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981. Time Based Key-Value Store

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:

class TimeMap {
    Map<String, List<Pair<String, Integer>>> map;
  // "foo", {["bar",  1]}
  // "foo", {["bar", 1], ["bar2", 4]}

    public TimeMap() {
        map = new HashMap<String, List<Pair<String, Integer>>>();
    }

    public void set(String key, String value, int timestamp) {
        map.putIfAbsent(key, new ArrayList<Pair<String, Integer>>());

        // Store '(timestamp, value)' pair in 'key' bucket;
        map.get(key).add(new Pair(value, timestamp));
    }
  // TC: O(n)

    public String get(String key, int timestamp) {
        String res = "";

        if (map.containsKey(key)){
            List<Pair<String, Integer>> temp = map.get(key);

            // Using binary search on the list of pairs.
            int left = 0;
            int right = temp.size() - 1;

            while(left <= right){
                int mid = left + (right - left)/2;

                if (temp.get(mid).getValue() <= timestamp){
                    res = temp.get(mid).getKey();
                    left = mid + 1;
                }else{
                    right = mid - 1;
                }
            }
        }
                // If the 'key' does not exist in map we will return empty string
        return res;
    }

  // TC: O(nlogk) // k = we use binary search on the key's bucket which 
  // have at most M elelements and to hash the string it takes O(L) time.
}

/**
 * Your TimeMap object will be instantiated and called as such:
 * TimeMap obj = new TimeMap();
 * obj.set(key,value,timestamp);
 * String param_2 = obj.get(key,timestamp);
 */

// TC: O(nlogk)
// SC: O(n)