Leetcode - Unique Paths

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.

Solution #1: (Recursive, O(n^2), Time Limit Exceeded...)

// recursive
public class Solution {
    public int uniquePaths(int m, int n) {
        return backtrack(0, 0, m, n);
    }
    
    private int backtrack(int row, int col, int m, int n) {
        // to the target
        if(row == m - 1 && col == n - 1)
            return 1;
        // out of range
        if(row > m - 1 || col > n - 1)
            return 0;
            
        // move down + move right
        return backtrack(row + 1, col, m, n) + backtrack(row, col + 1, m, n);
    }
}

Solution #2: loop, O(n^2) space & time

// loop, similar as min path sum, O(n^2) time & space
public class Solution {
    public int uniquePaths(int m, int n) {
        int[][] res = new int[m][n];
        
        // init left
        for(int i = 0; i < m; i++) {
            res[i][0] = 1;
        }
        // init top
        for(int j = 0; j < n; j++) {
            res[0][j] = 1;
        }
        
        // add values
        for(int i = 1; i < m; i++) {
            for(int j = 1; j < n; j++) {
                res[i][j] = res[i - 1][j] + res[i][j - 1];
            }
        }
        
        return res[m - 1][n - 1];
    }
}

Solution #3: loop, O(n^2) time, O(n) space

// loop, similar as min path sum, O(n^2) time, O(n) space
public class Solution {
    public int uniquePaths(int m, int n) {
        int[] res = new int[n];
        
        // init array
        for(int j = 0; j < n; j++) {
            res[j] = 1;
        }
        
        // add values
        for(int i = 1; i < m; i++) {
            // reset the head to 1 (simulate the next row head)
            // similar to set all left most elements in a 2D array to 1
            res[0] = 1;
            for(int j = 1; j < n; j++) {
                res[j] = res[j - 1] + res[j];
            }
        }
        
        return res[n - 1];
    }
}

Many thanks to:
http://cuijing.org/interview/leetcode/summary-of-dynamic-programming-in-leetcode.html
&
http://yucoding.blogspot.com/2013/04/leetcode-question-116-unique-path-i.html