64. Minimum Path Sum

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right, which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

Example 1:

Input: grid = [[1,3,1],[1,5,1],[4,2,1]]
Output: 7
Explanation: Because the path 1 → 3 → 1 → 1 → 1 minimizes the sum.

Example 2:

Input: grid = [[1,2,3],[4,5,6]]
Output: 12

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 200
• 0 <= grid[i][j] <= 200

Solution:

class Solution {
    public int minPathSum(int[][] grid) {
        int rows = grid.length;
        int cols = grid[0].length;

        // Create a 2D array to store the minimum path sums.
        int[][] dp = new int[rows][cols];

        // Initialize the starting point with the first cell's value.
        dp[0][0] = grid[0][0];

        // Fill in the first row (can only come from the left).
        for (int col = 1; col < cols; col++) {
            dp[0][col] = dp[0][col - 1] + grid[0][col];
        }

        // Fill in the first column (can only come from above).
        for (int row = 1; row < rows; row++) {
            dp[row][0] = dp[row - 1][0] + grid[row][0];
        }

        // Fill in the rest of the dp array.
        for (int row = 1; row < rows; row++) {
            for (int col = 1; col < cols; col++) {
                dp[row][col] = Math.min(dp[row - 1][col], dp[row][col - 1]) + grid[row][col];
            }
        }

        // Return the value at the bottom-right corner, which is the minimum path sum.
        return dp[rows - 1][cols - 1];
    }
}