# Unique Paths LeetCode Solution | Easy Approach Share:

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`

Time: O(m*n)
Space: O(m*n)

### C++

``````class Solution {
public:
int uniquePaths(int m, int n) {
// dp[i][j] := unique paths from (0, 0) to (i, j)
vector<vector<int>> dp(m, vector<int>(n, 1));

for (int i = 1; i < m; ++i)
for (int j = 1; j < n; ++j)
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];

return dp[m - 1][n - 1];
}
};
``````

### Java

`````` class Solution {
public int uniquePaths(int m, int n) {
// dp[i][j] := unique paths from (0, 0) to (i, j)
int[][] dp = new int[m][n];
Arrays.stream(dp).forEach(row -> Arrays.fill(row, 1));

for (int i = 1; i < m; ++i)
for (int j = 1; j < n; ++j)
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];

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

``````

### Python

`````` class Solution:
def uniquePaths(self, m: int, n: int) -> int:
# dp[i][j] := unique paths from (0, 0) to (i, j)
dp = [ * n for _ in range(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[-1][-1]
``````

#### Watch Tutorial

Checkout more Solutions here