Unique Paths 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 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 Problem – Unique Paths– LeetCode Problem

Unique Paths– LeetCode Problem

Problem:

There is a robot on an m x n grid. The robot is 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.

Given the two integers m and n, return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The test cases are generated so that the answer will be less than or equal to 2 * 109.

Example 1:

Input: m = 3, n = 7
Output: 28

Example 2:

Input: m = 3, n = 2
Output: 3
Explanation: From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Down -> Down
2. Down -> Down -> Right
3. Down -> Right -> Down

Constraints:

• 1 <= m, n <= 100

Unique Paths– LeetCode Solutions

Unique Paths in C++:

class Solution {
 public:
  int uniquePaths(int m, int n) {
    vector<int> dp(n, 1);

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

    return dp[n - 1];
  }
};

Unique Paths in Java:

class Solution {
  public int uniquePaths(int m, int n) {
    int[] dp = new int[n];
    Arrays.fill(dp, 1);

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

    return dp[n - 1];
  }
}

Unique Paths in Python:

class Solution:
  def uniquePaths(self, m: int, n: int) -> int:
    # dp[i][j] := unique paths from (0, 0) to (i, j)
    dp = [[1] * n for _ in range(m)]

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

    return dp[-1][-1]