Solution Explanation: Online Election

This problem requires designing a data structure that efficiently handles queries about the leading candidate in an election at various points in time. The solution utilizes a combination of preprocessing and binary search for optimal performance.

Approach:

The core idea is to preprocess the input data to create a lookup table that stores the leading candidate at each time point. This preprocessing step allows us to answer queries in logarithmic time, significantly improving efficiency compared to repeatedly processing the entire vote history for each query.

1. Preprocessing:

   • We iterate through the persons and times arrays simultaneously.
   • For each vote, we maintain a count of votes for each candidate using a Counter (Python) or a similar data structure (other languages).
   • After processing each vote, we determine the current leading candidate. In case of a tie, the most recently voted candidate wins.
   • We store the leading candidate at each time point in a list or array called wins. This forms our lookup table.

2. Querying (q function):

   • Given a query time t, we use binary search on the times array to find the index i such that times[i] <= t and times[i+1] > t (or times[i] is the last element if t is greater than all times).
   • The leading candidate at time t is then simply wins[i].

Time Complexity Analysis:

• __init__ (Constructor): The preprocessing step iterates through the persons and times arrays once, which takes O(n) time, where n is the number of votes.
• q (Query): The binary search on the times array takes O(log n) time.

Space Complexity Analysis:

• The wins array (or list) stores the leading candidate for each time point, requiring O(n) space.
• The cnt array (or similar data structure) used for vote counting during preprocessing also requires O(n) space in the worst case (all candidates are unique).

Code Implementation (Python):

from collections import Counter
import bisect
 
class TopVotedCandidate:
    def __init__(self, persons: List[int], times: List[int]):
        self.times = times
        self.wins = []
        cnt = Counter()
        leader = -1  # Initialize leader
        for i, person in enumerate(persons):
            cnt[person] += 1
            if leader == -1 or cnt[person] >= cnt[leader]:
                leader = person  #Tie goes to the most recent
            self.wins.append(leader)
 
    def q(self, t: int) -> int:
        index = bisect_right(self.times, t) -1 #Find the index of the greatest time less than or equal to t
        return self.wins[index]
 

Other Languages: The implementations in Java, C++, Go, and TypeScript follow the same logic, adapting the data structures and binary search implementations to their respective language features. The core algorithm remains consistent. See the original response for those implementations.

This solution efficiently handles the online election problem by leveraging preprocessing and binary search, achieving a time complexity of O(n) for initialization and O(log n) for each query. The space complexity is O(n).