` `-` `i` `-` `1` `] :` ` ` `res ` `=` `res ` `+` `1` ` ` `return` `res ` `# Driver Code` `mat ` `=` `[[` `1` `, ` `2` `, ` `3` `],` ` ` `[` `4` `, ` `5` `, ` `6` `],` ` ` `[` `7` `, ` `8` `, ` `9` `]]` `print` `(countCommon(mat, ` `3` `))` `# This code is contributed by Anant Agarwal.` |

## C#

`// C# program to find common` `// elements in two diagonals.` `using` `System;` `class` `GFG {` ` ` ` ` `// Returns count of row wise same` ` ` `// elements in two diagonals of` ` ` `// mat[n][n]` ` ` `static` `int` `countCommon(` `int` `[,]mat, ` `int` `n)` ` ` `{` ` ` `int` `res = 0;` ` ` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `if` `(mat[i,i] == mat[i,n - i - 1])` ` ` `res++;` ` ` ` ` `return` `res;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `[,]mat = {{1, 2, 3},` ` ` `{4, 5, 6},` ` ` `{7, 8, 9}};` ` ` `Console.WriteLine(countCommon(mat, 3));` ` ` `}` `}` `// This code is contributed by vt_m.` |

## PHP

`<?php` `// PHP program to find common` `// elements in two diagonals.` `$MAX` `= 100;` `// Returns count of row wise` `// same elements in two` `// diagonals of mat[n][n]` `function` `countCommon(` `$mat` `, ` `$n` `)` `{` ` ` `global` `$MAX` `;` ` ` `$res` `= 0;` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `if` `(` `$mat` `[` `$i` `][` `$i` `] == ` `$mat` `[` `$i` `][` `$n` `- ` `$i` `- 1])` ` ` `$res` `++;` ` ` `return` `$res` `;` `}` `// Driver Code` `$mat` `= ` `array` `(` `array` `(1, 2, 3),` ` ` `array` `(4, 5, 6),` ` ` `array` `(7, 8, 9));` `echo` `countCommon(` `$mat` `, 3);` `// This code is contributed by aj_36` `?>` |

## Javascript

`<script>` `// Javascript program to find common elements in` `// two diagonals.` `let MAX = 100;` `// Returns count of row wise same` `// elements in two diagonals of` `// mat[n][n]` `function` `countCommon(mat, n)` `{` ` ` `let res = 0;` ` ` `for` `(let i = 0; i < n; i++)` ` ` `if` `(mat[i][i] == mat[i][n - i - 1])` ` ` `res++;` ` ` `return` `res;` `}` `// Driver Code` ` ` `let mat = [[1, 2, 3],` ` ` `[4, 5, 6],` ` ` `[7, 8, 9]];` ` ` `document.write(countCommon(mat, 3));` `// This code is contributed by subham348.` `</script>` |

**Output :**

1

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