Maximum sum path in a matrix from top-left to bottom-right

• Difficulty Level : Easy
• Last Updated : 07 May, 2021

Given a matrix mat[][] of dimensions N * M, the task is to find the path from the top-left cell (0, 0) to the bottom-right cell (N – 1, M – 1) of the given matrix such that sum of the elements in the path is maximum. The only moves allowed from any cell (i, j) of the matrix are (i + 1, j) or (i, j + 1).

Examples:

Input: mat[][] = {{3, 7}, {9, 8}}
Output: 20
Explanation:
Path with maximum sum is 3 => 9 => 8 as 20.

Input: mat[][] = {{1, 2}, {3, 5}}
Output: 9
Explanation:
Path with maximum sum is 1 => 3 => 5 as 9

Approach 1 (Bottom-Up): The idea is to use Dynamic Programming to solve this problem. The key observation is that the cell grid[i][j] can only be reached from grid[i – 1][j] or grid[i][j – 1]. Therefore, the recurrence relation for this problem is given by the equation:

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

1. Initialize an auxiliary matrix sum[][] of dimensions N * M.
2. Iterate over the matrix elements and update each cell of the auxiliary matrix sum[][] using the above recurrence relation formed.
3. After completing the above steps, the value sum[N][M] will contain the maximum sum possible for a path from the top-left corner to the bottom-right corner of the given matrix. Print that sum.

Below is the implementation of the above approach:

C++

 // C++ program for the above approach #include using namespace std; // Function to find the maximum sum// path in the gridint MaximumPath(vector >& grid){    // Dimensions of grid[][]    int N = grid.size();    int M = grid.size();     // Stores maximum sum at each cell    // sum[i][j] from cell sum    vector > sum;    sum.resize(N + 1,               vector(M + 1));     // Iterate to compute the maximum    // sum path in the grid    for (int i = 1; i <= N; i++) {         for (int j = 1; j <= M; j++) {             // Update the maximum path sum            sum[i][j] = max(sum[i - 1][j],                            sum[i][j - 1])                        + grid[i - 1][j - 1];        }    }     // Return the maximum sum    return sum[N][M];} // Driver Codeint main(){    vector > grid        = { { 1, 2 }, { 3, 5 } };     cout << MaximumPath(grid);     return 0;}

Java

 // Java program for//the above approachimport java.util.*;class GFG{ // Function to find the maximum sum// path in the gridstatic int MaximumPath(int [][]grid){    // Dimensions of grid[][]    int N = grid.length;    int M = grid.length;     // Stores maximum sum at each cell    // sum[i][j] from cell sum    int [][]sum = new int[N + 1][M + 1];     // Iterate to compute the maximum    // sum path in the grid    for (int i = 1; i <= N; i++)    {        for (int j = 1; j <= M; j++)        {            // Update the maximum path sum            sum[i][j] = Math.max(sum[i - 1][j],                                 sum[i][j - 1]) +                                 grid[i - 1][j - 1];        }    }     // Return the maximum sum    return sum[N][M];} // Driver Codepublic static void main(String[] args){  int [][]grid = {{1, 2}, {3, 5}};  System.out.print(MaximumPath(grid));}} // This code is contributed by shikhasingrajput

Python3

 # Python3 program for the above approach # Function to find the maximum sum# path in the griddef MaximumPath(grid):     # Dimensions of grid[][]    N = len(grid)    M = len(grid)     # Stores maximum sum at each cell    # sum[i][j] from cell sum    sum = [[0 for i in range(M + 1)]              for i in range(N + 1)]     # Iterate to compute the maximum    # sum path in the grid    for i in range(1, N + 1):        for j in range(1, M + 1):             # Update the maximum path sum            sum[i][j] = (max(sum[i - 1][j],                             sum[i][j - 1]) +                        grid[i - 1][j - 1])     # Return the maximum sum    return sum[N][M] # Driver Codeif __name__ == '__main__':     grid = [ [ 1, 2 ], [ 3, 5 ] ]     print(MaximumPath(grid)) # This code is contributed by mohit kumar 29

C#

 // C# program for the above approachusing System; class GFG{ // Function to find the maximum sum// path in the gridstatic int MaximumPath(int [,]grid){         // Dimensions of grid[,]    int N = grid.GetLength(0);    int M = grid.GetLength(1);     // Stores maximum sum at each cell    // sum[i,j] from cell sum[0,0]    int [,]sum = new int[N + 1, M + 1];     // Iterate to compute the maximum    // sum path in the grid    for(int i = 1; i <= N; i++)    {        for(int j = 1; j <= M; j++)        {            // Update the maximum path sum            sum[i, j] = Math.Max(sum[i - 1, j],                                 sum[i, j - 1]) +                                grid[i - 1, j - 1];        }    }     // Return the maximum sum    return sum[N, M];} // Driver Codepublic static void Main(String[] args){    int [,]grid = { { 1, 2 }, { 3, 5 } };         Console.Write(MaximumPath(grid));}} // This code is contributed by Amit Katiyar

Javascript


Output:
9

Time Complexity: O(N * M)
Auxiliary Space: O(N * M)

Approach 2 (Top-Down): We will solve the problem recursively in a top-down manner. We formulate the recurrence based on the two ways of reaching the cell grid[i][j] as follows:

1. If we move one step towards the right, from the cell grid[i][j-1] or,
2. If we move one step downwards, from the cell grid[i-1][j].

dp[i][j] = max(dp[i][j-1], dp[i-1][j]) + grid[i][j]

Thus, we need to select the step (between the above two) which gives us the maximum value. Also, we need to add the value present in the cell we step into, i.e. grid[i][j].  As this problem has got the property of overlapping subproblems, we can store the result (memoize) of the subproblems in a 2D matrix (let’s call it dp), in order to avoid repeated computations of the same subproblems. Initially, we set all the cells in the dp table with the value -1, and every time we find an answer to a subproblem we overwrite its result in the respective cell in the dp table. Thus, before computing any subproblem we once check that – if that particular subproblem has been previously solved or not, if it has been solved (i.e. its corresponding cell the dp matrix is not -1) we simple return that value, else we solve it and store the result in the dp table.

Below is the implementation of the above approach:

C++

 #include using namespace std; vector > dp; // Function to find the maximum sum path in the gridint MaximumPathUtil(int i, int j, vector >& grid){    // Base condition      if (i == 0 || j == 0)        return 0;       // If current subproblem is already computed,      // we simply return its result from the dp table      if (dp[i][j] != -1)        return dp[i][j];       // Computing the current subproblem and      // store the result in the dp table for future use      return dp[i][j] = max(MaximumPathUtil(i, j-1, grid), MaximumPathUtil(i - 1, j, grid)) +                        grid[i-1][j-1];} int MaximumPath(vector >& grid){    // Dimensions of grid[][]    int n = grid.size();    int m = grid.size();       // dp table to memoize the subproblem results      dp.resize(n+1, vector (m+1, -1));       // dp[n][m] gives the max. path sum      // from grid to grid[n-1][m-1]      return MaximumPathUtil(n, m, grid);} // Driver Codeint main(){    vector > grid = {{3, 7, 9, 2, 7},                                 {9, 8, 3, 5, 5},                                 {1, 7, 9, 8, 6},                                 {3, 8, 6, 4, 9},                                 {6, 3, 9, 7, 8}};     cout << MaximumPath(grid);    return 0;} // This code is contributed by tridib_samanta

Output:

67

Time Complexity: O(n*m)
Space Complexity: O(n*m)

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