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Maximum path sum in matrix

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.

C++

// CPP prorgam for finding max path in matrix
#include <bits/stdc++.h>
#define N 4
#define M 6
using namespace std;
  
// To calculate max path in matrix
int 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 findMaxPath
int 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 prorgam for finding max path in matrix
  
import 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

Python 3

# Python 3 prorgam for finding max path in matrix
# To calculate max path in matrix
  
def findMaxPath(mat):
  
    # To find max val in first row
    res = -1
    for i in range(M):
        res = max(res, mat[0][i])
   
    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
            elif (j > 0):
                mat[i][j] += max(mat[i - 1][j],
                                 mat[i - 1][j - 1])
   
            # When diagonal left is not possible
            elif (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 findMaxPath
N=4
M=6
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 ]])
                
print(findMaxPath(mat))
  
# This code is contributed by Azkia Anam.

C#

// C# prorgam for finding
// max path in matrix
using 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 prorgam 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)," "
  
// This code is contributed by Sach_Code
?>


Output:

74

Time Complexity: O(N*M)

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This article is attributed to GeeksforGeeks.org

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