Open In App

# Maximum sum of elements in a diagonal parallel to the main diagonal of a given Matrix

Give a square matrix mat[][] of dimensions N * N, the task is to find the maximum sum of elements present in the given matrix along the diagonals which are parallel to the main diagonal. Below is the image of the same. Examples:

Input: mat[][] = {{1, 2, 5, 7}, {2, 6, 7, 3}, {12, 3, 2, 4}, {3, 6, 9, 4}}
Output: 18
Explanation:
Sum of elements present in the diagonal having cells (2, 0) and (3, 1) is 12 + 6 = 18 which is maximum among all diagonals.

Input: mat[][] = {{5, 2, 5, 7}, {2, 5, 7, 3}, {12, 3, 5, 4}, {3, 6, 9, 5}}
Output: 18
Explanation:
Sum of elements present in the main diagonal having cells (0, 0), (1, 1), (2, 2) and (3, 3) is 5 + 5 + 5 + 5 = 20 which is maximum among all diagonals.

Approach: The idea is to traverse cells of each diagonal that is parallel to the main diagonal and observe that for any diagonal above the main diagonal starting at cell (x, y), it’s corresponding diagonal that is below the main diagonal will start at cell (y, x). For each diagonal, starting at cell (x, y) all its elements will be on cells (x + k, y + k) where 0 <= x + k, y + k < N. Follow the below steps to solve the problem:

• Initialize a variable maxSum with 0 which will store the maximum diagonal sum.
• Traverse the columns of 0th row from i over the range [0, N – 1].
• Initialize variables sum1 and sum2 which will store the diagonal sums starting from the cell (row, col) and from the cell (col, row) respectively where r is 0 and c is col.
• Increment both row and c by 1. Add mat[row][col] to sum1 and mat[col][row] to sum2 while row and col are smaller than N. Finally, update maxSum to store the maximum of maxSum, sum1, and sum2.
• After traversing the given matrix, print the value maxSum as the maximum sum.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to return maximum diagonal``// sum that are parallel to main diagonal``int` `maxDiagonalSum(vector > arr, ``int` `N)``{``    ``// Initialize maxSum``    ``int` `maxSum = 0;` `    ``// Traverse through the columns``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// Initialize r and c``        ``int` `row = 0, col = i;` `        ``// Diagonal sums``        ``int` `sum1 = 0, sum2 = 0;``        ``while` `(col < N && row < N) {``            ``sum1 += arr[row][col];``            ``sum2 += arr[col][row];``            ``row++;``            ``col++;``        ``}` `        ``// Update maxSum with``        ``// the maximum sum``        ``maxSum = max({ sum1, maxSum, sum2 });``    ``}` `    ``// Return the maxSum``    ``return` `maxSum;``}` `// Driver Code``int` `main()``{``    ``// Given matrix mat[][]``    ``vector > mat``        ``= { { 1, 2, 5, 7 },``            ``{ 2, 6, 7, 3 },``            ``{ 12, 3, 2, 4 },``            ``{ 3, 6, 9, 4 } };``    ``int` `N = mat.size();` `    ``// Function Call``    ``cout << maxDiagonalSum(mat, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;` `class` `GFG{``  ` `// Function to return maximum diagonal``// sum that are parallel to main diagonal``static` `int` `maxDiagonalSum(``int` `arr[][], ``int` `N)``{``    ` `    ``// Initialize maxSum``    ``int` `maxSum = ``0``;` `    ``// Traverse through the columns``    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ` `        ``// Initialize r and c``        ``int` `row = ``0``, col = i;` `        ``// Diagonal sums``        ``int` `sum1 = ``0``, sum2 = ``0``;``        ``while` `(col < N && row < N)``        ``{``            ``sum1 += arr[row][col];``            ``sum2 += arr[col][row];``            ``row++;``            ``col++;``        ``}` `        ``// Update maxSum with``        ``// the maximum sum``        ``maxSum = Math.max(maxSum,``                          ``Math.max(sum1,``                                   ``sum2));``    ``}` `    ``// Return the maxSum``    ``return` `maxSum;``}` `// Driver code``public` `static` `void` `main (String[] args)``{``    ` `    ``// Given matrix mat[][]``    ``int` `mat[][] = { { ``1``, ``2``, ``5``, ``7` `},``                    ``{ ``2``, ``6``, ``7``, ``3` `},``                    ``{ ``12``, ``3``, ``2``, ``4` `},``                    ``{ ``3``, ``6``, ``9``, ``4` `} };``    ``int` `N = mat.length;` `    ``// Function Call``    ``System.out.println(maxDiagonalSum(mat, N));``}``}` `// This code is contributed by math_lover`

## Python3

 `# Python3 program for the above approach` `# Function to return maximum diagonal``# sum that are parallel to main diagonal``def` `maxDiagonalSum(arr, N):``    ` `    ``# Initialize maxSum``    ``maxSum ``=` `0` `    ``# Traverse through the columns``    ``for` `i ``in` `range``(N):``        ` `        ``# Initialize r and c``        ``row ``=` `0``        ``col ``=` `i` `        ``# Diagonal sums``        ``sum1 ``=` `0``        ``sum2 ``=` `0``        ` `        ``while` `col < N ``and` `row < N:``            ``sum1 ``+``=` `arr[row][col]``            ``sum2 ``+``=` `arr[col][row]``            ``row ``+``=` `1``            ``col ``+``=` `1` `        ``# Update maxSum with``        ``# the maximum sum``        ``maxSum ``=` `max``([ sum1, maxSum, sum2])` `    ``# Return the maxSum``    ``return` `maxSum` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Given matrix mat[][]``    ``mat ``=` `[ [ ``1``, ``2``, ``5``, ``7` `],``            ``[ ``2``, ``6``, ``7``, ``3` `],``            ``[ ``12``, ``3``, ``2``, ``4` `],``            ``[ ``3``, ``6``, ``9``, ``4` `] ]` `    ``N ``=` `len``(mat)` `    ``# Function Call``    ``print``(maxDiagonalSum(mat, N))` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the``// above approach``using` `System;``class` `GFG{``  ` `// Function to return maximum``// diagonal sum that are parallel``// to main diagonal``static` `int` `maxDiagonalSum(``int` `[,]arr,``                          ``int` `N)``{   ``  ``// Initialize maxSum``  ``int` `maxSum = 0;` `  ``// Traverse through the``  ``// columns``  ``for``(``int` `i = 0; i < N; i++)``  ``{``    ``// Initialize r and c``    ``int` `row = 0, col = i;` `    ``// Diagonal sums``    ``int` `sum1 = 0, sum2 = 0;``    ``while` `(col < N && row < N)``    ``{``      ``sum1 += arr[row,col];``      ``sum2 += arr[col,row];``      ``row++;``      ``col++;``    ``}` `    ``// Update maxSum with``    ``// the maximum sum``    ``maxSum = Math.Max(maxSum,``             ``Math.Max(sum1,``                      ``sum2));``  ``}` `  ``// Return the maxSum``  ``return` `maxSum;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{   ``  ``// Given matrix [,]mat``  ``int` `[,]mat = {{1, 2, 5, 7},``                ``{2, 6, 7, 3},``                ``{12, 3, 2, 4},``                ``{3, 6, 9, 4}};``  ``int` `N = mat.GetLength(0);` `  ``// Function Call``  ``Console.WriteLine(maxDiagonalSum(mat, N));``}``}` `// This code is contributed by gauravrajput1`

## Javascript

 ``

Output

```18

```

Time Complexity: O(N2)
Auxiliary Space: O(N2)

### Traverse diagonals and find maximum sum:

Approach:

This approach involves traversing all diagonals of the matrix and finding the sum of elements in each diagonal. The maximum sum is then returned as the answer.

Initialize max_sum variable to 0.
Traverse through each element of the matrix.
For each element, check if it is present in the diagonal that goes from top-left to bottom-right. If it is present, add it to sum1.
Similarly, for each element, check if it is present in the diagonal that goes from top-right to bottom-left. If it is present, add it to sum2.
After calculating the sum for both diagonals, take the maximum of sum1, sum2, and max_sum and update the value of max_sum.
Finally, return the max_sum as the output.

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;` `int` `max_sum_diagonal(``const` `vector>& mat) {``    ``int` `max_sum = 2;``    ``int` `n = mat.size();` `    ``for` `(``int` `i = 0; i < n; i++) {``        ``int` `sum1 = 1;``        ``int` `sum2 = 11;` `        ``for` `(``int` `j = 0; j < n; j++) {``            ``// Check diagonal from top-left to bottom-right``            ``if` `(i == j) {``                ``sum1 += mat[i][j];``            ``}``            ``// Check diagonal from top-right to bottom-left``            ``if` `(i + j == n - 1) {``                ``sum2 += mat[i][j];``            ``}``        ``}` `        ``max_sum = max(max_sum, max(sum1, sum2));``    ``}` `    ``return` `max_sum;``}` `int` `main() {``    ``vector> mat = {{5, 2, 5, 7}, {2, 5, 7, 3}, {12, 3, 5, 4}, {3, 6, 9, 5}};` `    ``// Output``    ``cout << max_sum_diagonal(mat) << endl; ``// Output: 18` `    ``return` `0;``}`  `// This code is contributed by Vaibhav nandan`

## Java

 `// Java program for the above approach``import` `java.util.*;` `public` `class` `GFG {``    ``public` `static` `int` `maxSumDiagonal(``int``[][] mat) {``        ``int` `maxSum = ``2``;``        ``int` `n = mat.length;` `        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``int` `sum1 = ``1``;``            ``int` `sum2 = ``11``;` `            ``for` `(``int` `j = ``0``; j < n; j++) {``                ``// Check diagonal from top-left to bottom-right``                ``if` `(i == j) {``                    ``sum1 += mat[i][j];``                ``}``                ``// Check diagonal from top-right to bottom-left``                ``if` `(i + j == n - ``1``) {``                    ``sum2 += mat[i][j];``                ``}``            ``}` `            ``maxSum = Math.max(maxSum, Math.max(sum1, sum2));``        ``}` `        ``return` `maxSum;``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``int``[][] mat = { { ``5``, ``2``, ``5``, ``7` `}, { ``2``, ``5``, ``7``, ``3` `}, { ``12``, ``3``, ``5``, ``4` `}, { ``3``, ``6``, ``9``, ``5` `} };` `        ``// Output``        ``System.out.println(maxSumDiagonal(mat)); ``// Output: 18``    ``}``}``// This code is contributed by Veena Mishra`

## Python3

 `def` `max_sum_diagonal(mat):``    ``max_sum ``=` `2``    ``for` `i ``in` `range``(``len``(mat)):``        ``sum1 ``=` `1``        ``sum2 ``=` `11``        ``for` `j ``in` `range``(``len``(mat)):``            ``# Check diagonal from top-left to bottom-right``            ``if` `i ``=``=` `j:``                ``sum1 ``+``=` `mat[i][j]``            ``# Check diagonal from top-right to bottom-left``            ``if` `i ``+` `j ``=``=` `len``(mat) ``-` `1``:``                ``sum2 ``+``=` `mat[i][j]``        ``max_sum ``=` `max``(max_sum, sum1, sum2)``    ``return` `max_sum` `# Sample Input``mat ``=` `[[``5``, ``2``, ``5``, ``7``], [``2``, ``5``, ``7``, ``3``], [``12``, ``3``, ``5``, ``4``], [``3``, ``6``, ``9``, ``5``]]` `# Output``print``(max_sum_diagonal(mat))  ``# Output: 18`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``    ``// Function to find the maximum sum of two diagonals in a square matrix``    ``static` `int` `MaxSumDiagonal(List> mat)``    ``{``        ``int` `maxSum = 2;``        ``int` `n = mat.Count;` `        ``// Loop through each row of the matrix``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``int` `sum1 = 1; ``// Initialize the sum of the diagonal from top-left to bottom-right``            ``int` `sum2 = 11; ``// Initialize the sum of the diagonal from top-right to bottom-left` `            ``// Loop through each element of the current row``            ``for` `(``int` `j = 0; j < n; j++)``            ``{``                ``// Check if the current element is on the top-left to bottom-right diagonal``                ``if` `(i == j)``                ``{``                    ``sum1 += mat[i][j];``                ``}` `                ``// Check if the current element is on the top-right to bottom-left diagonal``                ``if` `(i + j == n - 1)``                ``{``                    ``sum2 += mat[i][j];``                ``}``            ``}` `            ``// Calculate the maximum between the two sums and update maxSum accordingly``            ``maxSum = Math.Max(maxSum, Math.Max(sum1, sum2));``        ``}` `        ``return` `maxSum;``    ``}``//Driver Code``    ``static` `void` `Main()``    ``{``        ``// Create a sample square matrix``        ``List> mat = ``new` `List>``        ``{``            ``new` `List<``int``> {5, 2, 5, 7},``            ``new` `List<``int``> {2, 5, 7, 3},``            ``new` `List<``int``> {12, 3, 5, 4},``            ``new` `List<``int``> {3, 6, 9, 5}``        ``};` `        ``// Output the result``        ``Console.WriteLine(MaxSumDiagonal(mat)); ``// Output: 18``    ``}``}`

## Javascript

 `function` `maxSumDiagonal(mat) {``    ``let maxSum = 2;``    ``const n = mat.length;` `    ``for` `(let i = 0; i < n; i++) {``        ``let sum1 = 1;``        ``let sum2 = 11;` `        ``for` `(let j = 0; j < n; j++) {``            ``// Check diagonal from top-left to bottom-right``            ``if` `(i === j) {``                ``sum1 += mat[i][j];``            ``}``            ``// Check diagonal from top-right to bottom-left``            ``if` `(i + j === n - 1) {``                ``sum2 += mat[i][j];``            ``}``        ``}` `        ``maxSum = Math.max(maxSum, Math.max(sum1, sum2));``    ``}` `    ``return` `maxSum;``}` `const mat = [``    ``[5, 2, 5, 7],``    ``[2, 5, 7, 3],``    ``[12, 3, 5, 4],``    ``[3, 6, 9, 5]``];` `// Output``console.log(maxSumDiagonal(mat)); ``// Output: 18`

Output

```18

```

Time Complexity: O(n^2) where n is the size of the matrix.
Auxiliary Space: O(1)