Introduction

In this blog post, we will explore how to solve the LeetCode problem "911" 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

You are given two integer arrays persons and times. In an election, the ith vote was cast for persons[i] at time times[i]. For each query at a time t, find the person that was leading the election at time t. Votes cast at time t will count towards our query. In the case of a tie, the most recent vote (among tied candidates) wins. Implement the TopVotedCandidate class: TopVotedCandidate(int[] persons, int[] times) Initializes the object with the persons and times arrays. int q(int t) Returns the number of the person that was leading the election at time t according to the mentioned rules. Example 1: Input ["TopVotedCandidate", "q", "q", "q", "q", "q", "q"] [[[0, 1, 1, 0, 0, 1, 0], [0, 5, 10, 15, 20, 25, 30]], [3], [12], [25], [15], [24], [8]] Output [null, 0, 1, 1, 0, 0, 1] Explanation TopVotedCandidate topVotedCandidate = new TopVotedCandidate([0, 1, 1, 0, 0, 1, 0], [0, 5, 10, 15, 20, 25, 30]); topVotedCandidate.q(3); // return 0, At time 3, the votes are [0], and 0 is leading. topVotedCandidate.q(12); // return 1, At time 12, the votes are [0,1,1], and 1 is leading. topVotedCandidate.q(25); // return 1, At time 25, the votes are [0,1,1,0,0,1], and 1 is leading (as ties go to the most recent vote.) topVotedCandidate.q(15); // return 0 topVotedCandidate.q(24); // return 0 topVotedCandidate.q(8); // return 1 Constraints: 1 <= persons.length <= 5000 times.length == persons.length 0 <= persons[i] < persons.length 0 <= times[i] <= 109 times is sorted in a strictly increasing order. times[0] <= t <= 109 At most 104 calls will be made to q.

Explanation

• Preprocessing: Precompute the leader at each time in the times array during initialization. Store these leaders in a separate array leaders. This allows for efficient querying.
  • Binary Search: Use binary search on the times array to find the latest time less than or equal to the query time t. The leader at that time is the answer.

• Runtime Complexity: O(N) for initialization, O(log N) for each query, where N is the length of the input arrays. Storage Complexity: O(N)

Code

    class TopVotedCandidate:

    def __init__(self, persons: list[int], times: list[int]):
        self.times = times
        self.leaders = []
        votes = {}
        leader = -1
        max_votes = -1

        for person, time in zip(persons, times):
            votes[person] = votes.get(person, 0) + 1
            if votes[person] >= max_votes:
                max_votes = votes[person]
                leader = person
            self.leaders.append(leader)

    def q(self, t: int) -> int:
        left, right = 0, len(self.times) - 1
        ans = -1
        while left <= right:
            mid = (left + right) // 2
            if self.times[mid] <= t:
                ans = mid
                left = mid + 1
            else:
                right = mid - 1
        return self.leaders[ans]