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[8/1] 62. Unique Paths

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  • 학부유학생 작성
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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:

robot_maze.png
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
Accepted
1,048,365
Submissions
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관련자료

댓글 2

학부유학생님의 댓글

  • 학부유학생
  • 작성일
Runtime: 34 ms, faster than 90.49% of Python3 online submissions for Unique Paths.
Memory Usage: 13.9 MB, less than 73.91% of Python3 online submissions for Unique Paths.
class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        prev_path = [1]*n
        next_path = []
        
        i = 1
        while i < m:
            for j in range(n):
                from_top = prev_path[j]
                from_left = next_path[j-1] if j > 0 else 0
                next_path.append(from_top + from_left)
            prev_path = next_path
            next_path = []
            i += 1
        
        return prev_path[n-1]

mingki님의 댓글

  • mingki
  • 작성일
C++
Runtime: 0 ms, faster than 100.00% of C++ online submissions for Unique Paths.
Memory Usage: 6.4 MB, less than 53.20% of C++ online submissions for Unique Paths.
class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
        
        for (int i = 0; i < m; ++i) {
            dp[i][0] = 1;
        }
        for (int j = 0; j < n; ++j) {
            dp[0][j] = 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];
    }
};
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