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
Unique Paths Solution
✅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 = [[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]
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