LC 981 - Time Based Key-Value Store

Question

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.

Links

Question here and solution here

Solution

concept

The key concept is store the data in a dict where key is the key in the question and val is a list that is [(val_1, ts),(val_2, ts), ...]. We can then do binary search on the ts since it is sorted, the main point we have to take note is when the mid is less than target, we should store that as answer since the question asked for the value with next largest ts if the target ts is not found.

code

class TimeMap:

    def __init__(self):
        self.kv_store = {} # key: key, val: [(value_1, ts). (value_2, ts)]

    def set(self, key: str, value: str, timestamp: int) -> None:
        if key in self.kv_store:
            self.kv_store[key].append((value, timestamp))
        else:
            self.kv_store[key] = [(value, timestamp)]
        
    def get(self, key: str, timestamp: int) -> str:
        ans = ""
        if key not in self.kv_store:
            return ans
        val_arr = self.kv_store[key]
        
        l, r = 0, len(val_arr) - 1
        while l <= r:
            mid = (l + r)//2
            if val_arr[mid][1] == timestamp:
                ans = val_arr[mid][0]
                return ans
            if val_arr[mid][1] > timestamp:
                r = mid - 1
            else:
                ans = val_arr[mid][0]
                l = mid + 1
            
        return ans
        
class NeetCode_TimeMap:

    def __init__(self):
        self.keyStore = {}  # key : list of [val, timestamp]

    def set(self, key: str, value: str, timestamp: int) -> None:
        if key not in self.keyStore:
            self.keyStore[key] = []
        self.keyStore[key].append([value, timestamp])

    def get(self, key: str, timestamp: int) -> str:
        res, values = "", self.keyStore.get(key, [])
        l, r = 0, len(values) - 1
        while l <= r:
            m = (l + r) // 2
            if values[m][1] <= timestamp:
                res = values[m][0]
                l = m + 1
            else:
                r = m - 1
        return res

Complexity

time: $O(\log n)$
space: $O(n)$