Unique Paths LeetCode Solution | Easy Approach

Unique Paths: 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 Solution

✅Time: O(m*n)
✅Space: O(m*n)

C++

class Solution {
 public:
  int uniquePaths(int m, int n) {
    // dp[i][j] := unique paths from (0, 0) to (i, j)
    vector<vector<int>> dp(m, vector<int>(n, 1));

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

    return dp[m - 1][n - 1];
  }
};

Java

 class Solution {
  public int uniquePaths(int m, int n) {
    // dp[i][j] := unique paths from (0, 0) to (i, j)
    int[][] dp = new int[m][n];
    Arrays.stream(dp).forEach(row -> Arrays.fill(row, 1));

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

    return dp[m - 1][n - 1];
  }
}

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]

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