432 All O`one Data Structure All O(1) data structure

Implement a data structure to support the following operations:
   1. Inc(key) - Inserts a new key with a value of 1. Or increment an existing key by one, ensuring that key is not an empty string.
   2.Dec(key) - If the value of this key is 1, then remove it from the data structure. Otherwise, an existing key value is decremented by one. If the key does not exist, this function does nothing. key is guaranteed not to be an empty string.
   3.GetMaxKey() - Returns any one of the keys with the largest value. If no element exists, returns an empty string "".
   4.GetMinKey() - Returns any one of the keys with the smallest value. If no element exists, returns an empty string "".
Challenge: Implement all operations in O(1) time complexity.
See: https://leetcode.com/problems/all-oone-data-structure/description/

C++：

class AllOne {
public:
    /** Initialize your data structure here. */
    AllOne() {}
    
    /** Inserts a new key <Key> with value 1. Or increments an existing key by 1. */
    void inc(string key) {
        if (!m.count(key)) {
            if (buckets.empty() || buckets.back().val != 1) {
                auto newBucket = buckets.insert(buckets.end(), {1, {key}});
                m[key] = newBucket;
            } else {
                auto newBucket = --buckets.end();
                newBucket->keys.insert(key);
                m[key] = newBucket;
            }
        } else {
            auto curBucket = m[key], lastBucket = (--m[key]);
            if (lastBucket == buckets.end() || lastBucket->val != curBucket->val + 1) {
                auto newBucket = buckets.insert(curBucket, {curBucket->val + 1, {key}});
                m[key] = newBucket;
            } else {
                lastBucket->keys.insert(key);
                m[key] = lastBucket;
            }
            curBucket->keys.erase(key);
            if (curBucket->keys.empty()) buckets.erase(curBucket);
        }
    }
    
    /** Decrements an existing key by 1. If Key's value is 1, remove it from the data structure. */
    void dec(string key) {
        if (!m.count(key)) return;
        auto curBucket = m[key];
        if (curBucket->val == 1) {
            curBucket->keys.erase(key);
            if (curBucket->keys.empty()) buckets.erase(curBucket);
            m.erase(key);
            return;
        }
        auto nextBucket = ++m[key];
        if (nextBucket == buckets.end() || nextBucket->val != curBucket->val - 1) {
            auto newBucket = buckets.insert(nextBucket, {curBucket->val - 1, {key}});
            m[key] = newBucket;
        } else {
            nextBucket->keys.insert(key);
            m[key] = nextBucket;
        }
        curBucket->keys.erase(key);
        if (curBucket->keys.empty()) buckets.erase(curBucket);
    }
    
    /** Returns one of the keys with maximal value. */
    string getMaxKey() {
        return buckets.empty() ? "" : *(buckets.begin()->keys.begin());
    }
    
    /** Returns one of the keys with Minimal value. */
    string getMinKey() {
        return buckets.empty() ? "" : *(buckets.rbegin()->keys.begin());
    }
private:
    struct Bucket { int val; unordered_set<string> keys; };
    list<Bucket> buckets;
    unordered_map<string, list<Bucket>::iterator> m;
};

/**
 * Your AllOne object will be instantiated and called as such:
 * AllOne obj = new AllOne();
 * obj.inc(key);
 * obj.dec(key);
 * string param_3 = obj.getMaxKey();
 * string param_4 = obj.getMinKey();
 */

 Reference: https://www.cnblogs.com/grandyang/p/6012229.html