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 = 9408Input: 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:
- We need to find out the principal diagonal and secondary diagonal elements of the matrix. Please refer to this article for this [ Program to print the Diagonals of a Matrix ]
- In this method, we use one loop i.e. a loop for calculating product of both the principal and secondary diagonals
- Divide the answer by middle element for matrix of odd size
Below is the implementation of the above approach:
CPP
// 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; } |
Java
// 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 |
Python3
# 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 |
C#
// 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 |
Javascript
<script> // JavaScript Program to find the Product // of diagonal elements of a matrix // Function to find the product of diagonals function productDiagonals(arr, n) { var product = 1; // loop for calculating product of both // the principal and secondary diagonals for ( var 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[parseInt(n / 2)][parseInt(n / 2)]; } return product; } // Driver code var arr1 = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 7, 4, 2 ], [ 2, 2, 2, 1 ] ]; // Function calling document.write( productDiagonals(arr1, 4) + "<br>" ); var 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 document.write( productDiagonals(arr2, 5)); </script> |
9408 512
Time Complexity: O(N), traversing a loop from 0 to N.
Auxiliary Space: O(1) because constant extra space is required.