Time Based key-Value Store

Medium

LeetCode ↗

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"

Constraints:

1 <= key.length, value.length <= 100
key and value consist of lowercase English letters and digits.
1 <= timestamp <= 107
All the timestamps timestamp of set are strictly increasing.
At most 2 * 105 calls will be made to set and get.

Explain Idea

Explanation

For each key, histories[key] is already sorted by timestamp because set calls come in strictly increasing time order. In get, we need the value with the largest timestamp that is <= queryTimestamp.

That is an "upper bound" binary search: we keep a variable pos for the best valid index seen so far, starting at -1. At each mid: if histories[key][mid][0] <= timestamp, this index is valid, so we save pos = mid and move right (left = mid + 1) to try finding a later valid timestamp. If histories[key][mid][0] > timestamp, this index is too large, so we move left (right = mid - 1).

When the loop ends, pos is the rightmost valid timestamp. If pos == -1, there is no timestamp <= queryTimestamp, so return "". Otherwise return histories[key][pos][1].

Implementation

class TimeMap(object):

    def __init__(self):
        self.histories = dict()

    def set(self, key, value, timestamp):
        if not key in self.histories:
            self.histories[key] = []
        self.histories[key].append([timestamp, value])


    def get(self, key, timestamp):
        if not key in self.histories: return ""
        left, right, pos = 0, len(self.histories[key])-1, -1
        while left <= right:
            mid = (left+right) // 2
            if self.histories[key][mid][0] <= timestamp:
                  left = mid + 1
                  pos = mid
            else:
                  right = mid - 1
        if pos == -1: return ""
        return self.histories[key][pos][1]

Intuition

We want to only store the changes of values for each key chronologically in its own array.

We use a map of string to array histories such that histories[key] is an array that stores pairs (timestamp, value) indicating that key stores value at timestamp.

For the get function, we want to find the value stored in key at timestamp. In histories[key], it is the entry with the greatest timestamp less or equal to the given timestamp.

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