062-unique-paths

Question

https://leetcode.com/problems/unique-paths/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:

Thought Process

1. DP
   1. For an one dimensional array, there is only one way to reach there, straight right or straight down
   2. Now for two dimensional array, the ways to get there will be way[left] and way[top]
   3. Time complexity O(mn)
   4. Space complexity O(mn)

2. DP - Optimized Space 1. W

   e can save the space by using one dimensional array

   1. We can choose the smaller one to save the space, for now we just stick with columns n
   2. Time complexity O(mn)
   3. Space complexity O(n)

3. Math formula

   1. There are total of m + n - 2 steps, m -1 down and n - 1 right
   2. This question is essentially asking how many ways we can get by performing m - 1 down steps out of total m + n - 2 moves
   3. The answer is simply Combination(n, k) = n! / (n - k)! k!
   4. Time complexity O(n)
   5. Space complexity O(1)

Solution

class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        for (int c = 0; c < n; c++) dp[0][c] = 1;
        for (int r = 0; r < m; r++) dp[r][0] = 1;
        for (int r = 1; r < m; r++) {
            for (int c = 1; c < n; c++) {
                dp[r][c] = dp[r - 1][c] + dp[r][c - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
}

class Solution {
    public int uniquePaths(int m, int n) {
        int[] dp = new int[n];
        dp[0] = 1;
        for (int r = 0; r < m; r++) {
            for (int c = 1; c < n; c++) {
                dp[c] += dp[c- 1];
            }
        }
        return dp[n - 1];
    }
}

class Solution {
    public int uniquePaths(int m, int n) {
        int num = m + n - 2, den = Math.min(m, n) - 1;
        long res = 1;
        for (int i = 0; i < den; i++) {
            res = res * (num - i) / (i + 1);
        }
        return (int) res;
    }
}

Additional