Given a matrix of N * M. Find the maximum path sum in matrix. The maximum path is sum of all elements from first row to last row where you are allowed to move only down or diagonally to left or right. You can start from any element in first row.
Examples:
Input : mat[][] = 10 10 2 0 20 4
1 0 0 30 2 5
0 10 4 0 2 0
1 0 2 20 0 4
Output : 74
The maximum sum path is 20-30-4-20.
Input : mat[][] = 1 2 3
9 8 7
4 5 6
Output : 17
The maximum sum path is 3-8-6.
We are given a matrix of N * M. To find max path sum first we have to find max value in first row of matrix. Store this value in res. Now for every element in matrix update element with max value which can be included in max path. If the value is greater then res then update res. In last return res which consists of max path sum value.
Implementation:
C++
// CPP program for finding max path in matrix#include <bits/stdc++.h>#define N 4#define M 6using namespace std;// To calculate max path in matrixint findMaxPath(int mat[][M]){ for (int i = 1; i < N; i++) { for (int j = 0; j < M; j++) { // When all paths are possible if (j > 0 && j < M - 1) mat[i][j] += max(mat[i - 1][j], max(mat[i - 1][j - 1], mat[i - 1][j + 1])); // When diagonal right is not possible else if (j > 0) mat[i][j] += max(mat[i - 1][j], mat[i - 1][j - 1]); // When diagonal left is not possible else if (j < M - 1) mat[i][j] += max(mat[i - 1][j], mat[i - 1][j + 1]); // Store max path sum } } int res = 0; for (int j = 0; j < M; j++) res = max(mat[N-1][j], res); return res;}// Driver program to check findMaxPathint main(){ int mat1[N][M] = { { 10, 10, 2, 0, 20, 4 }, { 1, 0, 0, 30, 2, 5 }, { 0, 10, 4, 0, 2, 0 }, { 1, 0, 2, 20, 0, 4 } }; cout << findMaxPath(mat1) << endl; return 0;} |
Java
// Java program for finding max path in matriximport static java.lang.Math.max;class GFG { public static int N = 4, M = 6; // Function to calculate max path in matrix static int findMaxPath(int mat[][]) { // To find max val in first row int res = -1; for (int i = 0; i < M; i++) res = max(res, mat[0][i]); for (int i = 1; i < N; i++) { res = -1; for (int j = 0; j < M; j++) { // When all paths are possible if (j > 0 && j < M - 1) mat[i][j] += max(mat[i - 1][j], max(mat[i - 1][j - 1], mat[i - 1][j + 1])); // When diagonal right is not possible else if (j > 0) mat[i][j] += max(mat[i - 1][j], mat[i - 1][j - 1]); // When diagonal left is not possible else if (j < M - 1) mat[i][j] += max(mat[i - 1][j], mat[i - 1][j + 1]); // Store max path sum res = max(mat[i][j], res); } } return res; } // driver program public static void main (String[] args) { int mat[][] = { { 10, 10, 2, 0, 20, 4 }, { 1, 0, 0, 30, 2, 5 }, { 0, 10, 4, 0, 2, 0 }, { 1, 0, 2, 20, 0, 4 } }; System.out.println(findMaxPath(mat)); }}// Contributed by Pramod Kumar |
Python3
# Python 3 program for finding max path in matrix# To calculate max path in matrixdef findMaxPath(mat): for i in range(1, N): res = -1 for j in range(M): # When all paths are possible if (j > 0 and j < M - 1): mat[i][j] += max(mat[i - 1][j], max(mat[i - 1][j - 1], mat[i - 1][j + 1])) # When diagonal right is not possible else if (j > 0): mat[i][j] += max(mat[i - 1][j], mat[i - 1][j - 1]) # When diagonal left is not possible else if (j < M - 1): mat[i][j] += max(mat[i - 1][j], mat[i - 1][j + 1]) # Store max path sum res = max(mat[i][j], res) return res# Driver program to check findMaxPathN=4M=6mat = ([[ 10, 10, 2, 0, 20, 4 ], [ 1, 0, 0, 30, 2, 5 ], [ 0, 10, 4, 0, 2, 0 ], [ 1, 0, 2, 20, 0, 4 ]]) print(findMaxPath(mat))# This code is contributed by Azkia Anam. |
C#
// C# program for finding// max path in matrixusing System;class GFG { static int N = 4, M = 6; // find the max element static int max(int a, int b) { if(a > b) return a; else return b; } // Function to calculate // max path in matrix static int findMaxPath(int [,]mat) { // To find max val // in first row int res = -1; for (int i = 0; i < M; i++) res = max(res, mat[0, i]); for (int i = 1; i < N; i++) { res = -1; for (int j = 0; j < M; j++) { // When all paths are possible if (j > 0 && j < M - 1) mat[i, j] += max(mat[i - 1, j], max(mat[i - 1, j - 1], mat[i - 1, j + 1])); // When diagonal right // is not possible else if (j > 0) mat[i, j] += max(mat[i - 1, j], mat[i - 1, j - 1]); // When diagonal left // is not possible else if (j < M - 1) mat[i, j] += max(mat[i - 1, j], mat[i - 1, j + 1]); // Store max path sum res = max(mat[i, j], res); } } return res; } // Driver code static public void Main (String[] args) { int[,] mat = {{10, 10, 2, 0, 20, 4}, {1, 0, 0, 30, 2, 5}, {0, 10, 4, 0, 2, 0}, {1, 0, 2, 20, 0, 4}}; Console.WriteLine(findMaxPath(mat)); }}// This code is contributed // by Arnab Kundu |
PHP
<?php// PHP program for finding max// path in matrix $N = 4; $M = 6; // To calculate max path in matrix function findMaxPath($mat) { global $N; global $M; for ($i = 1; $i < $N; $i++) { for ( $j = 0; $j < $M; $j++) { // When all paths are possible if ($j > 0 && $j < ($M - 1)) $mat[$i][$j] += max($mat[$i - 1][$j], max($mat[$i - 1][$j - 1], $mat[$i - 1][$j + 1])); // When diagonal right is // not possible else if ($j > 0) $mat[$i][$j] += max($mat[$i - 1][$j], $mat[$i - 1][$j - 1]); // When diagonal left is // not possible else if ($j < ($M - 1)) $mat[$i][$j] += max($mat[$i - 1][$j], $mat[$i - 1][$j + 1]); // Store max path sum } } $res = 0; for ($j = 0; $j < $M; $j++) $res = max($mat[$N - 1][$j], $res); return $res; } // Driver Code$mat1 = array( array( 10, 10, 2, 0, 20, 4 ), array( 1, 0, 0, 30, 2, 5 ), array( 0, 10, 4, 0, 2, 0 ), array( 1, 0, 2, 20, 0, 4 )); echo findMaxPath($mat1),"\n"; // This code is contributed by Sach_Code?> |
Javascript
<script>// Javascript program for finding max path in matrixlet N = 4, M = 6; // Function to calculate max path in matrixfunction findMaxPath(mat){ // To find max val in first row let res = -1; for(let i = 0; i < M; i++) res = Math.max(res, mat[0][i]); for(let i = 1; i < N; i++) { res = -1; for(let j = 0; j < M; j++) { // When all paths are possible if (j > 0 && j < M - 1) mat[i][j] += Math.max(mat[i - 1][j], Math.max(mat[i - 1][j - 1], mat[i - 1][j + 1])); // When diagonal right is not possible else if (j > 0) mat[i][j] += Math.max(mat[i - 1][j], mat[i - 1][j - 1]); // When diagonal left is not possible else if (j < M - 1) mat[i][j] += Math.max(mat[i - 1][j], mat[i - 1][j + 1]); // Store max path sum res = Math.max(mat[i][j], res); } } return res;}// Driver Codelet mat = [ [ 10, 10, 2, 0, 20, 4 ], [ 1, 0, 0, 30, 2, 5 ], [ 0, 10, 4, 0, 2, 0 ], [ 1, 0, 2, 20, 0, 4 ] ];document.write(findMaxPath(mat)); // This code is contributed by sravan kumar</script> |
Output
74
Time Complexity: O(N*M), where N and M are the dimensions of the matrix
Space Complexity: O(1), since no extra space has been taken.
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