# Time Based Key-Value Store Problem

## Description

LeetCode Problem 981.

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:

``````1
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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.

## Sample C++ Code

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class TimeMap {
public:
map<string, map<int, string>> m;
/** Initialize your data structure here. */
TimeMap() {
}

void set(string key, string value, int timestamp) {
m[key][timestamp] = value;
}

string get(string key, int timestamp) {
if (m.find(key) == m.end()) {
return "";
}
auto it = m[key].lower_bound(timestamp);
if (it->first == timestamp) {
return it->second;
}
else {
it --;
if (it->first > timestamp)
return "";
return it->second;
}

}
};

/**
* 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);
*/
``````