Time Based Key-Value Store Python Solution

lcpy gen -n 981   # by problem number
lcpy gen -s time_based_key_value_store   # by problem name

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"]

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"

class TimeMap:
    # Time: O(1)
    # Space: O(n)
    def __init__(self) -> None:
        self.store: dict[str, list[tuple[int, str]]] = {}

    # Time: O(1)
    # Space: O(1)
    def set(self, key: str, value: str, timestamp: int) -> None:
        if key not in self.store:
            self.store[key] = []
        self.store[key].append((timestamp, value))

    # Time: O(log n)
    # Space: O(1)
    def get(self, key: str, timestamp: int) -> str:
        if key not in self.store:
            return ""

        values = self.store[key]
        left, right = 0, len(values) - 1
        result = ""

        while left <= right:
            mid = (left + right) // 2
            if values[mid][0] <= timestamp:
                result = values[mid][1]
                left = mid + 1
            else:
                right = mid - 1

        return result