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 with timestamp1, with timestamp2. Both values are stored. The second value is NOT overriding first value.

Have a HashMap> 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:

 1 class TimeMap {
 2     HashMap> hm;
 3     
 4     /** Initialize your data structure here. */
 5     public TimeMap() {
 6         hm = new HashMap<>();
 7     }
 8     
 9     public void set(String key, String value, int timestamp) {
10         hm.putIfAbsent(key, new TreeMap<>());
11         hm.get(key).put(timestamp, value);
12     }
13     
14     public String get(String key, int timestamp) {
15         if(!hm.containsKey(key)){
16             return "";
17         }
18         
19         TreeMap item = hm.get(key);
20         Integer time = item.floorKey(timestamp);
21         return time == null ? "" : item.get(time);
22     }
23 }
24 
25 /**
26  * Your TimeMap object will be instantiated and called as such:
27  * TimeMap obj = new TimeMap();
28  * obj.set(key,value,timestamp);
29  * String param_2 = obj.get(key,timestamp);
30  */