Time Based Key-Value Store

Published: Oct 6, 2022

Medium Binary Search Hash Table Design

Introduction

A design problem always requires a consideration on what data structure(s) to save input data. This problem needs a key-value store (hash table) since get method needs key access. However, even though the same key is used, the get result depends on the timestamp. Given that, it needs another hash table to save timestamps. Conveniently, the timestamp is given chronologically to set method, which means it is sorted. If the array is sorted, the binary search can be used to find the answer. In this problem, the same index of timestamps can be used to get a value.

Problem Description

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 “”.

Constraints:

  • 1 <= key.length, value.length <= 100
  • key and value consist of lowercase English letters and digits.
  • 1 <= timestamp <= 10**7
  • All the timestamps timestamp of set are strictly increasing.
  • At most 2 * 10**5 calls will be made to set and get.

https://leetcode.com/problems/time-based-key-value-store/

Examples

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"
Example 2
Input
["TimeMap","set","set","get","get","get","get","get"]
[[],["love","high",10],["love","low",20],["love",5],["love",10],["love",15],["love",20],["love",25]]
Output
[null,null,null,"","high","high","low","low"]

Analysis

The solution here uses two dictionaries to save values and timestamps. The index is the same for both values and timestamps. When a timestamp is given to get method and whose index is i, the corresponding value’s index is also i. The timestamps associated to the key is sorted. If the timestamp is given to get method, use the binary search to find the index in timestamps array. The same index in values array is the answer.

Solution

class TimeMap:

    def __init__(self):
        self.values = collections.defaultdict(list)
        self.timestamps = collections.defaultdict(list)


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

    def get(self, key: str, timestamp: int) -> str:
        if key not in self.timestamps:
            return ''
        ts = self.timestamps[key]
        if not ts or timestamp < ts[0]:
            return ''
        idx = bisect.bisect_right(ts, timestamp)
        return self.values[key][idx - 1] if idx else ''

Complexities

  • Time: set – O(1), get – O(log(n)): n is a number of timestamps associated to a key
  • Space: O(mn): m is a number of keys