[LeetCode] 120. Triangle

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
For example, given the following triangle
[
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]
The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).
Note:
Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

Thought process:
  1. Sub-problem: for each number in the triangle, find its minimum path sum to bottom.
  2. Formula: f[i][j] = min(f[i + 1][j], f[i + 1][j + 1]) + triangle[i][j].
  3. Initialization: for every number in the bottom level, f[i][j] = triangle[i][j].
  4. Answer: f[0][0].
The bottom-up approach seems a little unnatural, because the question asks for the minimum path sum from top to bottom. We could do the top-down approach too. The formula will become f[i][j] = min(f[i - 1][j - 1], f[i - 1][j]) + triangle[i][j]. However, not every number has its corresponding triangle[i - 1][j] and triangle[i][j]. So this approach requires checking if the numbers exist. For the sake of convenience, we will use the bottom-up approach.

Solution 1:
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int rows = triangle.size();
        if (rows == 0) {
            return 0;
        }
        
        int[][] f = new int[rows][rows];
        
        for (int i = 0; i < rows; i++) {
            f[rows - 1][i] = triangle.get(rows - 1).get(i);
        }
        
        for (int i = rows - 2; i >= 0; i--) {
            for (int j = 0; j <= i; j++) {
                f[i][j] = Math.min(f[i + 1][j], f[i + 1][j + 1]) + triangle.get(i).get(j);
            }
        }
        
        return f[0][0];
    }
}

Time complexity: 
Say n = number of rows, the overall time complexity is O(n^2).
Space complexity: O(n^2).

 The space complexity can be reduced to O(n). After we have computed f[i][j] for level i, the results from level i + 1 will not be used again, so we can simplify the formula to f[j] = min(f[j], f[j + 1]) + triangle[i][j].

 Solution 2:
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int rows = triangle.size();
        if (rows == 0) {
            return 0;
        }
        
        int[]f = new int[rows];
        
        for (int i = 0; i < rows; i++) {
            f[i] = triangle.get(rows - 1).get(i);
        }
        
        for (int i = rows - 2; i >= 0; i--) {
            for (int j = 0; j <= i; j++) {
                f[j] = Math.min(f[j], f[j + 1]) + triangle.get(i).get(j);
            }
        }
        
        return f[0];
    }
}

Time complexity: O(n^2). Space complexity: O(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