Q62 Unique Paths

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 -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right

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

class Solution:
    # DP
    # Time: O(n^2)
    def uniquePaths(self, m, n):
        """
        :type m: int
        :type n: int
        :rtype: int
        """
        dp = [[1 for col in range(m)] for row in range(n)]  # n行m列的数组
        if m == 1 or n == 1:
            return 1
        for i in range(1, n):
            for j in range(1, m):
                dp[i][j] = dp[i][j-1] + dp[i-1][j]
        return dp[i][j]

    # Math 
    # C(m+n-2, m-1) or C(m+n-2, n-1)
    # C(N, M) = N!/(M!*(N-M)!)
    # Time: O(n)
    def uniquePaths2(self, m, n):
        """
        :type m: int
        :type n: int
        :rtype: int
        """
        return self.calFactorial(m+n-2) // self.calFactorial(m-1) // self.calFactorial(n-1)

    def calFactorial(self, num):
        if num <= 1:
            return 1
        return num * self.calFactorial(num - 1)

m = 3
n = 7
print(Solution().uniquePaths(1, 1))  # 1
print(Solution().uniquePaths(m, n))  # 28
print(Solution().uniquePaths2(m, n)) # 28