LeetCode 981. Time Based Key-Value Store

原题链接在这里：https://leetcode.com/problems/time-based-key-value-store/

题目：

Create a timebased key-value store class TimeMap, that supports two operations.

1. set(string key, string value, int timestamp)

• Stores the key and value, along with the given timestamp.

2. get(string key, int timestamp)

• Returns a value such that set(key, value, timestamp_prev) was called previously, with timestamp_prev <= timestamp.
• If there are multiple such values, it returns the one with the largest timestamp_prev.
• If there are no values, it returns the empty string ("").

Example 1:

Input: inputs = ["TimeMap","set","get","get","set","get","get"], inputs = [[],["foo","bar",1],["foo",1],["foo",3],["foo","bar2",4],["foo",4],["foo",5]]
Output: [null,null,"bar","bar",null,"bar2","bar2"]
Explanation:  
TimeMap kv;  
kv.set("foo", "bar", 1); // store the key "foo" and value "bar" along with timestamp = 1  
kv.get("foo", 1);  // output "bar"  
kv.get("foo", 3); // output "bar" since there is no value corresponding to foo at timestamp 3 and timestamp 2, then the only value is at timestamp 1 ie "bar"  
kv.set("foo", "bar2", 4);  
kv.get("foo", 4); // output "bar2"  
kv.get("foo", 5); //output "bar2"  

Example 2:

Input: inputs = ["TimeMap","set","set","get","get","get","get","get"], inputs = [[],["love","high",10],["love","low",20],["love",5],["love",10],["love",15],["love",20],["love",25]]
Output: [null,null,null,"","high","high","low","low"]

Note:

1. All key/value strings are lowercase.
2. All key/value strings have length in the range [1, 100]
3. The timestamps for all TimeMap.set operations are strictly increasing.
4. 1 <= timestamp <= 10^7
5. TimeMap.set and TimeMap.get functions will be called a total of 120000 times (combined) per test case.

题解：

For the same key, if timestamp is different, value could be different.

There could be cases <key, value1> with timestamp1, <key, value2> with timestamp2. Both values are stored. The second value is NOT overriding first value.

Have a HashMap<String, TreeMap<Integer, String>> hm to store keys. The value is TreeMap sorted based on timestamp.

set, update hm. get, first get the TreeMap based on key, then use floorKey to find the largest key <= timestamp.

Time Complexity: set, O(logn). get, O(logn). n = max(TreeMap size).

Space: O(m*n). m = hm.size().

AC Java:

 class TimeMap {
     HashMap<String, TreeMap<Integer, String>> hm;

     /** Initialize your data structure here. */
     public TimeMap() {
         hm = new HashMap<>();
     }

     public void set(String key, String value, int timestamp) {
         hm.putIfAbsent(key, new TreeMap<>());
         hm.get(key).put(timestamp, value);
     }

     public String get(String key, int timestamp) {
         if(!hm.containsKey(key)){
             return "";
         }

         TreeMap<Integer, String> item = hm.get(key);
         Integer time = item.floorKey(timestamp);
         return time == null ? "" : item.get(time);
     }
 }

 /**
  * Your TimeMap object will be instantiated and called as such:
  * TimeMap obj = new TimeMap();
  * obj.set(key,value,timestamp);
  * String param_2 = obj.get(key,timestamp);
  */