# Modify a matrix by replacing each element with the maximum of its left or right diagonal sum

Given a matrix mat[][] with dimensions M * N, the task is to replace each matrix elements with the maximum sum of its left or right diagonal.

Examples:

Input: mat[][] = {{5, 2, 1}, {7, 2, 6}, {3, 1, 9}}
Output:
16  9   6
9 16   8
6   8 16
Explanation:
Replace each element with max(sum of right diagonal, sum of left diagonal).
Follow the diagram below to understand more clearly.

Input: mat[][] = {{1, 2}, {3, 4}}
Output:
5 5
5 5

Approach: The main idea is based on the facts based on the observation stated below:

• Sum of row and column indices for the right diagonal elements are equal.
• Difference between the row and column indices for left diagonal elements is equal.
• Using the above two properties to store the sum of the left and right diagonals of each element using a Map.
• Traverse the matrix and replace each element with the maximum of left diagonal sum or right diagonal sum.
• Print the final matrix obtained.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to update given matrix with` `// maximum of left and right diagonal sum` `void` `updateMatrix(``int` `mat[][3])` `{`   `  ``// Stores the total sum` `  ``// of right diagonal` `  ``map<``int``, ``int``> right;`   `  ``// Stores the total sum` `  ``// of left diagonal` `  ``map<``int``, ``int``> left;`   `  ``for` `(``int` `i = 0; i < 3; i++) {` `    ``for` `(``int` `j = 0; j < 3; j++) {`   `      ``// Update the map storing` `      ``// right diagonal sums` `      ``if` `(right.find(i + j) == right.end())` `        ``right[i + j] = mat[i][j];` `      ``else` `        ``right[i + j] += mat[i][j];`   `      ``// Update the map storing` `      ``// left diagonal sums` `      ``if` `(left.find(i - j) == left.end())` `        ``left[i - j] = mat[i][j];` `      ``else` `        ``left[i - j] += mat[i][j];` `    ``}` `  ``}` `  `  `  ``// Traverse the matrix` `  ``for` `(``int` `i = 0; i < 3; i++) {` `    ``for` `(``int` `j = 0; j < 3; j++) {`   `      ``// Update the matrix` `      ``mat[i][j] = max(right[i + j], left[i - j]);` `    ``}` `  ``}`   `  ``// Print the matrix` `  ``for` `(``int` `i = 0; i < 3; i++) {` `    ``for` `(``int` `j = 0; j < 3; j++) {` `      ``cout << mat[i][j] << ``" "``;` `    ``}` `    ``cout << endl;` `  ``}` `}`   `// Driver code` `int` `main()` `{` `  ``int` `mat[][3]` `    ``= { { 5, 2, 1 }, { 7, 2, 6 }, { 3, 1, 9 } };` `  ``updateMatrix(mat);` `  ``return` `0;` `}`   `// This code is contributed by ukasp.`

## Java

 `// Java program for the above approach` ` `  `// Function to update given matrix with` `// maximum of left and right diagonal sum` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG {` `    `  `    ``static` `void` `updateMatrix(``int` `mat[][])` `    ``{` `        ``Map right` `            ``= ``new` `HashMap();  ` `            `  `        ``Map left` `            ``= ``new` `HashMap();      ` `            `  `        ``for``(``int` `i = ``0``; i < ``3``; i++)` `    ``{` `        ``for``(``int` `j = ``0``; j < ``3``; j++)` `        ``{` `             `  `            ``// Update the map storing` `            ``// right diagonal sums` `            ``if` `(!right.containsKey(i + j))` `                ``right.put(i + j, mat[i][j]);` `            ``else` `                ``right.put(i + j,` `                ``right.get(i + j) + mat[i][j]);` ` `  `            ``// Update the map storing` `            ``// left diagonal sums` `            ``if` `(!left.containsKey(i - j))` `                ``left.put(i - j, mat[i][j]);` ` `  `            ``else` `                ``left.put(i - j,` `                ``left.get(i - j) + mat[i][j]);` `        ``}` `    ``}` ` `  `    ``// Traverse the matrix` `    ``for``(``int` `i = ``0``; i < ``3``; i++)` `    ``{` `        ``for``(``int` `j = ``0``; j < ``3``; j++)` `        ``{` `             `  `            ``// Update the matrix` `            ``mat[i][j] = Math.max(right.get(i + j),` `                                  ``left.get(i - j));` `        ``}` `    ``}` ` `  `    ``// Print the matrix` `    ``for``(``int` `i = ``0``; i < ``3``; i++)` `    ``{` `        ``for``(``int` `j = ``0``; j < ``3``; j++)` `        ``{` `            ``System.out.print(mat[i][j] + ``" "``);` `        ``}` `        ``System.out.print(``"\n"``);` `    ``}    ` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `main (String[] args) {` `        ``int``[][] mat = {{ ``5``, ``2``, ``1` `},` `            ``{ ``7``, ``2``, ``6` `},` `            ``{ ``3``, ``1``, ``9` `}};` `            ``updateMatrix(mat); ` `    ``}` `}`   `// This code is contributed by avanitrachhadiya2155`

## Python3

 `# Python3 program for the above approach`   `# Function to update given matrix with` `# maximum of left and right diagonal sum` `def` `updateMatrix(mat):`   `    ``# Stores the total sum` `    ``# of right diagonal` `    ``right ``=` `{}`   `    ``# Stores the total sum` `    ``# of left diagonal` `    ``left ``=` `{}`   `    ``for` `i ``in` `range``(``len``(mat)):` `        ``for` `j ``in` `range``(``len``(mat[``0``])):`   `            ``# Update the map storing` `            ``# right diagonal sums` `            ``if` `i ``+` `j ``not` `in` `right:` `                ``right[i ``+` `j] ``=` `mat[i][j]` `            ``else``:` `                ``right[i ``+` `j] ``+``=` `mat[i][j]`   `            ``# Update the map storing` `            ``# left diagonal sums` `            ``if` `i``-``j ``not` `in` `left:` `                ``left[i``-``j] ``=` `mat[i][j]` `            ``else``:` `                ``left[i``-``j] ``+``=` `mat[i][j]`   `    ``# Traverse the matrix` `    ``for` `i ``in` `range``(``len``(mat)):` `        ``for` `j ``in` `range``(``len``(mat[``0``])):`   `            ``# Update the matrix` `            ``mat[i][j] ``=` `max``(right[i ``+` `j], left[i``-``j])`   `    ``# Print the matrix` `    ``for` `i ``in` `mat:` `        ``print``(``*``i)`   `# Given matrix` `mat ``=` `[[``5``, ``2``, ``1``], [``7``, ``2``, ``6``], [``3``, ``1``, ``9``]]` `updateMatrix(mat)`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections;` `using` `System.Collections.Generic;` `class` `GFG` `{` `    ``// Function to update given matrix with` `    ``// maximum of left and right diagonal sum` `    ``static` `void` `updateMatrix(``int``[,] mat)` `    ``{` `     `  `      ``// Stores the total sum` `      ``// of right diagonal` `      ``Dictionary<``int``, ``int``> right = ``new` `Dictionary<``int``, ``int``>(); ` `     `  `      ``// Stores the total sum` `      ``// of left diagonal` `      ``Dictionary<``int``, ``int``> left = ``new` `Dictionary<``int``, ``int``>(); ` `     `  `      ``for` `(``int` `i = 0; i < 3; i++) {` `        ``for` `(``int` `j = 0; j < 3; j++) {` `     `  `          ``// Update the map storing` `          ``// right diagonal sums` `          ``if` `(!right.ContainsKey(i + j))` `            ``right[i + j] = mat[i,j];` `          ``else` `            ``right[i + j] += mat[i,j];` `     `  `          ``// Update the map storing` `          ``// left diagonal sums` `          ``if` `(!left.ContainsKey(i - j))` `            ``left[i - j] = mat[i,j];` `          ``else` `            ``left[i - j] += mat[i,j];` `        ``}` `      ``}` `       `  `      ``// Traverse the matrix` `      ``for` `(``int` `i = 0; i < 3; i++) {` `        ``for` `(``int` `j = 0; j < 3; j++) {` `     `  `          ``// Update the matrix` `          ``mat[i,j] = Math.Max(right[i + j], left[i - j]);` `        ``}` `      ``}` `     `  `      ``// Print the matrix` `      ``for` `(``int` `i = 0; i < 3; i++) {` `        ``for` `(``int` `j = 0; j < 3; j++) {` `          ``Console.Write(mat[i,j] + ``" "``);` `        ``}` `        ``Console.WriteLine();` `      ``}` `    ``}` `    `  `  ``static` `void` `Main ()` `  ``{` `    ``int``[,] mat = { { 5, 2, 1 }, { 7, 2, 6 }, { 3, 1, 9 } };` `    ``updateMatrix(mat);` `  ``}` `}`   `// This code is contributed by suresh07.`

## Javascript

 ``

Output:

```16 9 6
9 16 8
6 8 16```

Time Complexity: O(N * M)
Auxiliary Space: O(N * M)

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