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Unique Paths II⚓︎

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Description⚓︎

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 * 10^9.

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:

  • Right -> Right -> Down -> Down

  • Down -> Down -> Right -> Right

Example 2:

  • 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.

Solution⚓︎

Way 1⚓︎

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class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
        if (obstacleGrid[0][0] == 1 || obstacleGrid[m - 1][n - 1] == 1) return 0;
        vector<vector<int>> dp(m, vector<int>(n, 0));
        for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) dp[i][0] = 1;
        for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) dp[0][j] = 1;

        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] == 0) dp[i][j] = dp[i][j - 1] + dp[i - 1][j];
                else continue;
            }
        }
        return dp[m - 1][n - 1];
    }
};
  • Time complexity: \(O(m\times n)\);
  • Space complexity: \(O(m\times n)\).

Way 2⚓︎

See reference.

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class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
        dp[0][1] = 1;
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (obstacleGrid[i - 1][j - 1] == 0)
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m][n];
    }
};

Way 3⚓︎

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class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
        vector<int> dp(n);
        dp[0] = (obstacleGrid[0][0] == 0) ? 1 : 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (obstacleGrid[i][j] == 1) {
                    dp[j] = 0;
                    continue;
                }
                if (j - 1 >= 0 && obstacleGrid[i][j - 1] == 0)
                    dp[j] += dp[j - 1];
            }
        }
        return dp.back();
    }
};
  • Time complexity: \(O(m\times n)\);
  • Space complexity: \(O(n)\).