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 <bits/stdc++.h>using namespace std;// Function to update given matrix with// maximum of left and right diagonal sumvoid 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 codeint 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 sumimport java.io.*;import java.util.*;class GFG { static void updateMatrix(int mat[][]) { Map<Integer, Integer> right = new HashMap<Integer, Integer>(); Map<Integer, Integer> left = new HashMap<Integer, Integer>(); 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 sumdef 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 matrixmat = [[5, 2, 1], [7, 2, 6], [3, 1, 9]]updateMatrix(mat) |
C#
// C# program for the above approachusing 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
<script>// Javascript program for the above approach// Function to update given matrix with// maximum of left and right diagonal sumfunction updateMatrix(mat) { // Stores the total sum // of right diagonal let right = new Map(); // Stores the total sum // of left diagonal let left = new Map(); for(let i = 0; i < 3; i++) { for(let j = 0; j < 3; j++) { // Update the map storing // right diagonal sums if (!right.has(i + j)) right.set(i + j, mat[i][j]); else right.set(i + j, right.get(i + j) + mat[i][j]); // Update the map storing // left diagonal sums if (!left.has(i - j)) left.set(i - j, mat[i][j]); else left.set(i - j, left.get(i - j) + mat[i][j]); } } // Traverse the matrix for(let i = 0; i < 3; i++) { for(let 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(let i = 0; i < 3; i++) { for(let j = 0; j < 3; j++) { document.write(mat[i][j] + " "); } document.write("<br>"); }}// Driver codelet mat = [ [ 5, 2, 1 ], [ 7, 2, 6 ], [ 3, 1, 9 ] ];updateMatrix(mat);// This code is contributed by gfgking.</script> |
Output:
16 9 6 9 16 8 6 8 16
Time Complexity: O(N * M)
Auxiliary Space: O(N * M)
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!
