• Time:O(mn)
• Space:O(mn)

## 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 = [[1] * 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]
``````

• Time:O(mn)
• Space:O(n)

## C++

``````class Solution {
public:
int uniquePaths(int m, int n) {
vector<int> dp(n, 1);

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

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

## JAVA

``````class Solution {
public int uniquePaths(int m, int n) {
int[] dp = new int[n];
Arrays.fill(dp, 1);

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

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

## Python

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

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

return dp[n - 1]
``````