981. Time Based Key-Value Store

1. 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 “”.

2. Example

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”

3. 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.

4. Solutions

Hash Table & Binary Search

n is the number of set
TimeMap: O(1)
set: O(1)
get Time complexity: O(logn):
get Space complexity: O(n)

class TimeMap {
private:
    unordered_map<string, vector<pair<string, int>>> time_value_pairs;

public:
    TimeMap() {}

    void set(const string &key, const string &value, int timestamp) {
        // don't need to sort, timestamps are strictly increasing
        time_value_pairs[key].emplace_back(value, timestamp);
    }

    string get(const string &key, int timestamp) {
        auto &pairs = time_value_pairs[key];
        string result;
        for (int left = 0, right = pairs.size(); left < right;) {
            int mid = (left + right) / 2;
            result = pairs[mid].second <= timestamp ? pairs[mid].first : string();
            pairs[mid].second <= timestamp ? left = mid + 1 : right = mid;
        }

        return result;
    }
};