Maximum weight path ending at any element of last row in a matrix

2.1

Given a matrix of integers where every element represents weight of the cell. Find the path having the maximum weight in matrix [N X N]. Path Traversal Rules are:

  • It should begin from top left element.
  • The path can end at any element of last row.
  • We can move to following two cells from a cell (i, j).
    1. Down Move : (i+1, j)
    2. Diagonal Move : (i+1, j+1)

Examples:

Input : N = 5
        mat[5][5] = {{ 4, 2 ,3 ,4 ,1 },
                     { 2 , 9 ,1 ,10 ,5 },
                     {15, 1 ,3 , 0 ,20 },
                     {16 ,92, 41, 44 ,1},
                     {8, 142, 6, 4, 8} };
Output : 255
Path with max weight : 4 + 2 +15 + 92 + 142 = 255                 

The above problem can be recursively defined.

Let maxCost(i, j) be the cost maximum cost to
reach mat[i][j].  Since end point can be any point
in last row, we finally return maximum of all values
maxCost(N-1, j) where j varies from 0 to N-1.

If i == 0 and j == 0
   maxCost(0, 0) = mat[0][0]

// We can traverse through first column only by
// down move
Else if j = 0
  maxCost(i, 0) = maxCost(i-1, 0) + mat[i][0]

// In other cases, a cell mat[i][j] can be reached
// through previous two cells ma[i-1][j] and 
// mat[i-1][j-1]
Else
  maxCost(i, j) = mat[i][j] + max(maxCost(i-1, j),
                                maxCost(i-1, j-1)),

If we draw recursion tree of above recursive solution, we can observe overlapping subproblems. Since the problem has overlapping subproblems, we can solve it efficiently using Dynamic Programming. Below is Dynamic Programming based solution.

C++

// C++ program to find the path having the
// maximum weight in matrix
#include<bits/stdc++.h>
using namespace std;
const int MAX = 1000;

/* Function which return the maximum weight
   path sum */
int maxCost(int mat[][MAX], int N)
{
    // creat 2D matrix to store the sum of the path
    int dp[N][N];
    memset(dp, 0, sizeof(dp));

    dp[0][0] = mat[0][0];

    // Initialize first column of total weight
    // array (dp[i to N][0])
    for (int i=1; i<N; i++)
        dp[i][0] = mat[i][0] + dp[i-1][0];

    // Calculate rest paht sum of weight matrix
    for (int i=1; i<N; i++)
       for (int j=1; j<i+1&&j<N; j++)
          dp[i][j] = mat[i][j] +
                    max(dp[i-1][j-1], dp[i-1][j]);

    // find the max weight path sum to rech
    // the last row
    int result = 0;
    for (int i=0; i<N; i++)
        if (result < dp[N-1][i])
            result = dp[N-1][i];

    // return maximum weight path sum
    return result;
}

// Driver program
int main()
{
    int mat[MAX][MAX] = {  { 4, 1 ,5 ,6 , 1 },
        { 2 ,9 ,2 ,11 ,10 },
        { 15,1 ,3 ,15, 2 },
        { 16, 92, 41,4,3},
        { 8, 142, 6, 4, 8 }
    };
    int N = 5;
    cout << "Maximum Path Sum : "
        << maxCost(mat, N)<<endl;
    return 0;
}

Java

// JAVA Code for Maximum weight path ending at 
// any element of last row in a matrix
import java.util.*;

class GFG {
    
    /* Function which return the maximum weight
       path sum */
    public static int maxCost(int mat[][], int N)
    {
        // create 2D matrix to store the sum of 
        // the path
        int dp[][]=new int[N][N];
        
        dp[0][0] = mat[0][0];
     
        // Initialize first column of total 
        // weight array (dp[i to N][0])
        for (int i = 1; i < N; i++)
            dp[i][0] = mat[i][0] + dp[i-1][0];
     
        // Calculate rest path sum of weight matrix
        for (int i = 1; i < N; i++)
           for (int j = 1; j < i + 1 && j < N; j++)
              dp[i][j] = mat[i][j] +
                        Math.max(dp[i-1][j-1], 
                                 dp[i-1][j]);
     
        // find the max weight path sum to reach
        // the last row
        int result = 0;
        for (int i = 0; i < N; i++)
            if (result < dp[N-1][i])
                result = dp[N-1][i];
     
        // return maximum weight path sum
        return result;
    }
    
    /* Driver program to test above function */
    public static void main(String[] args) 
    {
        int mat[][] = {  { 4, 1 ,5 ,6 , 1 },
                { 2 ,9 ,2 ,11 ,10 },
                { 15,1 ,3 ,15, 2 },
                { 16, 92, 41,4,3},
                { 8, 142, 6, 4, 8 }
            };
            int N = 5;
           System.out.println("Maximum Path Sum : "+ 
                               maxCost(mat, N));
    }
}
// This code is contributed by Arnav Kr. Mandal.  


Output:

Maximum Path Sum : 255

Time complexity : O(N*N)

This article is contributed by Nishant_singh (pintu). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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