Online Election

Solve the Online Election problem by implementing a class that tracks votes and queries the leading candidate at any given time.

array hash table binary search design

Problem Statement

You are tasked with implementing a system that tracks votes over time in an online election. Two integer arrays, persons and times, are given where persons[i] represents the candidate voted for at time times[i]. The election system should be able to return the leading candidate at any given query time t. In the event of a tie between candidates, the most recent vote among the tied candidates should determine the leader.

The solution requires implementing the TopVotedCandidate class which efficiently handles queries to return the person who is leading the election at any given time t. This class should be optimized for handling up to 10,000 queries and a vote history of up to 5,000 votes. The problem focuses on efficient array scanning and hash lookup strategies to manage the votes and determine the leader in constant time.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

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.

Solution Approach

Array Scanning and Hash Lookup

The core of the solution is efficiently scanning through the votes array while maintaining a count of votes per candidate. This can be done using a hash table that maps candidates to their vote counts. As we process each vote, we update the candidate’s vote count and track the current leader at each time point. For each query, a binary search can be used on the sorted times array to quickly locate the correct time period for the query.

Handling Ties with Latest Votes

In cases where multiple candidates have the same number of votes at a specific time, the most recent vote will decide the leader. This requires that for each vote, we store both the candidate and the time they were voted for. When handling queries, if there’s a tie, the latest vote among the tied candidates should determine the winner.

Optimized Query Responses with Binary Search

To ensure that queries can be answered in constant time, we perform a binary search on the times array to quickly locate the relevant vote history for the given time t. This allows us to determine the leading candidate without needing to iterate over the entire vote list for each query.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity for querying is O(log n) due to the binary search over the times array. The space complexity is O(n) for storing the vote history and the hash map of candidate votes. The approach is efficient enough to handle the upper bounds of the problem's constraints, with up to 5,000 votes and 10,000 queries.

What Interviewers Usually Probe

Look for an implementation that efficiently handles multiple queries with binary search and constant-time query results.
The candidate should correctly handle tie-breaking scenarios where the latest vote counts.
Check if the candidate optimizes the space complexity by storing only necessary data (votes and leader history).

Common Pitfalls or Variants

Common pitfalls

Not handling ties correctly by returning the most recent vote in case of equal votes.
Inefficient query response times, especially when failing to use binary search on the times array.
Not optimizing the space complexity, potentially storing unnecessary data or failing to use a hash map for vote counts.

Follow-up variants

Modify the problem to support multiple candidates in the persons array instead of just two.
Allow for multiple queries at the same time and return the leaders for all queries in a single batch.
Alter the times array to include non-integer values (e.g., floating point), testing edge cases with time formatting.

FAQ

How do I handle ties in the Online Election problem?

In case of a tie, the most recent vote among the tied candidates should be considered the leader. This can be implemented by keeping track of the last vote cast when the vote counts are the same.

Why is binary search used in this problem?

Binary search is used to quickly find the time range for a given query, optimizing the query response time from linear to logarithmic time.

What is the time complexity of this solution?

The time complexity for each query is O(log n), where n is the number of votes. This comes from using binary search to find the closest time to the query.

How can I optimize space complexity for this problem?

Space complexity can be optimized by storing only the necessary data, such as vote counts and leader history, while avoiding unnecessary duplication of data.

What are the main challenges of the Online Election problem?

The main challenges are efficiently handling ties in votes, optimizing query response times, and managing space complexity while tracking the leader history over time.

terminal

Solution

Solution 1: Binary Search

We can record the winner at each moment during initialization, and then use binary search to find the largest moment less than or equal to $t$ during the query, and return the winner at that moment.

class TopVotedCandidate:

    def __init__(self, persons: List[int], times: List[int]):
        cnt = Counter()
        self.times = times
        self.wins = []
        cur = 0
        for p in persons:
            cnt[p] += 1
            if cnt[cur] <= cnt[p]:
                cur = p
            self.wins.append(cur)

    def q(self, t: int) -> int:
        i = bisect_right(self.times, t) - 1
        return self.wins[i]


# Your TopVotedCandidate object will be instantiated and called as such:
# obj = TopVotedCandidate(persons, times)
# param_1 = obj.q(t)

Continue Practicing

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