0062. 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."""

# Math combination
class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        """一共需要移动 m+n−2 次，其中有 m−1 次向下移动，n−1 次向右移动。
        路径的总数等于从 m+n-2 次中选择 m-1 次向下移动的方案数，即组合数C((m+n-2)，(m-1))
        时间复杂度为O(n)，空间复杂度O(1). factorial=sum of the multiplication down to 1"""
        return round(math.factorial(m+n-2)/math.factorial(m-1)/math.factorial(n-1))


# comb() >P3.8
def uniquePaths(self, m: int, n: int) -> int:
        return math.comb(m + n - 2, n - 1)
        
        
# dp  T: O(N×M) S: O(N×M)
class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        d = [[1] * n for _ in range(m)]
        for col in range(1, m):
            for row in range(1, n):
                d[col][row] = d[col - 1][row] + d[col][row - 1]
        return d[m - 1][n - 1]


# recursion Time Limit Exceeded!
class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        if m == 1 or n == 1:
            return 1      
        return self.uniquePaths(m - 1, n) + self.uniquePaths(m, n - 1)