Row-wise Common Elements in Two Diagonals of a Square Matrix in C++

Given a 2D square matrix as input. The goal is to find the elements that are common in both its primary and secondary diagonals. If the input matrix is

1 2 3
2 2 4
1 4 7

Then its primary diagonal is 1 2 7 and the secondary diagonal is 3 2 1. Common element is 2.

There will always be at least one common element in both.

Examples

Input − Matrix[][5] = {{1, 2, 1}, {4, 1, 6}, {1, 8, 1}};

Output − Row-wise common elements in diagonals:3

Explanation − The matrix is:

1 2 1
4 1 6
1 8 1

Primary diagonal=1 1 1, Secondary diagonal= 1 1 1

All 3 values are common. count=3

Input − Matrix[][5] = {{1, 4, 4, 1}, {3, 4, 4, 6}, {1, 1, 1, 4}, {1, 9, 9, 2}};

Output − Row-wise common elements in diagonals:3

Explanation − The matrix is:

1 4 4 1
3 4 4 6
1 1 1 4
1 9 9 2

Primary diagonal=1 4 1 2, Secondary diagonal= 1 4 1 1

First 3 values are common. count=3

Approach used in the below program is as follows

In this approach we will first traverse the square matrix row-wise from row 0. For each row check if element M[ i ][ i ] is equal to M[ i ][ size-i-1 ]. If yes then its common element of both diagonals, increment count.

• Take the input matrix Matrix[][5].

• Set its size.

• Function countElement(int mt[][5], int size) takes the input matrix and its size and returns the count of common values in both primary and secondary diagonals.

• Take the initial count as 0.

• Traverse using for loop from i=0 to i

• If mt[i][i] == mt[i][size-i-1] then increment count.

• At the end return count as result.

• Print the result in main.

Example

#include <iostream>
using namespace std;
int countElement(int mt[][5], int size){
   int count = 0;
   for (int i=0;i<size;i++){
      if (mt[i][i] == mt[i][size-i-1]){
         count=count+1;
      }
   }
   return count;
}
int main(){
   int Matrix[][5] = {{1, 2, 1}, {4, 1, 6}, {1, 8, 1}};
   int size=3;
   cout<<"Row-wise common elements in diagonals:"<<countElement(Matrix, size);
   return 0;
}

Output

If we run the above code it will generate the following Output

Row-wise common elements in diagonals:3