[LeetCode] Time Based Key-Value Store

Problem: 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:

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 <= 10^7
• All the timestamps timestamp of set are strictly increasing.
• At most 2 * 10^5 calls will be made to set and get.

Approach: This is a binary search problem where we need to find the timestamp which is equal to or just smaller than the given timestamp. 

Given the constraint that timestamps are coming in strictly increasing order, we don't have to sort or maintain a sorted data structure. We can simply use a list.

Implementation in C#:

public class TimeMap

{

    public TimeMap()

    {

        this.kvMap = new Dictionary<string, List<Tuple<int, string>>>();

    }

   

    public void Set(string key, string value, int timestamp) {

        if (!this.kvMap.ContainsKey(key))

        {

            this.kvMap[key] = new List<Tuple<int, string>>();

        }

        this.kvMap[key].Add(new Tuple<int, string>(timestamp, value));

    }

   

    public string Get(string key, int timestamp)

    {

        if (!kvMap.ContainsKey(key))

        {

            return "";

        }

        int index = this.GetIndex(kvMap[key], timestamp);

        return index == -1 ?

               "" :

               kvMap[key][index].Item2;

    }

    private int GetIndex(List<Tuple<int, string>> sortedTimestamps,

                         int timestamp)

    {

        int start = 0, end = sortedTimestamps.Count - 1;

        if (start > end || timestamp < sortedTimestamps[0].Item1)

        {

            return -1;

        }

        while (start <= end)

        {

            int mid = start + (end - start) / 2;

            int currTimestamp = sortedTimestamps[mid].Item1;

            if (currTimestamp == timestamp)

            {

                return mid;

            }

            if (currTimestamp < timestamp)

            {

                start = mid + 1;

            }

            else

            {

                end = mid - 1;

            }

        }

        return end;

    }

    private Dictionary<string, List<Tuple<int, string>>> kvMap;

}

Complexity: Set - O(1), Get - O(logn)