Time Based Key Value Store
Detailed guide and Python implementation for the 'Time Based Key Value Store' problem.
1. Concept Overview
The 'Time Based Key Value Store' problem is a key challenge in the Binary Search section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Time Based Key Value Store.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Time Based Key Value Store carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
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.
- set(key: str, value: str, timestamp: int) Stores the key key with the value value at the given time timestamp.
- get(key: str, timestamp: int) -> str 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 "".
- •1 <= key.length, value.length <= 100
- •key and value consist of lowercase English letters and digits
- •1 <= timestamp <= 10^7
- •All timestamps of set are strictly increasing for each key
- •At most 2 * 10^5 calls will be made to set and get
Examples
["TimeMap", "set", "get", "get", "set", "get", "get"] [[], ["foo", "bar", 1], ["foo", 1], ["foo", 3], ["foo", "bar2", 4], ["foo", 4], ["foo", 5]]
[None, None, "bar", "bar", None, "bar2", "bar2"]
set("foo", "bar", 1): stores bar at time 1. get("foo", 1): returns "bar". get("foo", 3): returns "bar" (latest value at or before time 3). set("foo", "bar2", 4): stores bar2 at time 4. get("foo", 4): returns "bar2". get("foo", 5): returns "bar2".
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
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Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
class TimeMapOpt:
def __init__(self):
self.store = {}
def set(self, key: str, value: str, timestamp: int) -> None:
if key not in self.store: self.store[key] = []
self.store[key].append([value, timestamp])
def get(self, key: str, timestamp: int) -> str:
res = ""
values = self.store.get(key, [])
l, r = 0, len(values) - 1
while l <= r:
m = (l + r) // 2
if values[m][1] <= timestamp:
res = values[m][0]
l = m + 1
else:
r = m - 1
return resBrute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
class TimeMapBrute:
def __init__(self):
self.store = {}
def set(self, key: str, value: str, timestamp: int) -> None:
if key not in self.store: self.store[key] = []
self.store[key].append([value, timestamp])
def get(self, key: str, timestamp: int) -> str:
res = ""
values = self.store.get(key, [])
for v, t in values:
if t <= timestamp: res = v
return resAlgorithm Pattern Checklist
When dealing with Binary Search data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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