# Efficiently compute sums of diagonals of a matrix

• Difficulty Level : Basic
• Last Updated : 18 Jul, 2022

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.

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: 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 ``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;``}`

## Java

 `// 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.`

## 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#

 `// 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.`

## PHP

 ``

## Javascript

 ``

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 sum of both the principal and secondary diagonals:

Implementation:

## C++

 `// An efficient C++ program to find sum of diagonals``#include ``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;``}`

## Java

 `// 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.`

## 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#

 `// 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.`

## PHP

 ``

## Javascript

 ``

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 GeeksforGeeks 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 GeeksforGeeks main page and help other Geeks.

My Personal Notes arrow_drop_up