# LeetCode 981. Time Based Key-Value Store

## Description

https://leetcode.com/problems/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 “ba2r” 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`.

## Explanation

Create a map that uses arrays as values. If the timestamp associated with the key is not in the map, create a new list. Otherwise, add the value to the list. If the timestamp is larger than the list length, increase the list length. If the timestamp is within the array range, find the closest value close to the timestamp.

## Python Solution

``````class TimeMap:

def __init__(self):
"""
"""

self.timemap = {}

def set(self, key: str, value: str, timestamp: int) -> None:

if key not in self.timemap:

self.timemap[key] = ["" for i in range(timestamp + 1)]

self.timemap[key][timestamp] = value
else:
if timestamp < len(self.timemap[key]):
self.timemap[key][timestamp] = value
else:
for i in range(timestamp - len(self.timemap[key]) - 1):
self.timemap[key].append("")

self.timemap[key].append(value)

def get(self, key: str, timestamp: int) -> str:
if key not in self.timemap:
return ""

if len(self.timemap[key]) <= timestamp:
return self.timemap[key][-1]
else:
for i in range(timestamp, -1, -1):
if self.timemap[key][i] != "":
return self.timemap[key][i]

return ""

# Your TimeMap object will be instantiated and called as such:
# obj = TimeMap()
# obj.set(key,value,timestamp)
# param_2 = obj.get(key,timestamp)``````
• Time Complexity: O(N).
• Space Complexity: O(N).