Maximum sum path in a Matrix

Given an n*m matrix, the task is to find the maximum sum of elements of cell starting from the cell (0, 0) to cell (n-1, m-1).
However, the allowed moves are right, downwards or diagonally right, i.e, from location (i, j) next move can be (i+1, j), or, (i, j+1), or (i+1, j+1). Find the maximum sum of elements satisfying the allowed moves.

Examples:

Input:
mat[][] = {{100, -350, -200},
           {-100, -300, 700}}
Output: 500
Explanation: 
Path followed is 100 -> -300 -> 700

Input:
mat[][] = {{500, 100, 230},
           {1000, 300, 100},
           {200, 1000, 200}}
Output:
3000
Explanation:
Path followed is 500 -> 1000 -> 300 -> 1000 -> 200

Approach: Dynamic programming is used to solve the above problem in recursive way.



  1. Allot a position in the beginning of the matrix at (0, 0).
  2. Check each next allowed positions from the current position and select the path with maximum sum.
  3. Take care of the boundaries of the matrix, i.e, if the position reaches the last row or last column then the only possible choice will be right or downwards respectively.
  4. Use a map to store track of all the visiting positions and before visiting any (i, j), check whether or not the position is visited before.
  5. Update the maximum of all possible paths returned by each recursive calls made.
  6. Go till the position reaches the destination cell, i.e, (n-1.m-1).

Below is the implementation of the above approach:

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#include <bits/stdc++.h>
using namespace std;
#define N 100
  
// No of rows and columns
int n, m;
  
// Declaring the matrix of maximum
// 100 rows and 100 columns
int a[N][N];
  
// Variable visited is used to keep
// track of all the visited positions
// Variable dp is used to store
// maximum sum till current position
vector<vector<int> > dp(N, vector<int>(N)),
    visited(N, vector<int>(N));
  
// For storing current sum
int current_sum = 0;
  
// For continuous update of
// maximum sum required
int total_sum = 0;
  
// Function to Input the matrix of size n*m
void inputMatrix()
{
    n = 3;
    m = 3;
    a[0][0] = 500;
    a[0][1] = 100;
    a[0][2] = 230;
    a[1][0] = 1000;
    a[1][1] = 300;
    a[1][2] = 100;
    a[2][0] = 200;
    a[2][1] = 1000;
    a[2][2] = 200;
}
  
// Function to calculate maximum sum of path
int maximum_sum_path(int i, int j)
{
    // Checking boundary condition
    if (i == n - 1 && j == m - 1)
        return a[i][j];
  
    // Checking whether or not (i, j) is visited
    if (visited[i][j])
        return dp[i][j];
  
    // Marking (i, j) is visited
    visited[i][j] = 1;
  
    // Updating the maximum sum till
    // the current position in the dp
    int& total_sum = dp[i][j];
  
    // Checking whether the position hasn't
    // visited the last row or the last column.
    // Making recursive call for all the possible
    // moves from the current cell and then adding the
    // maximum returned by the calls and updating it.
    if (i < n - 1 & j < m - 1) {
        int current_sum = max(maximum_sum_path(i, j + 1),
                              max(
                                  maximum_sum_path(i + 1, j + 1),
                                  maximum_sum_path(i + 1, j)));
        total_sum = a[i][j] + current_sum;
    }
  
    // Checking whether
    // position has reached last row
    else if (i == n - 1)
        total_sum = a[i][j]
                    + maximum_sum_path(i, j + 1);
  
    // If the position is in the last column
    else
        total_sum = a[i][j]
                    + maximum_sum_path(i + 1, j);
  
    // Returning the updated maximum value
    return total_sum;
}
  
// Driver Code
int main()
{
    inputMatrix();
  
    // Calling the implemented function
    int maximum_sum = maximum_sum_path(0, 0);
  
    cout << maximum_sum;
    return 0;
}

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Output:


3000







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