Time Based Key-Value Store

 

  1. Clarify the problem:

    • The problem requires us to design a time-based key-value store that supports two operations: set and get.
    • The set operation sets the value of a key at a given timestamp.
    • The get operation retrieves the value of a key at a given timestamp or the closest previous timestamp.
    • If there is no value available for the key at or before the given timestamp, the get operation should return an empty string.
    • We need to implement a data structure that efficiently handles these operations.

  2. Analyze the problem:

    • Input: Operations in the form of key, value, and timestamp.
    • Output: Values retrieved by the get operation.
    • Constraints:
      • The number of operations is between 1 and 10^4.
      • The timestamp is a positive integer.

  3. Design an algorithm:

    • We can solve this problem using a hashmap and binary search.
    • We will use a hashmap to store the key-value pairs, where the value is a list of tuples representing the timestamp-value pairs.
    • For the set operation, we append the new timestamp-value pair to the list associated with the key.
    • For the get operation, we perform a binary search on the list of timestamp-value pairs to find the closest previous timestamp.
    • If we find a timestamp that is equal to the given timestamp, we return the corresponding value.
    • If we find a timestamp that is less than the given timestamp, we return the value associated with that timestamp.
    • If we don't find any timestamp less than or equal to the given timestamp, we return an empty string.

  4. Explain your approach:

    • We will implement the time-based key-value store using a hashmap and binary search.
    • We create a hashmap to store the key-value pairs, where the value is a list of tuples representing the timestamp-value pairs.
    • For the set operation, we append the new timestamp-value pair to the list associated with the key.
    • For the get operation, we perform a binary search on the list of timestamp-value pairs to find the closest previous timestamp.
    • We use a helper function to perform the binary search.
    • If we find a timestamp that is equal to the given timestamp, we return the corresponding value.
    • If we find a timestamp that is less than the given timestamp, we return the value associated with that timestamp.
    • If we don't find any timestamp less than or equal to the given timestamp, we return an empty string.

  5. Write clean and readable code:

    python
  6. class TimeMap:
    def __init__(self):
        self.store = {}

    def set(self, key, value, timestamp):
        if key in self.store:
            self.store[key].append((timestamp, value))
        else:
            self.store[key] = [(timestamp, value)]

    def get(self, key, timestamp):
        if key in self.store:
            values = self.store[key]
            left, right = 0, len(values) - 1

            while left <= right:
                mid = (left + right) // 2
                if values[mid][0] <= timestamp:
                    left = mid + 1
                else:
                    right = mid - 1

            if right >= 0:
                return values[right][1]
        return ""

  7. Test your code:

    python
  8. timeMap = TimeMap()

timeMap.set("key1", "value1", 1)
timeMap.set("key1", "value2", 2)
print(timeMap.get("key1", 1))  # Output: "value1"
print(timeMap.get("key1", 3))  # Output: "value2"

timeMap.set("key2", "value3", 3)
print(timeMap.get("key2", 4))  # Output: "value3"

print(timeMap.get("key3", 5))  # Output: ""

timeMap.set("key3", "value4", 4)
timeMap.set("key3", "value5", 5)
print(timeMap.get("key3", 3))  # Output: ""
print(timeMap.get("key3", 6))  # Output: "value5"

  9. Optimize if necessary:

    • The provided solution has a time complexity of O(log n) for both the set and get operations, where n is the number of timestamp-value pairs for a given key.
    • The space complexity is O(n), where n is the total number of timestamp-value pairs stored in the hashmap.
    • This solution is efficient and optimal for the given problem.

  10. Handle error cases:

    • The code assumes that the input for the get operation is a valid key and timestamp. If the key does not exist in the hashmap, the get operation will return an empty string.

  11. Discuss complexity analysis:

    • The time complexity of the solution is O(log n) for both the set and get operations, where n is the number of timestamp-value pairs for a given key.
    • The space complexity is O(n), where n is the total number of timestamp-value pairs stored in the hashmap.
    • The solution is efficient and provides the correct output for the given problem.
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