1023. Time Based Key Value Store¶

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1023. Time Based Key-Value Store

Medium

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¶

//Same Solution in java is giving TLE;
class TimeMap {
private:
    struct Pair {
        std::string value;
        int time;
        Pair(std::string val, int t) : value(val), time(t) {}
        bool operator<(const Pair& other) const {
            return time < other.time;
        }
    };
    std::map<std::string, std::set<Pair>> mp;
public:
    TimeMap() {
    }
    void set(std::string key, std::string value, int timestamp) {
        Pair p(value, timestamp);
        mp[key].insert(p); 
    }
    std::string get(std::string key, int timestamp) {
        if (mp.find(key) == mp.end()) return ""; 
        const std::set<Pair>& s = mp[key];
        Pair temp("", timestamp);
        auto it = s.upper_bound(temp);
        if (it == s.begin()) return ""; 
        --it;
        return it->value;
    }
};

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here