Winning Candidate - Solution & Explanation

Problem Overview: You are given two tables: Candidate and Vote. Each row in Vote stores a candidateId representing a vote cast for that candidate. The task is to return the name of the candidate who received the highest number of votes.

Approach 1: GROUP BY + COUNT Aggregation (O(n) time, O(k) space)

The most direct solution aggregates votes by candidate. Join the Vote table with Candidate using the candidate ID, group the rows by candidate, and compute the vote count using COUNT(*). Once counts are computed, sort them in descending order with ORDER BY and return the top row using LIMIT 1. The key idea is that SQL aggregation lets the database compute vote totals efficiently in a single scan. Time complexity is O(n) where n is the number of votes, and space complexity is O(k) for storing grouped counts of k candidates. This approach relies on concepts from database queries and SQL aggregation.

Approach 2: Aggregation Subquery + Join (O(n) time, O(k) space)

Another option computes vote counts in a derived table first. Use a subquery that groups the Vote table by candidateId and calculates COUNT(*). Then join this aggregated result with the Candidate table to fetch the candidate name. Finally, sort the aggregated counts and return the highest one. This approach separates counting logic from the final lookup step, which can improve readability when queries become more complex. Complexity remains O(n) time and O(k) space because the database still scans all votes and stores counts per candidate.

Approach 3: Window Function Ranking (O(n) time, O(k) space)

With modern MySQL versions, you can compute vote totals and rank candidates using a window function. First aggregate votes per candidate using GROUP BY. Then apply RANK() or DENSE_RANK() over the counts ordered descending. Finally, select the candidate whose rank equals 1. Window functions make the ranking logic explicit and are useful if the query later needs to return multiple top candidates or additional ranking information. The database still performs aggregation over all votes, so time complexity is O(n) and space complexity is O(k). This approach highlights advanced SQL features and analytical query patterns.

Recommended for interviews: The GROUP BY + COUNT + ORDER BY approach is what interviewers expect. It shows you understand relational aggregation and how to extract the maximum result directly in SQL. Mentioning alternative forms like subqueries or window functions demonstrates deeper familiarity with database query design.