README_EN.md

981. Time Based Key-Value Store

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Description

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.

Solutions

Python3

class TimeMap:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.ktv = defaultdict(list)

    def set(self, key: str, value: str, timestamp: int) -> None:
        self.ktv[key].append((timestamp, value))

    def get(self, key: str, timestamp: int) -> str:
        if key not in self.ktv:
            return ''
        tv = self.ktv[key]
        i = bisect.bisect_right(tv, (timestamp, chr(127)))
        return tv[i - 1][1] if i else ''



# Your TimeMap object will be instantiated and called as such:
# obj = TimeMap()
# obj.set(key,value,timestamp)
# param_2 = obj.get(key,timestamp)

Java

class TimeMap {
    private Map<String, TreeMap<Integer, String>> ktv;

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

    public void set(String key, String value, int timestamp) {
        ktv.computeIfAbsent(key, k -> new TreeMap<>()).put(timestamp, value);
    }

    public String get(String key, int timestamp) {
        if (!ktv.containsKey(key)) {
            return "";
        }
        TreeMap<Integer, String> tv = ktv.get(key);
        Integer t = tv.floorKey(timestamp);
        return t == null ? "" : tv.get(t);
    }
}

/**
 * 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);
 */

Go

Because timestamp is always increasing, you can use binary search to quickly find the value

type pair struct {
	timestamp int
	value     string
}

type TimeMap struct {
	data map[string][]pair
}

func Constructor() TimeMap {
	return TimeMap{data: make(map[string][]pair)}
}

func (m *TimeMap) Set(key string, value string, timestamp int) {
	m.data[key] = append(m.data[key], pair{timestamp, value})
}

func (m *TimeMap) Get(key string, timestamp int) string {
	pairs := m.data[key]
	// sort.Search return the smallest index i in [0, n) at which f(i) is true
	i := sort.Search(len(pairs), func(i int) bool {
		return pairs[i].timestamp > timestamp
	})
	if i > 0 {
		return pairs[i-1].value
	}
	return ""
}

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