Minimum Path Sum Leetcode Solution

Difficulty Level Medium
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Problem Statement

The Minimum Path Sum LeetCode Solution – “Minimum Path Sum” says that given a n x m grid consisting of non-negative integers and we need to find a path from top-left to bottom right, which minimizes the sum of all numbers along the path.

We can only move either down or right at any point in time.

Example:

Minimum Path Sum Leetcode SolutionPin

Input:  grid = [[1,3,1],[1,5,1],[4,2,1]]
Output:7

Explanation:

  • The optimal path that minimizes the sum from (0, 0) to (n – 1, m – 1) consists of the following cells.
  • (0, 0), (0, 1), (0, 2), (1, 2), (2, 2) and sum of all cells values in the minimum path is: 1 + 3 + 1 + 1 + 1 = 7
Input:  grid = [[1,2,3],[4,5,6]]
Output: 12

Explanation:

  • The Minimum Path Sum is 12 and the path corresponding to such value is: (0, 0), (0, 1), (0, 2), (1, 2).

Approach

Idea:

  1. The main idea to solve this problem is to use dynamic programming.
  2. To reach every cell (i, j), we can either come from the cell lying upwards (i – 1, j) or from the cell lying leftwards (i, j – 1).
  3. Iterate for each row and each column, and for every cell find the minimum of these two values:
    1. grid[i – 1][j] if (i – 1, j) exists otherwise INT_MAX.
    2. grid[i][j – 1] if (i, j – 1) exists otherwise INT_MAX.
  4. Hence, the minimum cost to reach the current cell will be grid[i][j] + min_value where min_value is the value obtained in the above step.
  5. Our answer will be grid[n – 1][m – 1].

Code

Minimum Path Sum Leetcode C++ Solution:

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[n - 1].size();
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                int curr = INT_MAX;
                if(i - 1 >= 0)
                    curr = min(curr, grid[i - 1][j]);
                if(j - 1 >= 0)
                    curr = min(curr, grid[i][j - 1]);
                if(curr == INT_MAX)
                    curr = 0;
                grid[i][j] += curr;
            }
        }
        return grid[n - 1][m - 1];
    }
};

Minimum Path Sum Leetcode Java Solution:

class Solution {
    public int minPathSum(int[][] grid) {
        int n = grid.length, m = grid[n - 1].length;
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                int curr = Integer.MAX_VALUE;
                if(i - 1 >= 0){
                    curr = Math.min(curr, grid[i - 1][j]);
                }
                if(j - 1 >= 0){
                    curr = Math.min(curr, grid[i][j - 1]);
                }
                if(curr == Integer.MAX_VALUE){
                    curr = 0;
                }
                grid[i][j] += curr;
            }
        }
        
        return grid[n - 1][m - 1];
    }
}

Complexity Analysis for Minimum Path Sum Leetcode Solution

Time Complexity

The time complexity of the above code is O(N*M) since we’re iterating for each cell, where N = number of rows and M = number of columns.

Space Complexity

The space complexity of the above code is O(1) since we are using constant extra space.

Reference: https://en.wikipedia.org/wiki/Shortest_path_problem

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