LeetCode Time Based key-Value Store Solution

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

1TimeMap timeMap = new TimeMap();
2
3timeMap.set("foo", "bar", 1);  // store the key "foo" and value "bar" along with timestamp = 1.
4
5timeMap.get("foo", 1);         // return "bar"
6
7timeMap.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".
8
9timeMap.set("foo", "bar2", 4); // store the key "foo" and value "bar2" along with timestamp = 4.
10
11timeMap.get("foo", 4);         // return "bar2"
12
13timeMap.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.

To look for the location pos of the timestamp entry, we must find the timestamp pair less than or equal to timestamp. Hence we repeatly update pos to mid, if the timestamp at histories[mid] is less than or equal to the given timestamp (histories[mid][0] <= timestamp), to find the greatest timestamp less than or equal to timestamp. In the binary search loop, we will continue to find the desired timestamp on the right side of the loop if histories[mid][0] <= timestamp, and search the left side otherwise.

Implementation

1class TimeMap(object):
2
3    def __init__(self):
4        self.histories = dict()
5
6    def set(self, key, value, timestamp):
7        if not key in self.histories:
8            self.histories[key] = []
9        self.histories[key].append([timestamp, value])
10        
11
12    def get(self, key, timestamp):
13        if not key in self.histories: return ""
14        left, right, pos = 0, len(self.histories[key])-1, -1
15        while left <= right:
16            mid = (left+right) // 2
17            if self.histories[key][mid][0] <= timestamp:
18                  left = mid + 1
19                  pos = mid
20            else:
21                  right = mid - 1
22        if pos == -1: return ""
23        return self.histories[key][pos][1]