# C++ Program to Efficiently Compute Sums of Diagonals of a Matrix

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 (Brute Force) :

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++

 `// 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; ` `} `

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:

## 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; ` `} `

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.
Please refer complete article on Efficiently compute sums of diagonals of a matrix for more details!

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