432. All Oone Data Structure¶

Difficulty: Hard

432. All O`one Data Structure

Hard

Design a data structure to store the strings' count with the ability to return the strings with minimum and maximum counts.

Implement the AllOne class:

AllOne() Initializes the object of the data structure.
inc(String key) Increments the count of the string key by 1. If key does not exist in the data structure, insert it with count 1.
dec(String key) Decrements the count of the string key by 1. If the count of key is 0 after the decrement, remove it from the data structure. It is guaranteed that key exists in the data structure before the decrement.
getMaxKey() Returns one of the keys with the maximal count. If no element exists, return an empty string "".
getMinKey() Returns one of the keys with the minimum count. If no element exists, return an empty string "".

Note that each function must run in O(1) average time complexity.

Example 1:

Input
["AllOne", "inc", "inc", "getMaxKey", "getMinKey", "inc", "getMaxKey", "getMinKey"]
[[], ["hello"], ["hello"], [], [], ["leet"], [], []]
Output
[null, null, null, "hello", "hello", null, "hello", "leet"]

Explanation
AllOne allOne = new AllOne();
allOne.inc("hello");
allOne.inc("hello");
allOne.getMaxKey(); // return "hello"
allOne.getMinKey(); // return "hello"
allOne.inc("leet");
allOne.getMaxKey(); // return "hello"
allOne.getMinKey(); // return "leet"

Constraints:

1 <= key.length <= 10
key consists of lowercase English letters.
It is guaranteed that for each call to dec, key is existing in the data structure.
At most 5 * 10⁴ calls will be made to inc, dec, getMaxKey, and getMinKey.

Solution¶

import java.util.*;
class AllOne {
    private HashMap<String, Integer> map;
    private TreeMap<Integer, HashSet<String>> freqMap;
    public AllOne() {
        map = new HashMap<>();
        freqMap = new TreeMap<>();
    }
    public void inc(String key) {
        if (map.containsKey(key)) {
            int freq = map.get(key);
            map.put(key, freq + 1);
            freqMap.get(freq).remove(key);
            if (freqMap.get(freq).isEmpty()) freqMap.remove(freq);
            freqMap.computeIfAbsent(freq + 1, k -> new HashSet<>()).add(key);
        } 
        else {
            map.put(key, 1);
            freqMap.computeIfAbsent(1, k -> new HashSet<>()).add(key);
        }
    }
    public void dec(String key) {
        if (!map.containsKey(key)) return;
        int freq = map.get(key);
        if (freq == 1) {
            map.remove(key);
            freqMap.get(1).remove(key);
            if (freqMap.get(1).isEmpty()) freqMap.remove(1);
        } 
        else {
            map.put(key, freq - 1);
            freqMap.get(freq).remove(key);
            if (freqMap.get(freq).isEmpty()) {
                freqMap.remove(freq);
            }
            freqMap.computeIfAbsent(freq - 1, k -> new HashSet<>()).add(key);
        }
    }
    public String getMaxKey() {
        if (freqMap.isEmpty()) return "";
        return freqMap.lastEntry().getValue().iterator().next();
    }
    public String getMinKey() {
        if (freqMap.isEmpty()) return "";
        return freqMap.firstEntry().getValue().iterator().next();
    }
}

/**
 * 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();
 */

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here