Leetcode: 349. Intersection of Two Arrays

Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1]nums2 = [2, 2], return [2].

Note:

  • Each element in the result must be unique.
  • The result can be in any order.

public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
    int count = 0;
    HashMap<Integer, Integer> twoSum1 = new HashMap<>();
    HashMap<Integer, Integer> twoSum2 = new HashMap<>();

    constructTwoPairMap(twoSum1, A, B);
    constructTwoPairMap(twoSum2, C, D);
    for (Integer pairSum : twoSum1.keySet()) {
        count += twoSum1.get(pairSum) * twoSum2.getOrDefault(pairSum * (-1), 0);
    }
    return count;
}

public void constructTwoPairMap(HashMap<Integer, Integer> map, int[] A, int[] B) {
    for (int i = 0; i < A.length; i++) {
        for (int j = 0; j < B.length; j++) {
            int sum = A[i] + B[j];
            map.put(sum, map.getOrDefault(sum, 0) + 1);
        }
    }
}

TCP VS UDP

UDP (User Datagram Protocol) – Light wight but not that reliable

Pros:

1. Packets size is smaller than TCP (udp header: 8 bytes vs tcp header 20bytes)

2. No connections to create and maintain

3. More control on when the data is sent

Cons:

1. No error recovery. lost packets are just discarded.

2. Packets can arrive out of order

3. No traffic control.

4. Optional checksum in IPv4, only mandatory in IPv6

TCP (Transmission Control Protocol) – Reliable but bigger communication overhead

3 way handshake

Pros:

1. Delivery acknowledgement. Receiver would send a respond confirming the packets has been received.

2. Retransmission. When the sender doesn’t get the confirmation from the receiver after a certain amount of time, it will resent the packets.

3. Traffic control. TCP delays transmission if the network is congested.

4. Checksum is now mandatory for both IPv6 and IPv4

Cons:

1. Bigger header

2. Data doesn’t always get sent out immediately (think of video conferencing)

3. Bigger traffic overhead (because of the retransmission and delivery acknowledgement)

Dungeon Game

Dungeon Game

The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.

The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.

Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0’s) or contain magic orbs that increase the knight’s health (positive integers).

In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.

Write a function to determine the knight’s minimum initial health so that he is able to rescue the princess.

For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.

-2 (K) -3 3
-5 -10 1
10 30 -5 (P)

Notes:

  • The knight’s health has no upper bound.
  • Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned.
public int calculateMinimumHP(int[][] dungeon) {
        if (dungeon == null || dungeon.length == 0 || dungeon[0].length == 0) {
            return 0;
        }
        int row = dungeon.length;
        int col = dungeon[0].length;
        //dp[i][j]: the minimum hp knight need to have at dungeon[i][j]
        int[][] dp = new int[row][col];
    
        //init bottom-right cell
        dp[row - 1][col - 1] = Math.max(1 - dungeon[row - 1][col - 1], 1);
    
        //init rightmost rooms
        for (int i = row - 2; i >= 0; i--) {
            dp[i][col - 1] =  Math.max(dp[i + 1][col - 1] - dungeon[i][col - 1], 1);
        }
        //init bottom rooms
        for (int j = col - 2; j >= 0; j--) {
            dp[row - 1][j] = Math.max(dp[row - 1][j + 1] - dungeon[row - 1][j], 1);
        }
        //dp
        for (int i = row - 2; i >= 0; i--) {
            for (int j = col - 2; j >= 0; j--) {
                dp[i][j] = Math.max(Math.min(dp[i][j + 1] - dungeon[i][j], dp[i + 1][j] - dungeon[i][j]), 1);
            }
        }
    
        return dp[0][0];
    }

Phone Screen @Uber Oct. 26th, 2015

Given a pattern and an input string, check if the input string matches the pattern.

Eg.
pattern: abab
input: howwhathowwhat
mapping: a => how, b => what
so match(abab, howwhathowwhat) -> true
match(abab, whathowhowwhat) -> false
match(“”, “”) -> true
match(“a”, “a”) -> true
match(“ab”, “abab”) -> true
match(“aaaa”, “howwhatwhathow”) -> false
match(“aaaa”, “whatwhatwhatwhat”) -> true

Each char in pattern string can match 1-n chars in input string. Same char can only match same string. Different chars can match same strings.

class Solution {
    public boolean match(String pattern, String input) {
        if (pattern == null || input == null || input.length() < pattern.length()) {
            return false;
        }
        Map<Character, String> map = new HashMap<>();
        return matchHelper(pattern, 0, input, 0, map);
    }

    public boolean matchHelper(String pattern, int index, String input, int inputIndex, Map<Character, String> map) {
        if (index == pattern.length()) {
            if (inputIndex == input.length()) {
                return true;
            }
            return false;
        }
        char c = pattern.charAt(index);

        if (map.containsKey(c)) {
            String val = map.get(c);
            if (inputIndex + val.length() <= input.length() && val.equals(input.substring(inputIndex, inputIndex + val.length()))) {
                return matchHelper(pattern, index + 1, input, inputIndex + val.length(), map);
            }
        } else {
            for (int i = inputIndex; i < input.length(); i++) {
                String val = input.substring(inputIndex, i + 1);
                map.put(c, val);
                if (matchHelper(pattern, index + 1, input, i + 1, map)) {
                    return true;
                }
                map.remove(c);
            }
        }
        return false;
    }
    public static void main(String[] args) {
        Solution s = new Solution();
        System.out.println(s.match("", "")); //true
        System.out.println(s.match("a", "a")); //true
        System.out.println(s.match("ab", "abab")); //true
        System.out.println(s.match("abab", "howwhathowwhat")); //true
        System.out.println(s.match("abab", "howwhatwhathow")); //false
        System.out.println(s.match("aaaa", "howwhatwhathow")); //false
        System.out.println(s.match("aaaa", "whatwhatwhatwhat")); //true
    }
}

Sqrt(x)

Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x.

Example

sqrt(3) = 1

sqrt(4) = 2

sqrt(5) = 2

sqrt(10) = 3

Challenge

O(log(x))

Solution: Binary search O(log(x)). Remember to use Long during computation

public int sqrt(int x) {
    // find the last number which square of it <= x
    long start = 1, end = x;
    while (start + 1 < end) {
        long mid = start + (end - start) / 2;
        if (mid * mid <= x) {
            start = mid;
        } else {
            end = mid;
        }
    }

    if (end * end <= x) {
        return (int) end;
    }
    return (int) start;
}

Shortest Word Distance I, II & III

Word Distance I – Only called once

Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list.

For example,
Assume that words = ["practice", "makes", "perfect", "coding", "makes"].

Given word1 = “coding”, word2 = “practice”, return 3.
Given word1 = "makes", word2 = "coding", return 1.

Note:
You may assume that word1 does not equal to word2, and word1 and word2 are both in the list.

Solution: One pass. O(n), Every time find an occurrence of word1 or word2, compare the distance of p1 and p2.

public int shortestDistance(String[] words, String word1, String word2) {
        int p1 = -1, p2 = -1, min = Integer.MAX_VALUE;
    
        for (int i = 0; i < words.length; i++) {
            if (words[i].equals(word1)) 
                p1 = i;
    
            if (words[i].equals(word2)) 
                p2 = i;
    
            if (p1 != -1 && p2 != -1)
                min = Math.min(min, Math.abs(p1 - p2));
        }
    
        return min;
    }

Shortest Word Distance II – API, will be called multiple times

This is a follow up of Shortest Word Distance. The only difference is now you are given the list of words and your method will be called repeatedly many times with different parameters. How would you optimize it?

Design a class which receives a list of words in the constructor, and implements a method that takes two words word1 and word2 and return the shortest distance between these two words in the list.

For example,
Assume that words = ["practice", "makes", "perfect", "coding", "makes"].

Given word1 = “coding”, word2 = “practice”, return 3.
Given word1 = "makes", word2 = "coding", return 1.

Note:
You may assume that word1 does not equal to word2, and word1 and word2 are both in the list.

Solution: First preprocess the word to dict, which is a <word, list of index>map.

Then every time just get the word only check two list of indexes.

public class WordDistance {
    Map<String, List<Integer>> dict;
    public WordDistance(String[] words) {
        dict = new HashMap<>();
        for(int i=0;i<words.length;i++){
            if(dict.containsKey(words[i])){
                dict.get(words[i]).add(i);
            }else{
                List<Integer> indexes = new ArrayList<>();
                indexes.add(i);
                dict.put(words[i], indexes);
            }
        }
    }

    public int shortest(String word1, String word2) {
        if(dict==null||!dict.containsKey(word1)||!dict.containsKey(word2)){
            return 0;
        }
        int res = Integer.MAX_VALUE;
        List<Integer> indexes1 = dict.get(word1);
        List<Integer> indexes2 = dict.get(word2);
        int index1 = 0;
        int index2 = 0;
        while(index1<indexes1.size()&&index2<indexes2.size()){
            res = Math.min(res, Math.abs(indexes1.get(index1)-indexes2.get(index2)));
            if(indexes1.get(index1)<indexes2.get(index2)){
                index1++;
            }else{
                index2++;
            }
        }
        return res;
        
    }
}

Shortest Word Distance III – word1 could be the same as word2

This is a follow up of Shortest Word Distance. The only difference is now word1 could be the same as word2.

Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list.

word1 and word2 may be the same and they represent two individual words in the list.

For example,
Assume that words = ["practice", "makes", "perfect", "coding", "makes"].

Given word1 = “makes”, word2 = “coding”, return 1.
Given word1 = "makes", word2 = "makes", return 3.

Note:
You may assume word1 and word2 are both in the list.

Solution: if word1==word2==words[], then p1=p2=i, and every time we update p1, we check the distance between p1 and p2.

public int shortestWordDistance(String[] words, String word1, String word2) {
    int p1 = -1, p2 = -1, min = Integer.MAX_VALUE;

    for (int i = 0; i < words.length; i++) {
        if (words[i].equals(word1)) {
            p1 = i;
            if (word1.equals(word2) && p1 != -1 && p2 != -1) {
                min = Math.min(min, Math.abs(p1 - p2));
            }
        }

        if (words[i].equals(word2)) {
            p2 = i;
        }

        if (!word1.equals(word2) && p1 != -1 && p2 != -1) {
            min = Math.min(min, Math.abs(p1 - p2));
        }
    }

    return min;
}

Binary Tree Maximum Path Sum

Binary Tree Maximum Path Sum

Given a binary tree, find the maximum path sum.

The path may start and end at any node in the tree.

Example

Given the below binary tree:

  1
 / \
2   3

return 6.

Solution: Divide and Conquer. for each node, calculate two maxPath:

1. maxSinglePath: start from root to any nodes, could be empty. 从root往下走到任意点的最大路径,这条路径可以不包含任何点.

2. maxPath: 从树中任意到任意点的最大路径,这条路径至少包含一个点. Max path in the tree, can not be empty, doesn’t have to include root.

So the result is root.maxPath

O(n) time complexity. Each node is visited once.

Version 1. cleaner version, with global variable

public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: An integer.
     */
     
    int maxPath = Integer.MIN_VALUE;
    
    public int maxPathSum(TreeNode root) {
        maxSinglePath(root);
        return maxPath;
    }
    
    public int maxSinglePath(TreeNode root){
        if(root==null){
            return 0;
        }
        //devide
        int left = Math.max(0, maxSinglePath(root.left));
        int right = Math.max(0, maxSinglePath(root.right));
        
        //conquer
        maxPath = Math.max(maxPath, left+right+root.val);
        
        return Math.max(left, right)+root.val;
    }
}

Version 2. jiuzhang, with result class

private class maxPathResult {
    int maxSinglePath;
    int maxPath;
    public maxPathResult(int maxSinglePath, int maxPath) {
        this.maxSinglePath = maxSinglePath;//start from root to any nodes, could be empty
        this.maxPath = maxPath;//max path in the tree, can not be empty, doesn't have to include root
    }
}
public int maxPathSum(TreeNode root) {
    // write your code here
    return helper(root).maxPath;
}

public maxPathResult helper(TreeNode root) {
    if (root == null) {
        return new maxPathResult(0, Integer.MIN_VALUE);
    }
    //devide
    maxPathResult left = helper(root.left);
    maxPathResult right = helper(root.right);

    //conquer
    int maxSinglePath = Math.max(left.maxSinglePath, right.maxSinglePath) + root.val;
    maxSinglePath = Math.max(maxSinglePath, 0);

    int maxDoublePath = left.maxSinglePath + right.maxSinglePath + root.val;

    int maxPath = Math.max(Math.max(left.maxPath, right.maxPath), maxDoublePath);
    return new maxPathResult(maxSinglePath, maxPath);
}