How do you solve Time Based Key-Value Store on LeetCode?

Time Based Key-Value Store (LeetCode #981) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(n) space complexity. Key insight: Using binary search significantly optimizes retrieval time.

What is the brute force solution for Time Based Key-Value Store?

The brute force approach for Time Based Key-Value Store is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves storing all key-value pairs in a list and searching through this list for the appropriate timestamp when retrieving a value. This method is straightforward but inefficient for larger datasets. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Time Based Key-Value Store?

Time Based Key-Value Store uses the following concepts: Hash Table, String, Binary Search, Design. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Time Based Key-Value Store?

Always consider edge cases, such as retrieving values for timestamps that have no associated values. Explain your thought process clearly when transitioning from brute force to optimal solutions.

What are common mistakes when solving Time Based Key-Value Store?

Not using binary search for the 'get' operation, leading to inefficient retrieval. Failing to handle cases where the key does not exist.

#981

Time Based Key-Value Store

Medium

Hash TableStringBinary SearchDesignHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log n)
Space	O(1)	O(n)

💡

Intuition

Time O(log n)Space O(n)

The optimal solution uses a dictionary to map keys to a list of timestamps and values. This allows us to perform binary search on the timestamps for efficient retrieval, reducing the time complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Create a dictionary to store keys, where each key maps to a list of tuples containing (timestamp, value).
2Step 2: For the 'set' operation, append the (timestamp, value) pair to the list for the given key.
3Step 3: For the 'get' operation, use binary search to find the largest timestamp less than or equal to the requested timestamp.

solution.py16 lines

1from collections import defaultdict
2import bisect
3
4class TimeMap:
5    def __init__(self):
6        self.store = defaultdict(list)
7
8    def set(self, key: str, value: str, timestamp: int) -> None:
9        self.store[key].append((timestamp, value))
10
11    def get(self, key: str, timestamp: int) -> str:
12        if key not in self.store:
13            return ""
14        values = self.store[key]
15        i = bisect.bisect_right(values, (timestamp,)) - 1
16        return values[i][1] if i >= 0 else ""

ℹ

Complexity note: The time complexity is O(log n) for the 'get' operation due to the binary search on the list of timestamps, while the 'set' operation is O(1) as we simply append to the list.

1Using binary search significantly optimizes retrieval time.
2Storing timestamps alongside values allows for efficient access to the most recent value.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.