Introduction

In this blog post, we will explore how to solve the LeetCode problem "981" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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 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 "bar2" 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

• Data Structure: Use a dictionary (hash map) where keys are the input keys (strings) and values are lists of (timestamp, value) pairs. The lists are maintained in ascending order of timestamps.
  • Set Operation: Append the (timestamp, value) pair to the list associated with the given key.
  • Get Operation: Use binary search on the list of (timestamp, value) pairs associated with the given key to find the largest timestamp that is less than or equal to the target timestamp.

• Runtime Complexity: set: O(1), get: O(log N) where N is the number of timestamps for a given key. Storage Complexity: O(M) where M is the total number of set operations.

Code

    class TimeMap:

    def __init__(self):
        self.data = {}  # key: list of (timestamp, value)

    def set(self, key: str, value: str, timestamp: int) -> None:
        if key not in self.data:
            self.data[key] = []
        self.data[key].append((timestamp, value))

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

        values = self.data[key]
        left, right = 0, len(values) - 1
        result = ""

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

        return result