Question

You are given an integer n. We say that two integers x and y form a prime number pair if:

• 1 <= x <= y <= n
• x + y == n
• x and y are prime numbers

Return the 2D sorted list of prime number pairs [xi, yi]. The list should be sorted in increasing order of xi. If there are no prime number pairs at all, return an empty array.

Note: A prime number is a natural number greater than 1 with only two factors, itself and 1.

Question

You are given an integer array cards where cards[i] represents the value of the ith card. A pair of cards are matching if the cards have the same value.

Return the minimum number of consecutive cards you have to pick up to have a pair of matching cards among the picked cards. If it is impossible to have matching cards, return -1.

Question

Given a positive integer n, generate an n x n matrix filled with elements from 1 to n2 in spiral order.

Question

Given an m x n matrix, return all elements of the matrix in spiral order.

Question

Given a string s and an integer k, return the maximum number of vowel letters in any substring of s with length k.

Vowel letters in English are 'a', 'e', 'i', 'o', and 'u'.

Question

In the world of Dota2, there are two parties: the Radiant and the Dire.

The Dota2 senate consists of senators coming from two parties. Now the Senate wants to decide on a change in the Dota2 game. The voting for this change is a round-based procedure. In each round, each senator can exercise one of the two rights:

• Ban one senator’s right: A senator can make another senator lose all his rights in this and all the following rounds.
• Announce the victory: If this senator found the senators who still have rights to vote are all from the same party, he can announce the victory and decide on the change in the game.

Given a string senate representing each senator’s party belonging. The character 'R' and 'D' represent the Radiant party and the Dire party. Then if there are n senators, the size of the given string will be n.

The round-based procedure starts from the first senator to the last senator in the given order. This procedure will last until the end of voting. All the senators who have lost their rights will be skipped during the procedure.

Suppose every senator is smart enough and will play the best strategy for his own party. Predict which party will finally announce the victory and change the Dota2 game. The output should be "Radiant" or "Dire".

Question

There are n bulbs that are initially off. You first turn on all the bulbs, then you turn off every second bulb.

On the third round, you toggle every third bulb (turning on if it’s off or turning off if it’s on). For the ith round, you toggle every i bulb. For the nth round, you only toggle the last bulb.

Return the number of bulbs that are on after n rounds.

Solution

Each position is only switched on its factor rounds, and factors appear in pairs except when squared, where the factor is the same. Therefore, the problem is equivalent to finding the number of times a perfect square appears.

每个位置只在其因子回合会被切换，除了平方时因子相同，其他时候因子都是成对出现的。因此改题目等价于求平方数出现次数。

Code

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class Solution {
public:
    int bulbSwitch(int n) {
        // bulbs will switch in factor round
        // since factor round will appear with pair
        // we just need to count square round

        return (int) sqrt(n);
    }
};

Problem

Alice has n balloons arranged on a rope. You are given a 0-indexed string colors where colors[i] is the color of the i<sup>th</sup> balloon.

Alice wants the rope to be colorful. She does not want two consecutive balloons to be of the same color, so she asks Bob for help. Bob can remove some balloons from the rope to make it colorful. You are given a 0-indexed integer array neededTime where neededTime[i] is the time (in seconds) that Bob needs to remove the i<sup>th</sup> balloon from the rope.

Return the minimum time Bob needs to make the rope colorful.

Solution

滑动窗口，遍历字符串。
common用来记录连续颜色相同的个数，初始化为1。
如果当前字符与下一个字符相同，则窗口向右侧扩展，common++。
遍历时记录替换气球需要的最大时间maxTime和替换掉所有同色气球的总时间deleteTime。

如果common大于1，则总时间加上需要删除的时间（刨除最大时间maxTime）。
更新i为i + common。

Code

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class Solution {
    public int minCost(String colors, int[] neededTime) {
        int i = 0, totalTime = 0;
        char[] c = colors.toCharArray();
        while(i < neededTime.length){
            int common = 1, maxTime = neededTime[i], deleteTime = neededTime[i];
            while(i + common < neededTime.length && c[i] == c[i + common]){
                maxTime = Math.max(maxTime, neededTime[i + common]);
                deleteTime += neededTime[i + common];
                common++;
            }
            if(common > 1){
                deleteTime -= maxTime;
                totalTime += deleteTime;
            }
            i += common;
        }
        return totalTime;
    }
}

Question

Given a list paths of directory info, including the directory path, and all the files with contents in this directory, return all the duplicate files in the file system in terms of their paths. You may return the answer in any order.

A group of duplicate files consists of at least two files that have the same content.

A single directory info string in the input list has the following format:

• "root/d1/d2/.../dm f1.txt(f1_content) f2.txt(f2_content) ... fn.txt(fn_content)"

It means there are n files (f1.txt, f2.txt ... fn.txt) with content (f1_content, f2_content ... fn_content) respectively in the directory “root/d1/d2/.../dm". Note that n >= 1 and m >= 0. If m = 0, it means the directory is just the root directory.

The output is a list of groups of duplicate file paths. For each group, it contains all the file paths of the files that have the same content. A file path is a string that has the following format:

• "directory_path/file_name.txt"

Solution

使用StringBuffer对字符串进行处理。
采用HashMap建立内容和对应列表的映射。

遍历处理字符串，并添加到map中。
最后遍历map，如果对应的列表size大于1，则加入结果。

Code

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class Solution {
    public List<List<String>> findDuplicate(String[] paths) {
        List<List<String>> res = new ArrayList<>();
        HashMap<String, List<String>> map = new HashMap<>();
        
        for(String path : paths){
            StringBuffer folder = new StringBuffer();
            
            int i = 0;
            while(path.charAt(i) != ' '){
                folder.append(path.charAt(i));
                i++;
            }
            while(i < path.length()){
                while(path.charAt(i) == ' ') i++;
                StringBuffer filename = new StringBuffer();
                while(i < path.length() && path.charAt(i) != '('){
                    filename.append(path.charAt(i));
                    i++;
                }

                StringBuffer content = new StringBuffer();
                while(i < path.length() && path.charAt(i) != ')'){
                    content.append(path.charAt(i));
                    i++;
                }
                i++;
                List<String> arr = map.getOrDefault(content.toString(), new ArrayList<String>());
                arr.add(folder.toString() + '/' + filename);
                map.put(content.toString(), arr);
            }
        }
        
        for(String content : map.keySet()){
            if(map.get(content).size() > 1) res.add(map.get(content));
        }
        
        return res;
    }
}

Question

You are given two integer arrays nums and multipliers** **of size n and m respectively, where n >= m. The arrays are 1-indexed.

You begin with a score of 0. You want to perform exactly m operations. On the i<sup>th</sup> operation (1-indexed), you will:

• Choose one integer x from **either the start or the end **of the array nums.
• Add multipliers[i] * x to your score.
• Remove x from the array nums.

Return *the maximum score after performing *m operations.

Solution

动态规划，dp[][]数组记录左边取的个数和右边取的个数。

从取1开始到取multipliers的长度位置开始遍历。
然后从left取0个开始，直到left取i个为止遍历。
计算对应的right指针位置。

注意访问数组时需要访问left和right的上一个位置。

如果left为0，则只能取右侧的上一个位置加上右侧的上一个数值乘以mul。
如果right为0，则只能取左侧的上一个位置加上左侧的上一个数值乘以mul。
否则取两者之间的最大值。

最后遍历数组中left + right和为m的位置，并返回最大值。

Code

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class Solution {
    public int maximumScore(int[] nums, int[] multipliers) {
        int n = nums.length, m = multipliers.length;
        int[][] dp = new int[m+1][m+1];
        
        for(int i = 1; i <= m; i++){
            int mul = multipliers[i-1];
            for(int l = 0; l <= i; l++){
                int r = i - l;
                int iL = l - 1, iR = n - r;
                if(l == 0) dp[l][r] = dp[l][r-1] + mul * nums[iR];
                else if(r == 0) dp[l][r] = dp[l-1][r] + mul * nums[iL];
                else dp[l][r] = Math.max(dp[l-1][r] + mul * nums[iL], dp[l][r-1] + mul * nums[iR]);
            }
        }
        int ans = Integer.MIN_VALUE;
        for(int l = 1; l <= m; l++){
            int r = m - l;
            ans = Math.max(ans, dp[l][r]);
        }
        return ans;
    }
}

Question

Given a binary tree where node values are digits from 1 to 9. A path in the binary tree is said to be pseudo-palindromic if at least one permutation of the node values in the path is a palindrome.

Return the number of pseudo-palindromic paths going from the root node to leaf nodes.

Solution

用数组bin[]记录一个树枝上的节点。

回溯，遇到根节点则判断数组bin[]中是否只有0个或1个奇数的数字。

Code

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int res;
    int[] bin;
    public int pseudoPalindromicPaths (TreeNode root) {
        res = 0;
        bin = new int[10];
        backtrack(root);
        return res;
    }
    
    private void backtrack(TreeNode root){
        if(root.left == null && root.right == null){
            bin[root.val]++;
            boolean flag = false;
            for(int i = 0; i < bin.length; i++){
                if(flag && (bin[i] & 1) == 1){
                    bin[root.val]--;
                    return;
                }
                else if((bin[i] & 1) == 1) flag = true;
            }
            res++;
            bin[root.val]--;
            return;
        }
        bin[root.val]++;
        if(root.left != null) backtrack(root.left);
        if(root.right != null) backtrack(root.right);
        bin[root.val]--;
    }
}

Question

Given an integer array data representing the data, return whether it is a valid UTF-8 encoding (i.e. it translates to a sequence of valid UTF-8 encoded characters).

A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:

1. For a 1-byte character, the first bit is a 0, followed by its Unicode code.
2. For an n-bytes character, the first n bits are all one’s, the n + 1 bit is 0, followed by n - 1 bytes with the most significant 2 bits being 10.

x denotes a bit in the binary form of a byte that may be either 0 or 1.

**Note: **The input is an array of integers. Only the least significant 8 bits of each integer is used to store the data. This means each integer represents only 1 byte of data.

Solution

位运算，循环更新UTF头部的位置start。
getByte()方法用来计算位数。
check()方法用来检查从start+1开始直到位数结束是否二进制位数以10开头。
isStartWith10()方法使用掩码判断二进制是否以10开头。

Code

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class Solution {
    int[] dt;
    final int MASK = 0b11000000;
    public boolean validUtf8(int[] data) {
        dt = data;
        int start = 0;
        
        while(start < data.length){
            int bit = getByte(start);
            if(bit == -1 || start + bit > data.length) return false;
            if(!check(start, bit)) return false;
            start += bit;
        }
        return true;
    }
    
    private int getByte(int start){
        int bit = 5;
        for(int i = 0; i < 8; i++){
            if((dt[start] & 1) == 0) bit = 7 - i;
            dt[start] >>= 1;
        }
        if(bit == 0) return 1;
        else if(bit == 1 || bit > 4) return -1;
        else return bit;
    }
    
    private boolean check(int start, int bit){
        for(int i = start + 1; i < start + bit; i++) if(!isStartWith10(dt[i])) return false;
        return true;
    }
    
    private boolean isStartWith10(int num){
        return (num & MASK) == 0b10000000;
    }
}

Question

You are playing a game that contains multiple characters, and each of the characters has two main properties: attack and defense. You are given a 2D integer array properties where properties[i] = [attack<sub>i</sub>, defense<sub>i</sub>] represents the properties of the i<sup>th</sup> character in the game.

A character is said to be weak if any other character has both attack and defense levels strictly greater than this character’s attack and defense levels. More formally, a character i is said to be weak if there exists another character j where attack<sub>j</sub><span> </span>> attack<sub>i</sub> and defense<sub>j</sub><span> </span>> defense<sub>i</sub>.

Return the number of weak characters.

Solution

本题与354. Russian Doll Envelopes相近。

首先对数组进行排序，根据角色的attack数值降序排序，如果两者的attack数值相等，则根据两者的defence升序排序。

记录一个遍历过的最大defence数值maxDefence，遍历数组，如果当前maxDef大于角色的防御值，则此时当前遍历角色的attack与defence均严格小于上一个计算的角色，因此count加一。
否则更新maxDef。

*由于attack是降序的，因此可以确定遍历时下一组数组的attack一定更弱。（由于相同attack的角色是根据defence升序排列，因此记录maxDef时会逐个更新maxDef的值。）

Code

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class Solution {
    public int numberOfWeakCharacters(int[][] properties) {
        int count = 0, maxDef = Integer.MIN_VALUE;
            
        Arrays.sort(properties, new Comparator<int[]>(){
            public int compare(int[] a, int[] b){
                if(a[0] == b[0]) return a[1] - b[1];
                return b[0] - a[0];
            }
        });
        
        for(int[] character : properties){
            if(maxDef > character[1]) count++;
            else maxDef = character[1];
        }
        return count;
    }
}

Question

Given the root of a binary tree, return the same tree where every subtree (of the given tree) not containing a 1 has been removed.

A subtree of a node node is node plus every node that is a descendant of node.

Solution 1

DFS搜索，首先递归两个子节点。
在搜索时如果节点为0且两个子节点均为null，则返回null。否则返回节点本身。

Code

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode pruneTree(TreeNode root) {
        if(root == null) return null;
        root.left = pruneTree(root.left);
        root.right = pruneTree(root.right);
        if(root.left == null && root.right == null && root.val == 0) return null;
        else return root;
    }
}

Solution 2

DFS搜索，返回子节点和自己中是否包含1。
如果节点为null，则返回false。
如果自己为1，则返回true。
否则返回两个子节点中是否有true。

BFS搜索，根据每个节点的子节点是否包含1来决定左右子节点是否保存。
如果没有1，则将对应的子节点修改为null。

Code

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode pruneTree(TreeNode root) {
        if(!hasOne(root)) return null;
        Queue<TreeNode> q = new LinkedList<>();
        q.add(root);
        while(!q.isEmpty()){
            TreeNode curr = q.poll();
            if(hasOne(curr.left)) q.add(curr.left);
            else curr.left = null;
            if(hasOne(curr.right)) q.add(curr.right);
            else curr.right = null;
        }
        return root;
    }
    
    private boolean hasOne(TreeNode root){
        if(root == null) return false;
        if(root.val == 1) return true;
        else return hasOne(root.left) || hasOne(root.right);
    }
}

Question

Given an n-ary tree, return the level order traversal of its nodes’ values.

Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).

Solution

BFS搜索，层序遍历。
用level记录上一层的个数。
当队列不为空时，每次挤出level个节点，并将其子节点加入队列。
将当前层次的节点值加入列表。
最后更新level的数量。

Code

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/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null) return res;
        Queue<Node> q = new LinkedList<>();
        q.add(root);
        int level = 1;
        
        while(!q.isEmpty()){
            List<Integer> arr = new ArrayList<>();
            for(int i = 0; i < level; i++){
                Node curr = q.poll();
                for(Node child : curr.children){
                    q.offer(child);
                }
                arr.add(curr.val);
            }
            res.add(arr);
            level = q.size();
        }
        return res;
    }
}