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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0063_unique_paths_ii;

// #Medium #Array #Dynamic_Programming #Matrix #Dynamic_Programming_I_Day_15
// #2023_08_11_Time_0_ms_(100.00%)_Space_40.6_MB_(73.18%)

/**
 * 63 - Unique Paths II\.
 *
 * Medium
 *
 * A robot is located at the top-left corner of a `m x n` grid (marked 'Start' in the diagram below).
 *
 * The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
 *
 * Now consider if some obstacles are added to the grids. How many unique paths would there be?
 *
 * An obstacle and space is marked as `1` and `0` respectively in the grid.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/11/04/robot1.jpg)
 *
 * **Input:** obstacleGrid = \[\[0,0,0],[0,1,0],[0,0,0]]
 *
 * **Output:** 2
 *
 * **Explanation:** There is one obstacle in the middle of the 3x3 grid above. There are two ways to reach the bottom-right corner: 1. Right -> Right -> Down -> Down 2. Down -> Down -> Right -> Right 
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/11/04/robot2.jpg)
 *
 * **Input:** obstacleGrid = \[\[0,1],[0,0]]
 *
 * **Output:** 1 
 *
 * **Constraints:**
 *
 * *   `m == obstacleGrid.length`
 * *   `n == obstacleGrid[i].length`
 * *   `1 <= m, n <= 100`
 * *   `obstacleGrid[i][j]` is `0` or `1`.
**/
public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        // if start point has obstacle, there's no path
        if (obstacleGrid[0][0] == 1) {
            return 0;
        }
        obstacleGrid[0][0] = 1;
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        for (int i = 1; i < m; i++) {
            if (obstacleGrid[i][0] == 1) {
                obstacleGrid[i][0] = 0;
            } else {
                obstacleGrid[i][0] = obstacleGrid[i - 1][0];
            }
        }
        for (int j = 1; j < n; j++) {
            if (obstacleGrid[0][j] == 1) {
                obstacleGrid[0][j] = 0;
            } else {
                obstacleGrid[0][j] = obstacleGrid[0][j - 1];
            }
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] == 1) {
                    obstacleGrid[i][j] = 0;
                } else {
                    obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
                }
            }
        }
        return obstacleGrid[m - 1][n - 1];
    }
}