[LeetCode] 63. Unique Paths II

Follow up for "Unique Paths":
Now consider if some obstacles are added to the grids. How many unique paths would there be?
An obstacle and empty space is marked as 1 and 0 respectively in the grid.
For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
The total number of unique paths is 2.
Note: m and n will be at most 100.

Thought process:
Sub-problem, initialization, and answer are the same as "62. Unique Paths".
Formula: f[i][j] = f[i - 1][j] + f[i][j - 1] if grid[i][j] == 0; 0 otherwise.

Solution:
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int height = obstacleGrid.length;
        if (height == 0) {
            return 0;
        }
        int width = obstacleGrid[0].length;
        if (width == 0) {
            return 0;
        }
        
        int[][] f = new int[height][width];
        
        for (int i = 0; i < height; i++) {
            if (obstacleGrid[i][0] == 1) {
                break;
            }
            f[i][0] = 1;
        }
        for (int i = 0; i < width; i++) {
            if (obstacleGrid[0][i] == 1) {
                break;
            }
            f[0][i] = 1;
        }
        
        for (int i = 1; i < height; i++) {
            for (int j = 1; j < width; j++) {
                if (obstacleGrid[i][j] == 1) {
                    f[i][j] = 0;
                } else {
                    f[i][j] = f[i - 1][j] + f[i][j - 1];
                }
            }
        }
        
        return f[height - 1][width - 1];
    }
}

Time complexity: O(mn).

Count and also print all unique paths:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        if (m == 0) {
            return 0;
        }
        int n = obstacleGrid[0].length;
        
        List<String> paths = new ArrayList<>();
        int result = uniquePathsWithObstacles(obstacleGrid, m - 1, n - 1, "", paths);
        System.out.println(paths);
        return result;
    }
    
    private int uniquePathsWithObstacles(int[][] obstacleGrid, int row, int col, String path, List<String> paths) {
        if (obstacleGrid[row][col] == 1) {
            return 0;
        }
        if (row == 0 && col == 0) {
            paths.add(path);
            return 1;
        }
        if (row == 0) {
            return uniquePathsWithObstacles(obstacleGrid, row, col - 1, "r" + path, paths);
        }
        if (col == 0) {
            return uniquePathsWithObstacles(obstacleGrid, row - 1, col, "b" + path, paths);
        }
        
        return uniquePathsWithObstacles(obstacleGrid, row - 1, col, "b" + path, paths) +
               uniquePathsWithObstacles(obstacleGrid, row, col - 1, "r" + path, paths);
    }
}

Time complexity: O(2^(m + n)).
Space complexity: O(2^(m + n)).
Location: San Jose, CA, USA

Comments

Post a Comment

Popular posts from this blog

[LeetCode] 714. Best Time to Buy and Sell Stock with Transaction Fee

Your are given an array of integers  prices , for which the  i -th element is the price of a given stock on day  i ; and a non-negative integer  fee  representing a transaction fee. You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.) Return the maximum profit you can make. Example 1: Input: prices = [1, 3, 2, 8, 4, 9], fee = 2 Output: 8 Explanation: The maximum profit can be achieved by: Buying at prices[0] = 1 Selling at prices[3] = 8 Buying at prices[4] = 4 Selling at prices[5] = 9 The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8. Note: 0 < prices.length <= 50000 . 0 < prices[i] < 50000 . 0 <= fee < 50000 . Thought process: Each day there are two options, buy or sell. Use two variables buy and sell to keep track of the max profit. Solutio...
Read more

[LeetCode] 269. Alien Dictionary

There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of  non-empty  words from the dictionary, where  words are sorted lexicographically by the rules of this new language . Derive the order of letters in this language. Example 1: Given the following words in dictionary, [ "wrt", "wrf", "er", "ett", "rftt" ] The correct order is:  "wertf" . Example 2: Given the following words in dictionary, [ "z", "x" ] The correct order is:  "zx" . Example 3: Given the following words in dictionary, [ "z", "x", "z" ] The order is invalid, so return  "" . Note: You may assume all letters are in lowercase. You may assume that if a is a prefix of b, then a must appear before b in the given dictionary. If the order is invalid, return an empty string...
Read more

[LeetCode] 631. Design Excel Sum Formula

Your task is to design the basic function of Excel and implement the function of sum formula. Specifically, you need to implement the following functions: Excel(int H, char W):  This is the constructor. The inputs represents the height and width of the Excel form.  H  is a positive integer, range from 1 to 26. It represents the height.  W  is a character range from 'A' to 'Z'. It represents that the width is the number of characters from 'A' to  W . The Excel form content is represented by a height * width 2D integer array  C , it should be initialized to zero. You should assume that the first row of  C  starts from 1, and the first column of  C  starts from 'A'. void Set(int row, char column, int val):  Change the value at  C(row, column)  to be val. int Get(int row, char column):  Return the value at  C(row, column) . int Sum(int row, char column, List of Strings : numbers):  This function...
Read more