力扣371周赛

力扣第371场周赛

找出强数对的最大异或值 I

枚举

class Solution {
public:
    int maximumStrongPairXor(vector<int>& a) {
        int n = a.size() , res = 0;
        for(int i = 0 ; i < n ; i ++){
            for(int j = 0 ; j < n ; j ++){
                if(abs(a[i]-a[j])<=min(a[i],a[j])){
                    int c = (a[i]^a[j]);
                    res = max(res , c);
                }
            }
        }
        return res;
    }
};
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高访问员工

排序后模拟枚举每个点模拟一下

class Solution {
public:
    vector<string> findHighAccessEmployees(vector<vector<string>>& as) {
        unordered_map<string,vector<string>>mp;
        vector<string>ans;
        int n = as.size();
        for(int i = 0 ; i < n ; i ++){
            mp[as[i][0]].push_back(as[i][1]);
        }
        for(auto x: mp){
            string c = x.first;
            vector<string>tmp = x.second;
            int m = tmp.size();
            if(m < 3)continue;
            ranges::sort(tmp);
            for(int i = 1 ; i < m - 1; i ++){
                int last=stoi(tmp[i-1]);
                int cur = stoi(tmp[i]);
                int next=stoi(tmp[i+1]);
                if(abs(cur - last) < 100 && abs(next - cur) < 100 && abs(next - last) < 100){
                    ans.push_back(c);
                    break;
                }
            }
        }
        return ans;
    }
};
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最大化数组末位元素的最少操作次数

枚举一下最后一个要不要交换

class Solution {
public:
    int check(int a ,int b ,int c1 ,int c2){
        if((a <= c1) && (b<=c2))return 0;
        if(a > c1){
            if(a > c2)return -1;
            if(b > c1)return -1;
            return 1;
        }
        if(b > c2){
            if(b > c1)return -1;
            if(a > c2)return -1;
            return 1;
        }
        return 0;
    }
    int minOperations(vector<int>& nums1, vector<int>& nums2) {
        int ans1 = 0,ans2 = 1, n = nums1.size();
        int c1 = 0 , c2 = 0;
        bool f1 = true , f2 = true;
        c1 = nums1[n-1],c2=nums2[n-1];
        for(int i = n - 2 ; i >= 0 ; i --){
            int s = check(nums1[i],nums2[i],c1,c2);
            if(s == -1){
                f1 = false;
                break;
            }
            ans1 += s;
        }
        c1 = nums2[n-1],c2=nums1[n-1];
        for(int i = n - 2 ; i >= 0 ; i --){
            int s = check(nums1[i],nums2[i],c1,c2);
            if(s == -1){
                f2 = false;
                break;
            }
            ans2+=s;
        }
        if(f1 && f2){
            return min(ans1,ans2);
        }
        if(f1 == true && f2 == false)return ans1;
        if(f1 == false && f2 == true)return ans2;
        return -1;
    }
};
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找出强数对的最大异或值 II

排序+01trie

// 代码来自灵神
class Node {
public:
    array<Node*, 2> children{};
    int cnt = 0; // 子树大小
};

class Trie {
    static const int HIGH_BIT = 19;
public:
    Node *root = new Node();

    // 添加 val
    void insert(int val) {
        Node *cur = root;
        for (int i = HIGH_BIT; i >= 0; i--) {
            int bit = (val >> i) & 1;
            if (cur->children[bit] == nullptr) {
                cur->children[bit] = new Node();
            }
            cur = cur->children[bit];
            cur->cnt++; // 维护子树大小
        }
    }

    // 删除 val，但不删除节点
    // 要求 val 必须在 trie 中
    void remove(int val) {
        Node *cur = root;
        for (int i = HIGH_BIT; i >= 0; i--) {
            cur = cur->children[(val >> i) & 1];
            cur->cnt--; // 维护子树大小
        }
    }

    // 返回 val 与 trie 中一个元素的最大异或和
    // 要求 trie 不能为空
    int max_xor(int val) {
        Node *cur = root;
        int ans = 0;
        for (int i = HIGH_BIT; i >= 0; i--) {
            int bit = (val >> i) & 1;
            // 如果 cur.children[bit^1].cnt == 0，视作空节点
            if (cur->children[bit ^ 1] && cur->children[bit ^ 1]->cnt) {
                ans |= 1 << i;
                bit ^= 1;
            }
            cur = cur->children[bit];
        }
        return ans;
    }
};

class Solution {
public:
    int maximumStrongPairXor(vector<int> &nums) {
        sort(nums.begin(), nums.end());
        Trie t{};
        int ans = 0, left = 0;
        for (int y: nums) {
            t.insert(y);
            while (nums[left] * 2 < y) {
                t.remove(nums[left++]);
            }
            ans = max(ans, t.max_xor(y));
        }
        return ans;
    }
};
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