981. Time Based Key-Value Store

Problem Description

The problem is about creating a data structure that allows storing multiple values for the same key but at different timestamps, similar to a versioning system. The TimeMap should support two operations:

1. set: This operation stores the key with the value at the given timestamp.
2. get: This operation retrieves the value associated with a key at a specific timestamp. The catch here is that if the exact timestamp doesn't exist for that key, we need to provide the value with the closest previous timestamp. If there are no earlier timestamps, it should return an empty string.

To understand it better, imagine you are building a history tracking system for document edits. Each time someone edits a document, you record the new version of the document with a timestamp. Later, someone might want to see the version at a particular time. If a version is not available at the exact time requested, you show the latest version before the requested time.

Intuition

The key to solving this problem lies in efficiently managing the history of each key's values along with their timestamps. The intuition is to use a hash map to store keys but with a twist: each key has a list of pairs, each containing a timestamp and the corresponding value at that timestamp.

Given this structure, implementing the set function becomes straightforward: simply append the (timestamp, value) pair to the list associated with the key.

However, the challenge is in the get function, where you're asked to retrieve a value that is the closest to, but not greater than, the given timestamp. This is where binary search comes into play. The list of (timestamp, value) pairs for each key can be considered as sorted by timestamp. Using binary search, we that the closest previous timestamp to the given timestamp without going over it.

The binary search is implemented using the bisect_right function from Python's bisect module, which returns the index where an element should be inserted to maintain the sorted order. We search for a tuple where the first element is the given timestamp, and the second element is a dummy character chosen to be larger than any possible value (in this case, chr(127) which is the last ASCII character). If bisect_right provides a non-zero index, we step one index back and return the associated value. If the list is empty or the index is 0, it means there are no valid timestamps before the given timestamp, and as per problem description, we return an empty string.

Learn more about Binary Search patterns.

Solution Approach

The solution is implemented with the following key ideas:

1. Use of Defaultdict: The TimeMap class leverages a defaultdict with lists as its default value. This is a choice of data structure from the collections module in Python which helps in automatically initializing a list for each new key without checking for the key's existence. The structure self.ktv is a dictionary where each key has a list of (timestamp, value) tuples.

2. Storing Values with TimeStamps:

   • The set method appends a tuple consisting of the timestamp and the value to the list associated with the key. Since set is always called with increasing timestamp order, the list of values for each key naturally remains sorted by timestamp.

3. Retrieving Values:

   • The get method uses a binary search algorithm to quickly find the correct value for a given timestamp.
   • It checks whether the key exists in self.ktv. If not, it returns an empty string since there are no entries for that key.
   • If the key exists, a binary search is performed using the bisect_right function, which finds the insertion point in the list of values to maintain the sorted order.
   • The target of the binary search is a tuple where the first element is the timestamp and the second is a very high-value character. chr(127) works as the dummy character because chr(127) will always be greater than any string value, ensuring bisect_right returns the position for the latest timestamp that is not greater than the given timestamp.
   • If bisect_right returns zero, it means there were no timestamps less than or equal to the timestamp that we passed, so we return an empty string.
   • Otherwise, we subtract one from the index that bisect_right returned to find the largest timestamp less than or equal to the timestamp requested. Then we return the value associated with this timestamp.

Here's the encapsulated code structure:

• Initialization: def __init__(self): self.ktv = defaultdict(list)
• set: def set(self, key: str, value: str, timestamp: int) -> None: self.ktv[key].append((timestamp, value))
• get:

  1def get(self, key: str, timestamp: int) -> str:
  2    if key not in self.ktv:
  3        return ''
  4    tv = self.ktv[key]
  5    i = bisect_right(tv, (timestamp, chr(127)))
  6    return tv[i - 1][1] if i else ''

Overall, the use of binary search ensures that the retrieval of values is efficient, even when the number of timestamps grows large. It allows the get operation to have a time complexity of O(log n), where n is the number of timestamps associated with a key, making the solution scalable.

Example Walkthrough

Let's walk through the process with a simple example to illustrate the solution approach.

Suppose we are using the TimeMap to record the status of a project at different times.

• At time 1, the status is "Started"
• At time 3, the status updates to "In Progress"
• At time 5, the status finally changes to "Completed"

We would perform the following operations on our TimeMap:

1. set("projectStatus", "Started", 1)
2. set("projectStatus", "In Progress", 3)
3. set("projectStatus", "Completed", 5)

Now, our TimeMap data structure, self.ktv, has an entry for "projectStatus", which is a list of timestamp and value tuples like the following:

1"projectStatus": [(1, "Started"), (3, "In Progress"), (5, "Completed")]

Next, let's retrieve the status of the project at different times using the get operation:

• get("projectStatus", 2): This should return "Started" since it's the latest status before time 2.
• get("projectStatus", 4): This should return "In Progress" for the same reason.
• get("projectStatus", 5): This should return "Completed" as it matches the timestamp exactly.
• get("projectStatus", 0): This should return an empty string since there are no statuses recorded before time 1.

Let's walk through the first get operation step-by-step:

1. Look up "projectStatus" in self.ktv and find the list [(1, "Started"), (3, "In Progress"), (5, "Completed")].
2. Using bisect_right, find the index where we would place (2, chr(127)) to maintain sorted order.
   • The sorted position would be index 1 (between (1, "Started") and (3, "In Progress")).
3. Subtract one from the index: 1 - 1 = 0.
4. We return the value associated with this index, so ("projectStatus", 2) returns "Started".

Now, imagine checking for get("projectStatus", 0):

1. Look up "projectStatus" and find the same list as above.
2. With bisect_right, we find the index where (0, chr(127)) would fit, which gives us index 0 because 0 is less than any existing timestamps in the list.
3. Since the index is 0, it would be incorrect to subtract 1 to find the previous timestamp, as this would lead to a negative index. Instead, we return an empty string, as there's no status before time 1.

Therefore, ("projectStatus", 0) returns an empty string, which aligns with the functionality described in the problem. This example confirms that the binary search method provides an efficient and effective way to retrieve the correct value, complying with the problem's requirements.

Solution Implementation

Time and Space Complexity

Time Complexity

The set method has a time complexity of O(1) for each operation, as it appends the (timestamp, value) tuple to the list corresponding to the key in the defaultdict.

The get method's time complexity is O(log N) for each operation, where N is the number of entries associated with the specific key. This is because it performs a binary search (bisect_right) to find the position where the given timestamp would fit in the sorted order of timestamps stored.

Space Complexity

The space complexity is O(K + T), where K is the number of unique keys and T is the total number of set calls or the total number of timestamp and value pairs. This is because all values with their corresponding timestamps for all keys are stored in the defaultdict.

Learn more about how to find time and space complexity quickly using problem constraints.