Leetcode Weekly Contest 152

退役老人现在连leetcode都不会做了 = =
今天早上做了leetcode第三题题目看错了，加上比赛中间还在调投稿的实验，一心二用直接gg
总结下教训就是 本渣现在做题连题目都看不清就开始做。开始写题之前应当把样例过一遍，然后自己再造1-2个例子，然后再开始做

A题：统计素数的个数(素数筛或者sqrt(n)判断都可以)，然后分别计算count!

class Solution {
public:
    int numPrimeArrangements(int n) {
        vector<int> has(n + 5, 0);
        int cntPrime = 0;
        const int MOD = 1e9 + 7;
        for(int i = 2; i <= n; ++i) {
            if(has[i] == 0) cntPrime ++;
            for(int j = i * 2; j <= n; j += i) {
                has[j] = 1;
            }
        }
        int result = 1;
        for(int i = 1; i <= n - cntPrime; ++i) {
            result = 1ll * i * result % MOD;
        }
        for(int i = 1; i <= cntPrime; ++i) {
            result = 1ll * i * result % MOD;
        }
        return result;
    }
};

B题：滚动求和

class Solution {
public:
    int dietPlanPerformance(vector<int>& calories, int k, int lower, int upper) {
        long long sum = 0;
        for(int i = 0; i < k; ++i) {
            sum += calories[i];
        }
        int result = 0;
        for(int i = k - 1, len = calories.size(); i < len; ++i) {
            if(sum < lower) result --;
            else if(sum > upper) result ++;
            if(i != len - 1) {
                sum -= calories[i - k + 1];
                sum += calories[i + 1];
            }
        }
        return result;
    }
};

C题：我之前把题意看成整个字符串满足是回文，这个复杂很多。。。。
正确解法是判断区间的每种字母个数，我们知道偶数是可以直接配对的(放对称位置就好了)，统计字母个数为奇数的，奇数的需要改一半就好了

class Solution {
private:
    vector<int> has[26];
    int get(int id, int fr, int to) {
        if(fr > to) return 0;
        if(fr == 0) return has[id][to];
        else return has[id][to] - has[id][fr - 1];
    }
public:
    vector<bool> canMakePaliQueries(string s, vector<vector<int>>& queries) {
        
        int slen = s.length();
        for(int i = 0; i < 26; ++i) has[i].clear();
        
        /***statistic character count*****/
        for(int i = 0; i < 26; ++i) {
            for(int j = 0; j < slen; ++j) {
                has[i].push_back(0);
            }
        }
        for(int i = 0; i < slen; ++i) {
            has[s[i] - 'a'][i] ++;
        }
        for(int i = 0; i < 26; ++i) {
            for(int j = 1; j < slen; ++j) {
                has[i][j] += has[i][j-1];
            }
        }

        
        vector<bool> result;
        for(int i = 0, len = queries.size(); i < len; ++i) {
            int left = queries[i][0]; int right = queries[i][1]; int k = queries[i][2];

            int cnt = 0;
            for(int j = 0; j < 26; ++j) {
                int t1 = get(j, left, right);
                if(t1 % 2 == 1) {
                    cnt ++;
                }
            }
            cnt /= 2;
            if(cnt <= k) result.push_back(true);
            else result.push_back(false);
        }
        
        return result;
    }
};

D题：基本思想就是暴力查询，首先我们知道，字符串中的字母排列前后和出现次数都是无所谓的(除了puzzle的第一个字母的问题)，这个是可以进行处理的
除此之外有两种加速的方法
方法1: 位压缩为26位，直接通过数位判断word是否是满足puzzle
方法2: 对word建立字典树，让puzzle在字典树中进行dfs，这里每个puzzle大约会有7!的查询复杂度

下面两个代码分别对应这两种方案，我是直接discuss的大佬里面爬下来了

class Solution {
    vector<int> base;
    void born(vector<string>& words){
        for(auto& s: words){
            set<char> tmp;
            int bit = 0;
            for(auto c:s){
                tmp.insert(c);
                bit = bit | (1<<(c-'a'));
            }
            if(tmp.size() >7)continue;
            base.push_back(bit);
        }
    }
public:
    vector<int> findNumOfValidWords(vector<string>& words, vector<string>& puzzles) {
        vector<int> ans;
        born(words);
        
        for(auto& s: puzzles){
            int num = 0;
            int bit = 0;
            for(auto c: s){
                bit = bit | (1<<(c-'a'));
            }
            int firstBit = 1 << (s[0] - 'a');
            for(auto v: base){
                if((v & bit) == v && ((firstBit & v) == firstBit)){
                    num++;
                }
            }
            ans.push_back(num);
        }
        
        return ans;
    }
};

const int ALPHABET_SIZE = 26;

/* The structure of a trie node */
struct TrieNode
{
    struct TrieNode* children[ALPHABET_SIZE];
    int count = 0;
};

/* Creates a new trie node and returns the pointer */
struct TrieNode* getNode()
{
    struct TrieNode* newNode = new TrieNode;

    for(int i=0; i<ALPHABET_SIZE; i++)
        newNode->children[i] = nullptr;
    
    newNode->count = 0;

    return newNode;
}

/* Inserts the given string to the collection */
void insert(struct TrieNode* root, string str)
{
    struct TrieNode* pCrawl = root;

    for(int i=0; i<str.length(); i++)
    {
        int index = str[i]-'a';

        if(!pCrawl->children[index])
            pCrawl->children[index] = getNode();
        
        pCrawl = pCrawl->children[index];
    }

    pCrawl->count = (pCrawl->count + 1);
}

/* Returns the count of strings which are valid */
int search(struct TrieNode* root, string str, bool firstSeen, char firstLetter)
{
    if(!root)
        return 0;
    
    int count = 0;
    
    if(firstSeen)
        count += root->count;
    
    for(int i=0; i<str.length(); i++)
    {
        int index = str[i] - 'a';
        
        if(str[i] == firstLetter)
            count += search(root->children[index], str, true, firstLetter);
        else
            count += search(root->children[index], str, firstSeen, firstLetter);
    }

    return count;
}

class Solution
{
public:
    vector<int> findNumOfValidWords(vector<string>& words, vector<string>& puzzles);
};

vector<int> Solution :: findNumOfValidWords(vector<string>& words, vector<string>& puzzles)
{
    struct TrieNode* root = getNode();
    
    for(auto str : words)
    {
        set<char> temp;
        temp.insert(str.begin(), str.end());
        
        string sorted = "";
        for(auto ele : temp)
            sorted += ele;
        
        insert(root, sorted);
    }
    
    vector<int> count;
    for(auto puzzle : puzzles)
    {
        char firstLetter = puzzle[0];
        sort(puzzle.begin(), puzzle.end());
        count.push_back(search(root, puzzle, false, firstLetter));
    }
    
    return count;
    
}