Find maximum path sum in a 2D matrix when exactly two left moves are allowed


Given a 2D matrix arr[][] of dimensions N * M where N is number of rows and M is number of columns.The task is to find maximum path sum in this matrix satisfying some condition which are as follows :

  1. We can only start with arr[i][M] where 0 <= i <= N.
  2. We end the path on the same side, such that we can take exactly 2 left turns.
  3. Example:
    2D matrix showing path

    Examples:



    Input : N = 3, M = 3
            arr[][] = {{1, 2, 3},
                       {3, 3, 1},
                       {4, 1, 6}}
    Output : 20
    Explanation : 
    If we follow this path then we get the sum 20.
    
    Input : N = 3, M = 3
            arr[][] = {{3, 7, 4},
                       {1, 9, 6},
                       {1, 7, 7}}
    Output : 34
    Explanation : 
    If we follow this path then we get the sum 34.
    

    The idea is to use dynamic programming and select an optimal structure in the matrix i.e.
    C shaped structure as shown in the below image.

    Steps are as follows :

    1. First we calculate suffix sum in each row and store it in another 2D matrix call it b[][]so that at every valid index we get the sum of the entire row starting from that index.

      b[i][j] = arr[i][j] + b[i][j + 1]

    2. Now we check each consecutive two rows and find the sum of their corresponding columns and simultaneously updating the maximum sum variable. Till now we have found both horizontal lines from that above structure.

      sum = max(sum, b[i][j] + b[i – 1][j])

      We need to find that vertical line connecting these horizontal lines i.e. column.

    3. After traversing each row, for each valid index we have two choices either we link this index to corresponding index of upper row i.e. add in previous column or start a new column.
      Whichever value is maximum we retain that value and we update the value at this index.

      b[i][j] = max(b[i][j], b[i – 1][j] + arr[i][j])

    Below is the implementation of the above approach:

    C++

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // C++ program to find maximum path sum
    // in a 2D matrix when exactly two
    // left moves are allowed
    #include <bits/stdc++.h>
    #define N 3
    #define M 3
    using namespace std;
      
    // Function to return the maximum path sum
    int findMaxSum(int arr[][M])
    {
        int sum = 0;
        int b[N][M];
          
        // Copy last column i.e. starting and 
        // ending columns in another array
        for (int i = 0; i < N; i++) {
            b[i][M - 1] = arr[i][M - 1];
        }
          
        // Calculate suffix sum in each row
        for (int i = 0; i < N; i++) {
            for (int j = M - 2; j >= 0; j--) {
                b[i][j] = arr[i][j] + b[i][j + 1];
            }
        }
          
        // Select the path we are going to follow
        for (int i = 1; i < N; i++) {
            for (int j = 0; j < M; j++) {
                sum = max(sum, b[i][j] + b[i - 1][j]);
                  
                b[i][j] = max(b[i][j], b[i - 1][j] + arr[i][j]);
            }
        }
          
        return sum;
    }
      
    // Driver Code
    int main()
    {
        int arr[N][M] = {{ 3, 7, 4 }, 
                         { 1, 9, 6 }, 
                         { 1, 7, 7 }};
                           
        cout << findMaxSum(arr) << endl;
      
        return 0;
    }

    chevron_right

    
    

    Java

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // Java program to find maximum path sum
    // in a 2D matrix when exactly two
    // left moves are allowed
    import java.io.*;
      
    class GFG 
    {
          
    static int N = 3;
    static int M = 3;
      
    // Function to return the maximum path sum
    static int findMaxSum(int arr[][])
    {
        int sum = 0;
        int [][]b = new int [N][M];
          
        // Copy last column i.e. starting and 
        // ending columns in another array
        for (int i = 0; i < N; i++) 
        {
            b[i][M - 1] = arr[i][M - 1];
        }
          
        // Calculate suffix sum in each row
        for (int i = 0; i < N; i++) 
        {
            for (int j = M - 2; j >= 0; j--) 
            {
                b[i][j] = arr[i][j] + b[i][j + 1];
            }
        }
          
        // Select the path we are going to follow
        for (int i = 1; i < N; i++) 
        {
            for (int j = 0; j < M; j++) 
            {
                sum = Math.max(sum, b[i][j] + b[i - 1][j]);
                  
                b[i][j] = Math.max(b[i][j], b[i - 1][j] + arr[i][j]);
            }
        }
          
        return sum;
    }
      
    // Driver Code
    public static void main (String[] args) 
    {
      
        int arr[][] = {{ 3, 7, 4 }, 
                        { 1, 9, 6 }, 
                        { 1, 7, 7 }};
                      
        System.out.println (findMaxSum(arr));
    }
    }
      
    // This code is contributed by ajit.

    chevron_right

    
    

    Python3

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    # Python3 program to find maximum path sum 
    # in a 2D matrix when exactly two 
    # left moves are allowed 
    import numpy as np
    N = 3
    M = 3
      
    # Function to return the maximum path sum 
    def findMaxSum(arr) : 
      
        sum = 0
        b = np.zeros((N, M)); 
          
        # Copy last column i.e. starting and 
        # ending columns in another array 
        for i in range(N) : 
            b[i][M - 1] = arr[i][M - 1]; 
          
        # Calculate suffix sum in each row 
        for i in range(N) :
            for j in range(M - 2, -1, -1) : 
                b[i][j] = arr[i][j] + b[i][j + 1]; 
          
        # Select the path we are going to follow 
        for i in range(1, N) :
            for j in range(M) :
                sum = max(sum, b[i][j] + b[i - 1][j]); 
                  
                b[i][j] = max(b[i][j], 
                              b[i - 1][j] + arr[i][j]);
                  
        return sum
      
    # Driver Code 
    if __name__ == "__main__"
      
        arr = [[ 3, 7, 4 ], 
               [ 1, 9, 6 ], 
               [ 1, 7, 7 ]]; 
                          
        print(findMaxSum(arr)); 
      
    # This code is contributed by AnkitRai01

    chevron_right

    
    

    C#

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // C# program to find maximum path sum
    // in a 2D matrix when exactly two
    // left moves are allowed
    using System;
          
    class GFG 
    {
          
    static int N = 3;
    static int M = 3;
      
    // Function to return the maximum path sum
    static int findMaxSum(int [,]arr)
    {
        int sum = 0;
        int [,]b = new int [N, M];
          
        // Copy last column i.e. starting and 
        // ending columns in another array
        for (int i = 0; i < N; i++) 
        {
            b[i, M - 1] = arr[i, M - 1];
        }
          
        // Calculate suffix sum in each row
        for (int i = 0; i < N; i++) 
        {
            for (int j = M - 2; j >= 0; j--) 
            {
                b[i, j] = arr[i, j] + b[i, j + 1];
            }
        }
          
        // Select the path we are going to follow
        for (int i = 1; i < N; i++) 
        {
            for (int j = 0; j < M; j++) 
            {
                sum = Math.Max(sum, b[i, j] + b[i - 1, j]);
                  
                b[i, j] = Math.Max(b[i, j], b[i - 1, j] + arr[i, j]);
            }
        }
          
        return sum;
    }
      
    // Driver Code
    public static void Main () 
    {
      
        int [,]arr = {{ 3, 7, 4 }, 
                        { 1, 9, 6 }, 
                        { 1, 7, 7 }};
                      
        Console.WriteLine(findMaxSum(arr));
    }
    }
      
    /* This code contributed by PrinciRaj1992 */

    chevron_right

    
    

    Output:

    34
    

    Time Complexity : O(N * M)



    My Personal Notes arrow_drop_up

    Check out this Author's contributed articles.

    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.

    Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.