LeetCode 2087: Minimum Cost Homecoming of a Robot in a Grid Solution

Master LeetCode problem 2087 (Minimum Cost Homecoming of a Robot in a Grid), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2087. Minimum Cost Homecoming of a Robot in a Grid

Medium

Array Greedy

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. We will create a 2D array to store the minimum cost of reaching each cell in the grid. We start from the startPos and iteratively calculate the minimum cost to reach each cell based on the costs of moving up, down, left, and right. Finally, we return the cost of reaching the homePos.

Here are the steps:

Initialize a 2D array dp[][] of size (m x n) to store the minimum cost to reach each cell.
Initialize dp[startRow][startCol] as 0 since the robot is already at this position.
Iterate over all cells in the grid and update the minimum cost to reach each cell using the costs of moving up, down, left, and right.
Return dp[homeRow][homeCol] as the minimum total cost for the robot to return home.

The time complexity of this solution is O(m x n) where m is the number of rows and n is the number of columns in the grid. The space complexity is also O(m x n) for the 2D array dp.

Solution Code

class Solution {
    public int minCost(int[] startPos, int[] homePos, int[] rowCosts, int[] colCosts) {
        int m = rowCosts.length;
        int n = colCosts.length;
        
        int startRow = startPos[0];
        int startCol = startPos[1];
        int homeRow = homePos[0];
        int homeCol = homePos[1];
        
        int[][] dp = new int[m][n];
        dp[startRow][startCol] = 0;
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i > 0) {
                    dp[i][j] = Math.min(dp[i][j] + rowCosts[i], dp[i - 1][j] + rowCosts[i]);
                }
                if (j > 0) {
                    dp[i][j] = Math.min(dp[i][j] + colCosts[j], dp[i][j - 1] + colCosts[j]);
                }
            }
        }
        
        return dp[homeRow][homeCol];
    }
}

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