Unique Paths II LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Unique Paths II in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemUnique Paths II– LeetCode Problem

Unique Paths II– LeetCode Problem


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


  • m == obstacleGrid.length
  • n == obstacleGrid[i].length
  • 1 <= m, n <= 100
  • obstacleGrid[i][j] is 0 or 1.
Unique Paths II– LeetCode Solutions
Unique Paths II in C++:
class Solution {
  int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
    const int m = obstacleGrid.size();
    const int n = obstacleGrid[0].size();

    vector<long> dp(n);
    dp[0] = 1;

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (obstacleGrid[i][j])
          dp[j] = 0;
        else if (j > 0)
          dp[j] += dp[j - 1];

    return dp[n - 1];
Unique Paths II  in 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];
Unique Paths II in 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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