Program to find the Product of diagonal elements of a matrix

Given an N * N matrix, the task is to find the product of the elements of left and right diagonal.

Examples:

Input: arr[] = 1 2 3 4
               5 6 7 8 
               9 7 4 2
               2 2 2 1
Output: 9408
Explanation:
Product of left diagonal = 1 * 4 * 6 * 1 = 24
Product of right diagonal = 4 * 7 * 7 * 2 = 392
Total product = 24 * 392 = 9408

Input: arr[] = 2 1 2 1 2
               1 2 1 2 1
               2 1 2 1 2
               1 2 1 2 1
               2 1 2 1 2  
Output : 512
Explanation:
Product of left diagonal = 2 * 2 * 2 * 2 * 2 = 32
Product of right diagonal = 2 * 2 * 2 * 2 * 2 = 32
But we have a common element in this case so
Total product = (32 * 32)/2  = 512

Approach:



Below is the implementation of the above approach:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ Program to find the Product
// of diagonal elements of a matrix
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the product of diagonals
int productDiagonals(int arr[][100], int n)
{
  
    int product = 1;
    // loop for calculating product of both
    // the principal and secondary diagonals
    for (int i = 0; i < n; i++) {
  
        // For principal diagonal index of row
        // is equal to index of column
        product = product * arr[i][i];
  
        // For secondary diagonal index
        // of column is n-(index of row)-1
        product = product * arr[i][n - i - 1];
    }
  
    // Divide the answer by middle element for
    // matrix of odd size
    if (n % 2 == 1) {
        product = product / arr[n / 2][n / 2];
    }
  
    return product;
}
  
// Driver code
int main()
{
    int arr1[100][100] = { { 1, 2, 3, 4 },
                           { 5, 6, 7, 8 },
                           { 9, 7, 4, 2 },
                           { 2, 2, 2, 1 } };
    // Function calling
    cout << productDiagonals(arr1, 4) << endl;
  
    int arr2[100][100] = { { 2, 1, 2, 1, 2 },
                           { 1, 2, 1, 2, 1 },
                           { 2, 1, 2, 1, 2 },
                           { 1, 2, 1, 2, 1 },
                           { 2, 1, 2, 1, 2 } };
    // Function calling
    cout << productDiagonals(arr2, 5) << endl;
    return 0;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java Program to find the Product
// of diagonal elements of a matrix
import java.util.*;
  
class GFG
{
  
// Function to find the product of diagonals
static int productDiagonals(int arr[][], int n)
{
  
    int product = 1;
    // loop for calculating product of both
    // the principal and secondary diagonals
    for (int i = 0; i < n; i++) 
    {
  
        // For principal diagonal index of row
        // is equal to index of column
        product = product * arr[i][i];
  
        // For secondary diagonal index
        // of column is n-(index of row)-1
        product = product * arr[i][n - i - 1];
    }
  
    // Divide the answer by middle element for
    // matrix of odd size
    if (n % 2 == 1)
    {
        product = product / arr[n / 2][n / 2];
    }
  
    return product;
}
  
// Driver code
public static void main(String[] args)
{
    int arr1[][] = { { 1, 2, 3, 4 },
                        { 5, 6, 7, 8 },
                        { 9, 7, 4, 2 },
                        { 2, 2, 2, 1 } };
    // Function calling
    System.out.print(productDiagonals(arr1, 4) + "\n");
  
    int arr2[][] = { { 2, 1, 2, 1, 2 },
                        { 1, 2, 1, 2, 1 },
                        { 2, 1, 2, 1, 2 },
                        { 1, 2, 1, 2, 1 },
                        { 2, 1, 2, 1, 2 } };
    // Function calling
    System.out.print(productDiagonals(arr2, 5) + "\n");
}
}
  
// This code is contributed by PrinciRaj1992
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 Program to find the Product
# of diagonal elements of a matrix
  
# Function to find the product of diagonals
def productDiagonals(arr, n):
  
    product = 1;
      
    # loop for calculating product of both
    # the principal and secondary diagonals
    for i in range(n):
  
        # For principal diagonal index of row
        # is equal to index of column
        product = product * arr[i][i];
  
        # For secondary diagonal index
        # of column is n-(index of row)-1
        product = product * arr[i][n - i - 1];
      
    # Divide the answer by middle element for
    # matrix of odd size
    if (n % 2 == 1):
        product = product // arr[n // 2][n // 2];
  
    return product;
  
# Driver code
if __name__ == '__main__':
    arr1 = [[ 1, 2, 3, 4 ],[ 5, 6, 7, 8 ],
            [ 9, 7, 4, 2 ],[ 2, 2, 2, 1 ]];
  
    # Function calling
    print(productDiagonals(arr1, 4));
  
    arr2 = [[ 2, 1, 2, 1, 2 ],[ 1, 2, 1, 2, 1 ],
            [ 2, 1, 2, 1, 2 ],[ 1, 2, 1, 2, 1 ],
            [ 2, 1, 2, 1, 2 ]];
  
    # Function calling
    print(productDiagonals(arr2, 5));
      
# This code is contributed by 29AjayKumar
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# Program to find the Product
// of diagonal elements of a matrix
using System;
  
class GFG
{
  
// Function to find the product of diagonals
static int productDiagonals(int [,]arr, int n)
{
  
    int product = 1;
      
    // loop for calculating product of both
    // the principal and secondary diagonals
    for (int i = 0; i < n; i++) 
    {
  
        // For principal diagonal index of row
        // is equal to index of column
        product = product * arr[i,i];
  
        // For secondary diagonal index
        // of column is n-(index of row)-1
        product = product * arr[i,n - i - 1];
    }
  
    // Divide the answer by middle element for
    // matrix of odd size
    if (n % 2 == 1)
    {
        product = product / arr[n / 2,n / 2];
    }
  
    return product;
}
  
// Driver code
public static void Main(String[] args)
{
    int [,]arr1 = { { 1, 2, 3, 4 },
                    { 5, 6, 7, 8 },
                    { 9, 7, 4, 2 },
                    { 2, 2, 2, 1 } };
      
    // Function calling
    Console.Write(productDiagonals(arr1, 4) + "\n");
  
    int [,]arr2 = { { 2, 1, 2, 1, 2 },
                    { 1, 2, 1, 2, 1 },
                    { 2, 1, 2, 1, 2 },
                    { 1, 2, 1, 2, 1 },
                    { 2, 1, 2, 1, 2 } };
      
    // Function calling
    Console.Write(productDiagonals(arr2, 5) + "\n");
}
}
  
// This code is contributed by 29AjayKumar
chevron_right

Output:
9408
512

Time Complexity: O(N)





Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : princiraj1992, 29AjayKumar

Article Tags :
Practice Tags :