Question

Formatted question description: https://leetcode.ca/all/62.html

62	Unique Paths

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?

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

@tag-array

Algorithm

Using Dynamic Programming to solve, you can maintain a two-dimensional array dp, where dp[i][j] represents the number of different moves to the current position, and then the state transition equation can be obtained as: dp[i][j ] = dp[i-1][j] + dp[i][j-1], here in order to save space, use a one-dimensional array dp, refresh line by line.

Code

public class Unique_Paths {

    class Solution {
        public int uniquePaths(int m, int n) {

            if (m <= 0 || n <= 0) {
                return 0;
            }

            int[][] dp = new int[m][n];

            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++) {

                    if (i == 0 || j == 0) {
                        dp[i][j] = 1;
                    } else { // i,j both not 1
                        dp[i][j] = dp[i-1][j] + dp[i][j-1];
                    }
                }
            }

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


}

############

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];
    }
}

// OJ: https://leetcode.com/problems/unique-paths/
// Time: O(MN)
// Space: O(N)
class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<int> dp(n + 1, 0);
        dp[n - 1] = 1;
        for (int i = m - 1; i >= 0; --i) {
            for (int j = n - 1; j >= 0; --j) dp[j] += dp[j + 1];
        }
        return dp[0];
    }
};

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]

############

class Solution(object):
  def uniquePaths(self, m, n):
    """
    :type m: int
    :type n: int
    :rtype: int
    """
    dp = [1] * n

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

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]
}

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];
}

impl Solution {
    pub fn unique_paths(m: i32, n: i32) -> i32 {
        let (m, n) = (m as usize, n as usize);
        let mut dp = vec![1; n];
        for i in 1..m {
            for j in 1..n {
                dp[j] += dp[j - 1];
            }
        }
        dp[n - 1]
    }
}

62 - Unique Paths

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