Today I have done leetcode's October LeetCoding Challenge with cpp.

October LeetCoding Challenge 6

Description

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

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class TimeMap {
  unordered_map<string, map<int, string, std::greater<int>>> container;
public:
  TimeMap() {}
  
  void set(string key, string value, int timestamp) {
    container[key][timestamp] = value;
  }
  
  string get(string key, int timestamp) {
    if(!container.count(key)) return "";
    auto it = container[key].lower_bound(timestamp);
    if(it == container[key].end()) return "";
    return it->second;
  }
};

// Accepted
// 48/48 cases passed (618 ms)
// Your runtime beats 49.72 % of cpp submissions
// Your memory usage beats 6.52 % of cpp submissions (131.9 MB)