# [leetcode] 62. Unique Paths _ Algorithm Problem Solve for python

## 1. Problem

62. 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). 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

## 2. Solution

I solved this problem like this.

### 2.1. DP

If we use dp, we can solve this problem.

class Solution:
def uniquePaths(self, m: int, n: int) -> int:
dp = [[0 for x in range(n)] for _ in range(m)]
dp = 1
for r in range(m):
for c in range(n):
if r == 0 and c == 0:
continue
up = dp[r-1][c] if r - 1 >= 0 else 0
left = dp[r][c-1] if c - 1 >= 0 else 0
dp[r][c] = up + left

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


### 2.2. Combination

The way to corner is always same right and down count. So we can get answer combination.

class Solution:
def uniquePaths(self, m: int, n: int) -> int:
# nCk
k = n - 1
n = m + n - 2
ans = 1
k = min(k, n - k)
for x in range(n, n-k, -1):
ans *= x
for x in range(1, k+1):
ans /= x

return int(ans)