LeetCode 981 Time Based Key-Value Store

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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:

<pre>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" </pre>

Constraints:

1 <= key.length, value.length <= 100
key and value consist of lowercase English letters and digits.
1 <= timestamp <= 10<sup>7</sup>
All the timestamps timestamp of set are strictly increasing.
At most 2 * 10<sup>5</sup> calls will be made to set and get.

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题目大意：

实现带历史记录的HashMap。也就是同一个key记录所有赋过值的value

解题思路：

N/A

解题步骤：

N/A

注意事项：

Map to list的思路，list含两个，包括value和timestamp，用binary search搜索timestamp的下标，然后返回对应的value

Python代码：

class TimeMap(TestCases):

    def __init__(self):
        self.key_to_val = collections.defaultdict(list)
        self.key_to_timestamp = collections.defaultdict(list)

    def set(self, key: str, value: str, timestamp: int) -> None:
        self.key_to_val[key].append(value)
        self.key_to_timestamp[key].append(timestamp)

    def get(self, key: str, timestamp: int) -> str:
        index = bisect.bisect(self.key_to_timestamp[key], timestamp) - 1
        if index == -1:
            return ''
        else:
            return self.key_to_val[key][index]

算法分析：

get时间复杂度为O(logn)，空间复杂度O(n)