# 62. Unique Paths

## 29 Dec 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[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`

``````class Solution:
def uniquePaths(self, m: int, n: int) -> int:
# Initialize the first row and column to 1
# Since the combination for
# dp[i][0] = dp[i-1][0]
# dp[0][j] = dp[0][j-1]
# dp[i][j] = dp[i-1][j] + dp[i][j-1]

# df_table = [[1]*n] + [[1] + [0]*(n-1) ] * (m-1)
# Since dp[i][j] only depends on the calculated cells before,
# we can simply initialize the value of all cells to 1.
dp = [[1]*n] * m

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

return dp[-1][-1]``````
class Solution:
def uniquePaths(self, m: int, n: int) -> int:
dp = [[1]*n]*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[m-1][n-1]