0000 - 0099
0063 - Unique Paths II (Medium)
Problem Link
https://leetcode.com/problems/unique-paths-ii/
Problem Statement
You are given an m x n integer array grid. There is a robot initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.
An obstacle and space are marked as 1 or 0 respectively in grid. A path that the robot takes cannot include any square that is an obstacle.
Return the number of possible unique paths that the robot can take to reach the bottom-right corner.
The testcases are generated so that the answer will be less than or equal to 2 * 1e9.
Example 1:
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 -> RightExample 2:
Input: obstacleGrid = [[0,1],[0,0]]
Output: 1Constraints:
m == obstacleGrid.lengthn == obstacleGrid[i].length1 <= m, n <= 100obstacleGrid[i][j]is0or1.
Approach 1: Dynamic Programming
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
// DP - O(N * M)
int n = obstacleGrid.size(), m = obstacleGrid[0].size();
// dp[i][j]: the number of possible unique paths to reach grid[i][j]
vector<vector<int>> dp(n, vector<int>(m));
// base case - there is only one way to reach grid[0][0]
dp[0][0] = 1;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (obstacleGrid[i][j]) {
// if there is an obstacle at grid[i][j],
// then we couldn't reach grid[i][j],
// hence setting dp[i][j] to 0
dp[i][j] = 0;
} else {
// otherwise, we can either reach grid[i][j] from the left cell dp[i][j - 1]
if (j) dp[i][j] += dp[i][j - 1];
// or from the top cell dp[i - 1][j]
if (i) dp[i][j] += dp[i - 1][j];
}
}
}
// the number of possible unique paths
// that the robot can take to reach the bottom-right corner
return dp[n - 1][m - 1];
}
};