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[leetcode] problem 981 – time based key-value store

admin 11月 13, 2020 0

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

No.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”

No.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.

Code

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public class  {
    private int timestamp;
    private String value;

    public (int timestamp, String value) {
        this.timestamp = timestamp;
        this.value = value;
    }
}

private Map<String, List<Pair>> map;

public TimeMap() {
    map = new HashMap<>();
}

public void set(String key, String value, int timestamp) {
    map.putIfAbsent(key, new ArrayList<>());
    map.get(key).add(new Pair(timestamp, value));
}

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

    int idx = Collections.binarySearch(map.get(key), new Pair(timestamp, ""), new Comparator<Pair>() {
        
        public int compare(Pair o1, Pair o2) {
            return o1.timestamp - o2.timestamp;
        }
    });

    if (idx >= 0)
        return map.get(key).get(idx).value;
    else if (idx == -1)
        return "";
    else
        return map.get(key).get(-idx-2).value;
}
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