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; ` `} ` |

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## 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. ` |

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## 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. ` |

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## 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. ` |

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## 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 ` `?> ` |

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**Output :**

1

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