Given a 2D square matrix, find the 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.
- Condition for Principal Diagonal: The row-column condition is row = column.
The secondary diagonal is formed by the elements A03, A12, A21, A30. - 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: 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:
Implementation:
C
// A simple C program to // find sum of diagonals#include <stdio.h>const int M = 4;const int N = 4;void printDiagonalSums(int mat[M][N]){ int principal = 0, secondary = 0; for (int i = 0; i < M; 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]; } } printf("%s", "Principal Diagonal:"); printf("%d\n", principal); printf("%s", "Secondary Diagonal:"); printf("%d\n", secondary);}// Driver codeint main(){ int a[][4] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {1, 2, 3, 4}, {5, 6, 7, 8}}; printDiagonalSums(a); return 0;} |
C++
// 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 codeint 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;} |
Java
// A simple java program to find// sum of diagonalsimport 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. |
Python3
# A simple Python program to # find sum of diagonalsMAX = 100def 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 codea = [[ 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#
// A simple C# program to find sum// of diagonalsusing 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. |
PHP
<?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 contributed by ajit?> |
Javascript
<script>// A simple Javascript program to find sum of diagonalsconst MAX = 100;void printDiagonalSums(mat, n){ let principal = 0, secondary = 0; for (let i = 0; i < n; i++) { for (let 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]; } } document.write("Principal Diagonal:" + principal + "<br>"); document.write("Secondary Diagonal:" + secondary + "<br>");}// Driver code let 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 subhammahato348.</script> |
Output
Principal Diagonal:18 Secondary Diagonal:18
Time Complexity: O(N*N), as we are using nested loops to traverse N*N times.
Auxiliary Space: O(1), as we are not using any extra space.
Method 2( Efficient Approach): In this method, we use one loop i.e. a loop for calculating the sum of both the principal and secondary diagonals:
Implementation:
C
// An efficient C program to find // sum of diagonals#include <stdio.h>#include <stdlib.h>const int M = 4;const int N = 4;void printDiagonalSums(int mat[M][N]){ int principal = 0, secondary = 0; for (int i = 0; i < N; i++) { principal += mat[i][i]; secondary += mat[i][N - i - 1]; } printf("%s","Principal Diagonal:"); printf("%d\n", principal); printf("%s", "Secondary Diagonal:"); printf("%d\n", secondary);}// Driver codeint main(){ int a[4][4] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {1, 2, 3, 4}, {5, 6, 7, 8}}; printDiagonalSums(a); return 0;} |
C++
// 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 codeint 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;} |
Java
// An efficient java program to find// sum of diagonalsimport 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. |
Python3
# A simple Python3 program to find# sum of diagonalsMAX = 100def 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 codea = [[ 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#
// An efficient C#program to find// sum of diagonalsusing 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. |
PHP
<?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 ?> |
Javascript
<script>// An efficient Javascript program to find// sum of diagonalsfunction printDiagonalSums(mat,n) { let principal = 0, secondary = 0; for (let i = 0; i < n; i++) { principal += mat[i][i]; secondary += mat[i][n - i - 1]; } document.write("Principal Diagonal:" + principal+"<br>"); document.write("Secondary Diagonal:" + secondary); } // Driver code let a = [[ 1, 2, 3, 4 ], [5, 6, 7, 8 ], [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ]]; printDiagonalSums(a, 4); // This code is contributed Bobby</script> |
Output
Principal Diagonal:18 Secondary Diagonal:18
Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.
This article is contributed by Mohak Agrawal. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
