**Physical Address**

304 North Cardinal St.

Dorchester Center, MA 02124

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 * 10`

.^{9}

**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`

```
public class Solution {
public int uniquePaths(int m, int n) {
if(m == 1 || n == 1)
return 1;
m--;
n--;
if(m < n) { // Swap, so that m is the bigger number
m = m + n;
n = m - n;
m = m - n;
}
long res = 1;
int j = 1;
for(int i = m+1; i <= m+n; i++, j++){ // Instead of taking factorial, keep on multiply & divide
res *= i;
res /= j;
}
return (int)res;
}
}
```

```
class Solution {
public:
int uniquePaths(int m, int n) {
int N = n+m-2; // total steps = n-1 + m-1
int r = min(n,m) - 1; // will iterate on the minimum for efficiency = (total) C (min(right, down))
double res = 1;
// compute nCr
for(int i=1; i<=r; ++i, N--){
res = res * (N) / i;
}
return (int)res;
}
};
```

```
class Solution:
# @return an integer
def uniquePaths(self, m, n):
aux = [[1 for x in range(n)] for x in range(m)]
for i in range(1, m):
for j in range(1, n):
aux[i][j] = aux[i][j-1]+aux[i-1][j]
return aux[-1][-1]
```

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