README_EN.md

62. Unique Paths

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Description

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there?

 

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

Example 3:

Input: m = 7, n = 3
Output: 28

Example 4:

Input: m = 3, n = 3
Output: 6

 

Constraints:

• 1 <= m, n <= 100
• It's guaranteed that the answer will be less than or equal to 2 * 109.

Solutions

Dynamic programming.

Python3

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        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]

Java

class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        for (int i = 0; i < m; ++i) {
            Arrays.fill(dp[i], 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];
    }
}

TypeScript

function uniquePaths(m: number, n: number): number {
    let dp = Array.from({length: m}, v => new Array(n).fill(1));
    for (let i = 1; i < m; ++i) {
        for (let j = 1; j < n; ++j) {
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
        }
    }
    return dp[m-1][n-1];
};

C++

class Solution {
public:
    int uniquePaths(int m, int n) {
        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];
    }
};

Go

func uniquePaths(m int, n int) int {
	dp := make([][]int, m)
	for i := 0; i < m; i++ {
		dp[i] = make([]int, n)
	}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if i == 0 || j == 0 {
				dp[i][j] = 1
			} else {
				dp[i][j] = dp[i-1][j] + dp[i][j-1]
			}
		}
	}
	return dp[m-1][n-1]
}

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