Unique Paths II LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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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:

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:

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.

Unique Paths II Solutions

Time: O(m*n)
Space: O(m*n)

C++

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];
  }
};

Java

 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];
  }
}

Python

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]

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