Efficiently compute sums of diagonals of a matrix

Given a 2D square matrix, find sum of elements in Principal and Secondary diagonals. For example, consider the following 4 X 4 input matrix.

A00 A01 A02 A03
A10 A11 A12 A13
A20 A21 A22 A23
A30 A31 A32 A33

The primary diagonal is formed by the elements A00, A11, A22, A33.

  1. Condition for Principal Diagonal: The row-column condition is row = column.
    The secondary diagonal is formed by the elements A03, A12, A21, A30.
  2. Condition for Secondary Diagonal: The row-column condition is row = numberOfRows – column -1.

Examples :

Input : 
4
1 2 3 4
4 3 2 1
7 8 9 6
6 5 4 3
Output :
Principal Diagonal: 16
Secondary Diagonal: 20

Input :
3
1 1 1
1 1 1
1 1 1
Output :
Principal Diagonal: 3
Secondary Diagonal: 3

Method 1 (O(n ^ 2) :

In this method we use two loops i.e. a loop for columns and a loop for rows and in the inner loop we check for the condition stated above:

C++

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// A simple C++ program to find sum of diagonals
#include <bits/stdc++.h>
using namespace std;
  
const int MAX = 100;
  
void printDiagonalSums(int mat[][MAX], int n)
{
    int principal = 0, secondary = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
  
            // Condition for principal diagonal
            if (i == j)
                principal += mat[i][j];
  
            // Condition for secondary diagonal
            if ((i + j) == (n - 1))
                secondary += mat[i][j];
        }
    }
  
    cout << "Principal Diagonal:" << principal << endl;
    cout << "Secondary Diagonal:" << secondary << endl;
}
  
// Driver code
int main()
{
    int a[][MAX] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, 
                    { 1, 2, 3, 4 }, { 5, 6, 7, 8 } };
    printDiagonalSums(a, 4);
    return 0;
}

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// A simple java program to find 


// sum of diagonals 


import java.io.*; 


  


public class GFG { 


  


    static void printDiagonalSums(int [][]mat, 


                                         int n) 


    { 


        int principal = 0, secondary = 0; 


        for (int i = 0; i < n; i++) { 


            for (int j = 0; j < n; j++) { 


      


                // Condition for principal 


                // diagonal 


                if (i == j) 


                    principal += mat[i][j]; 


      


                // Condition for secondary 


                // diagonal 


                if ((i + j) == (n - 1)) 


                    secondary += mat[i][j]; 


            } 


        } 


      


        System.out.println("Principal Diagonal:"


                                    + principal); 


                                      


        System.out.println("Secondary Diagonal:"


                                    + secondary); 


    } 


  


    // Driver code 


    static public void main (String[] args) 


    { 


          


        int [][]a = { { 1, 2, 3, 4 }, 


                      { 5, 6, 7, 8 },  


                      { 1, 2, 3, 4 }, 


                      { 5, 6, 7, 8 } }; 


                      


        printDiagonalSums(a, 4); 


    } 


} 


  


// This code is contributed by vt_m. 



















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Python3





# A simple Python program to

# find sum of diagonals

MAX = 100


def printDiagonalSums(mat, n):


	principal = 0

	secondary = 0;

	for i in range(0, n):

		for j in range(0, n): 


			# Condition for principal diagonal

			if (i == j):

				principal += mat[i][j]


			# Condition for secondary diagonal

			if ((i + j) == (n – 1)):

				secondary += mat[i][j]


	print(“Principal Diagonal:”, principal)

	print(“Secondary Diagonal:”, secondary)


# Driver code

a = [[ 1, 2, 3, 4 ],

	 [ 5, 6, 7, 8 ],

	 [ 1, 2, 3, 4 ],

 	 [ 5, 6, 7, 8 ]]

printDiagonalSums(a, 4)


# This code is contributed

# by ihritik



C#














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// A simple C# program to find sum 


// of diagonals 


using System; 


  


public class GFG { 


  


    static void printDiagonalSums(int [,]mat, 


                                        int n) 


    { 


        int principal = 0, secondary = 0; 


        for (int i = 0; i < n; i++) { 


            for (int j = 0; j < n; j++) { 


      


                // Condition for principal 


                // diagonal 


                if (i == j) 


                    principal += mat[i,j]; 


      


                // Condition for secondary 


                // diagonal 


                if ((i + j) == (n - 1)) 


                    secondary += mat[i,j]; 


            } 


        } 


      


        Console.WriteLine("Principal Diagonal:"


                                  + principal); 


                                    


        Console.WriteLine("Secondary Diagonal:"


                                  + secondary); 


    } 


  


    // Driver code 


    static public void Main () 


    { 


        int [,]a = { { 1, 2, 3, 4 }, 


                     { 5, 6, 7, 8 },  


                     { 1, 2, 3, 4 }, 


                     { 5, 6, 7, 8 } }; 


                       


        printDiagonalSums(a, 4); 


    } 


} 


  


// This code is contributed by vt_m. 



















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<?php 


// A simple PHP program to 


// find sum of diagonals 


$MAX = 100; 


  


function printDiagonalSums($mat, $n) 


{ 


    global $MAX; 


    $principal = 0; 


    $secondary = 0; 


    for ($i = 0; $i < $n; $i++)  


    { 


        for ($j = 0; $j < $n; $j++)  


        { 


  


            // Condition for  


            // principal diagonal 


            if ($i == $j) 


                $principal += $mat[$i][$j]; 


  


            // Condition for 


            // secondary diagonal 


            if (($i + $j) == ($n - 1)) 


                $secondary += $mat[$i][$j]; 


        } 


    } 


  


    echo "Principal Diagonal:" ,  


               $principal ,"\n"; 


    echo "Secondary Diagonal:",  


              $secondary ,"\n"; 


} 


  


// Driver code 


$a = array (array ( 1, 2, 3, 4 ),  


            array ( 5, 6, 7, 8 ),  


            array ( 1, 2, 3, 4 ),  


            array ( 5, 6, 7, 8 )); 


printDiagonalSums($a, 4); 


  


// This code is contrbuted by ajit 


?> 



















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Output:


Principal Diagonal:18
Secondary Diagonal:18



This code takes O(n^2) time  and O(1) auxiliary space


 




 Method 2 (O(n) : 




In this method we use one loop i.e. a loop for calculating sum of both the principal and secondary diagonals:




C++















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// An efficient C++ program to find sum of diagonals 


#include <bits/stdc++.h> 


using namespace std; 


  


const int MAX = 100; 


  


void printDiagonalSums(int mat[][MAX], int n) 


{ 


    int principal = 0, secondary = 0;  


    for (int i = 0; i < n; i++) { 


        principal += mat[i][i]; 


        secondary += mat[i][n - i - 1];         


    } 


  


    cout << "Principal Diagonal:" << principal << endl; 


    cout << "Secondary Diagonal:" << secondary << endl; 


} 


  


// Driver code 


int main() 


{ 


    int a[][MAX] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 },  


                     { 1, 2, 3, 4 }, { 5, 6, 7, 8 } }; 


    printDiagonalSums(a, 4); 


    return 0; 


} 



















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// An efficient java program to find 


// sum of diagonals 


import java.io.*; 


  


public class GFG { 


  


    static void printDiagonalSums(int [][]mat, 


                                        int n) 


    { 


        int principal = 0, secondary = 0;  


        for (int i = 0; i < n; i++) { 


            principal += mat[i][i]; 


            secondary += mat[i][n - i - 1];  


        } 


      


        System.out.println("Principal Diagonal:"


                                   + principal); 


                                     


        System.out.println("Secondary Diagonal:"


                                   + secondary); 


    } 


      


    // Driver code 


    static public void main (String[] args) 


    { 


        int [][]a = { { 1, 2, 3, 4 }, 


                      { 5, 6, 7, 8 },  


                      { 1, 2, 3, 4 }, 


                      { 5, 6, 7, 8 } }; 


      


        printDiagonalSums(a, 4); 


    } 


} 


  


// This code is contributed by vt_m. 



















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Python3





# A simple Python3 program to find

# sum of diagonals

MAX = 100


def printDiagonalSums(mat, n):


	principal = 0

	secondary = 0

	for i in range(0, n):

		principal += mat[i][i]

		secondary += mat[i][n – i – 1]


	print(“Principal Diagonal:”, principal)

	print(“Secondary Diagonal:”, secondary)


# Driver code

a = [[ 1, 2, 3, 4 ],

	 [ 5, 6, 7, 8 ],

	 [ 1, 2, 3, 4 ],

	 [ 5, 6, 7, 8 ]]

printDiagonalSums(a, 4)


# This code is contributed

# by ihritik



C#














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// An efficient C#program to find 


// sum of diagonals 


using System; 


  


public class GFG { 


  


    static void printDiagonalSums(int [,]mat, 


                                       int n) 


    { 


        int principal = 0, secondary = 0;  


        for (int i = 0; i < n; i++) { 


            principal += mat[i,i]; 


            secondary += mat[i,n - i - 1];  


        } 


      


        Console.WriteLine("Principal Diagonal:"


                                  + principal); 


                                    


        Console.WriteLine("Secondary Diagonal:"


                                  + secondary); 


    } 


      


    // Driver code 


    static public void Main () 


    { 


        int [,]a = { { 1, 2, 3, 4 }, 


                     { 5, 6, 7, 8 },  


                     { 1, 2, 3, 4 }, 


                     { 5, 6, 7, 8 } }; 


                       


        printDiagonalSums(a, 4); 


    } 


} 


  


// This code is contributed by vt_m. 



















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PHP














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<?php 


// An efficient PHP program  


// to find sum of diagonals 


$MAX = 100; 


  


function printDiagonalSums($mat, $n) 


{ 


    global $MAX; 


    $principal = 0; $secondary = 0;  


    for ($i = 0; $i < $n; $i++)  


    { 


        $principal += $mat[$i][$i]; 


        $secondary += $mat[$i][$n - $i - 1];      


    } 


  


    echo "Principal Diagonal:" , 


               $principal ,"\n"; 


    echo "Secondary Diagonal:" ,  


               $secondary ,"\n"; 


} 


  


// Driver Code 


$a = array(array(1, 2, 3, 4), 


           array(5, 6, 7, 8),  


           array(1, 2, 3, 4), 


           array(5, 6, 7, 8)); 


printDiagonalSums($a, 4); 


  


// This code is contributed by aj_36  


?> 



















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Output :


Principal Diagonal:18
Secondary Diagonal:18



This code takes O(n) time  and O(1) auxiliary space


This article is contributed by Mohak Agrawal. 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 write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



          
          
          
          

            

    

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Improved By :  vt_m, jit_t, ihritik