Binary Search     C++     Design     Hash Table     Medium     String    

Problem Statement:

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.

Solution:

The underlying idea is to use HashMap for key-wise storing data and ordered map for (timestamp,value) pairs ordered by timestamp for each key. Then we can do get and set operations for any key using binary search in its corresponding ordered map. More details in comments:

 
class TimeMap 
{
	private:
    unordered_map<string, map<int,string>> data;

	public:
    TimeMap() {}
    
    void set(string key, string value, int timestamp) 
    {
        if (!data.count(key))    // New key insertion
        {
            data[key] = { {timestamp, value} };
            return;
        }
		// If the same timestamp repeats for any key, replace it. We dont really need this line because of constraint but no harm
        if (data[key].count(timestamp)) data[key].erase(timestamp);
		// Finally insert in ordered map
        data[key].insert({timestamp, value});
    }
    
    string get(string key, int timestamp) 
    {
        if (!data.count(key)) return "";    // key not found
        auto it = data[key].lower_bound(timestamp);
        if (it != data[key].end() && (*it).first==timestamp)    // Exact timestamp found
            return (*it).second;
        if (it == data[key].begin()) return "";    // timestamp asked is before first time set
        --it;    // it now points to the last time it was set before "timestamp"
        return (*it).second;    // finally return value
    }
};