Time Based Key-Value Store

Design a time-based key-value data structure that can store multiple values for the same key at different time stamps and retrieve the key’s value at a certain timestamp.

Implement the TimeMap class:

TimeMap() Initializes the object of the data structure.
void set(String key, String value, int timestamp) Stores the key key with the value value at the given time timestamp.
String get(String key, int timestamp) Returns a value such that set was called previously, with timestamp_prev <= timestamp. If there are multiple such values, it returns the value associated with the largest timestamp_prev. If there are no values, it returns "".

Example 1:

Input
["TimeMap", "set", "get", "get", "set", "get", "get"]
[[], ["foo", "bar", 1], ["foo", 1], ["foo", 3], ["foo", "bar2", 4], ["foo", 4], ["foo", 5]]
Output
[null, null, "bar", "bar", null, "bar2", "bar2"]

Explanation
TimeMap timeMap = new TimeMap();
timeMap.set("foo", "bar", 1);  // store the key "foo" and value "bar" along with timestamp = 1.
timeMap.get("foo", 1);         // return "bar"
timeMap.get("foo", 3);         // return "bar", since there is no value corresponding to foo at timestamp 3 and timestamp 2, then the only value is at timestamp 1 is "bar".
timeMap.set("foo", "bar2", 4); // store the key "foo" and value "bar2" along with timestamp = 4.
timeMap.get("foo", 4);         // return "bar2"
timeMap.get("foo", 5);         // return "bar2"

Solution

Create a defaultdictionary of SortedSets. For each key, we add its value and its timestamp, and we bisect_right (binary_search for the location after where it would end up).

We know that if that exists, we can return that value, otherwise, we return the empty string.

from sortedcontainers import SortedSet

class TimeMap:
    def __init__(self):
        self.map = defaultdict(lambda: SortedSet(key=lambda k: k[1]))

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

    def get(self, key: str, timestamp: int) -> str:
        index = self.map[key].bisect_right(("",timestamp))
        return self.map[key][index-1][0] if index else ""

Table of Contents

Time Based Key-Value Store

Solution