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LeetCode Solutions

63. Unique Paths II

Time: $O(mn)$

Space: $O(mn)$

			


class Solution {
 public:
  int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
    const int m = obstacleGrid.size();
    const int n = obstacleGrid[0].size();
    // dp[i][j] := unique paths from (0, 0) to (i - 1, j - 1)
    vector<vector<long>> dp(m + 1, vector<long>(n + 1, 0));
    dp[0][1] = 1;  // Can also set dp[1][0] = 1

    for (int i = 1; i <= m; ++i)
      for (int j = 1; j <= n; ++j)
        if (!obstacleGrid[i - 1][j - 1])
          dp[i][j] = dp[i - 1][j] + dp[i][j - 1];

    return dp[m][n];
  }
};




			


class Solution {
  public int uniquePathsWithObstacles(int[][] obstacleGrid) {
    final int m = obstacleGrid.length;
    final int n = obstacleGrid[0].length;
    // dp[i][j] := unique paths from (0, 0) to (i - 1, j - 1)
    long[][] dp = new long[m + 1][n + 1];
    dp[0][1] = 1; // Can also set dp[1][0] = 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 (int) dp[m][n];
  }
}




			


class Solution:
  def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
    m = len(obstacleGrid)
    n = len(obstacleGrid[0])
    # dp[i][j] := unique paths from (0, 0) to (i - 1, j - 1)
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    dp[0][1] = 1  # Can also set dp[1][0] = 1

    for i in range(1, m + 1):
      for j in range(1, n + 1):
        if obstacleGrid[i - 1][j - 1] == 0:
          dp[i][j] = dp[i - 1][j] + dp[i][j - 1]

    return dp[m][n]