LeetCode 574: Winning Candidate Solution in Python – A Step-by-Step Guide

What Is LeetCode 574: Winning Candidate?

In LeetCode 574: Winning Candidate, you’re given a list of strings votes where each string represents a vote for a candidate’s name, and your task is to return the name of the candidate with the most votes. If there’s a tie, return any one of the tied candidates (typically the lexicographically smallest in practice, but here we’ll assume the first encountered for simplicity). For example, with votes = ["Alice", "Bob", "Alice", "Charlie"], the result is "Alice" with 2 votes. This problem builds on LeetCode 347: Top K Frequent Elements for frequency counting but focuses on finding the single most frequent element.

Problem Statement

• Input: votes (List[str])—list of candidate names as votes.
• Output: String—name of the winning candidate (most votes).
• Rules: Each string is a vote; return candidate with highest count; ties resolved arbitrarily (e.g., first encountered).

Constraints

• 1 <= votes.length <= 10⁵
• 1 <= votes[i].length <= 10
• votes[i] consists of English letters (assume lowercase for simplicity)

Examples

• Input: votes = ["Alice","Bob","Alice","Charlie"]
• Output: "Alice"
• Counts: Alice: 2, Bob: 1, Charlie: 1; Alice wins.
• Input: votes = ["Alice","Bob","Bob","Charlie"]
• Output: "Bob"
• Counts: Alice: 1, Bob: 2, Charlie: 1; Bob wins.
• Input: votes = ["Alice","Alice","Bob","Bob"]
• Output: "Alice"
• Counts: Alice: 2, Bob: 2; Alice returned (first encountered).

Understanding the Problem: Declaring the Winner

To solve LeetCode 574: Winning Candidate in Python, we need to count the votes for each candidate in the votes list and identify the one with the highest count, returning their name as a string, handling ties by picking one (e.g., the first encountered). With up to 10⁵ votes, a brute-force approach counting each candidate’s occurrences by scanning the list repeatedly (O(n²)) could work but would be slow. Instead, a dictionary-based approach efficiently tallies votes in one pass, making it both fast and scalable. We’ll explore:

• Best Solution (Dictionary-Based Vote Counting): O(n) time, O(k) space—fast and optimal (n = number of votes, k = unique candidates).
• Alternative Solution (Brute-Force Frequency Traversal): O(n²) time, O(1) space—simple but inefficient.

Let’s dive into the dictionary solution with a friendly breakdown!

Best Solution: Dictionary-Based Vote Counting

Why Dictionary Wins

The dictionary-based vote counting solution is the best for LeetCode 574 because it determines the winner in O(n) time and O(k) space (k ≤ n, typically small) by using a dictionary to tally votes for each candidate in a single pass, then finding the candidate with the maximum count efficiently. It’s like keeping a vote tally sheet, marking each candidate’s count as votes come in, and spotlighting the leader—all in one quick, organized sweep!

How It Works

Think of this as a vote-tally champion:

• Step 1: Build Vote Dictionary:
• Use a dictionary to map candidate names to their vote counts.
• Step 2: Count Votes:
• Iterate votes, increment count for each candidate.
• Step 3: Find Winner:
• Scan dictionary for candidate with highest count.
• If tie, pick first encountered (or lexicographically smallest, adjustable).
• Step 4: Return Result:
• Name of winning candidate.
• Why It Works:
• Single pass counts all votes.
• Dictionary enables O(1) updates and lookups.

It’s like a vote-counting maestro!

Step-by-Step Example

Example: votes = ["Alice","Bob","Alice","Charlie"]

• Step 1: Initialize: vote_counts = {}.
• Step 2: Count votes:
• "Alice": vote_counts = {"Alice": 1}.
• "Bob": vote_counts = {"Alice": 1, "Bob": 1}.
• "Alice": vote_counts = {"Alice": 2, "Bob": 1}.
• "Charlie": vote_counts = {"Alice": 2, "Bob": 1, "Charlie": 1}.
• Step 3: Find winner:
• Max count = 2, candidate = "Alice".
• Result: "Alice".

Example: votes = ["Alice","Bob","Bob","Charlie"]

• Step 2: Count:
• vote_counts = {"Alice": 1, "Bob": 2, "Charlie": 1}.
• Step 3: Max count = 2, candidate = "Bob".
• Result: "Bob".

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codeclass Solution:
    def findWinner(self, votes: List[str]) -> str:
        # Step 1: Initialize dictionary for vote counts
        vote_counts = {}

        # Step 2: Count votes
        for vote in votes:
            vote_counts[vote] = vote_counts.get(vote, 0) + 1

        # Step 3: Find candidate with maximum votes
        max_votes = 0
        winner = ""
        for candidate, count in vote_counts.items():
            if count > max_votes:
                max_votes = count
                winner = candidate

        # Step 4: Return winning candidate
        return winner

• Line 4: Initialize empty dictionary for counts.
• Lines 7-8: Iterate votes, increment counts using get for default 0.
• Lines 11-16: Find max count and corresponding candidate.
• Line 19: Return winner’s name.
• Time Complexity: O(n)—single pass over votes.
• Space Complexity: O(k)—dictionary size (k = unique candidates).

It’s like a vote-tallying wizard!

Alternative Solution: Brute-Force Frequency Traversal

Why an Alternative Approach?

The brute-force frequency traversal counts votes for each unique candidate by scanning the entire list repeatedly, running in O(n²) time and O(1) space (excluding minimal variables). It’s straightforward but inefficient for large lists, making it a good baseline for small datasets or understanding.

How It Works

Picture this as a vote-scanning seeker:

• Step 1: Get unique candidates (or assume list scan).
• Step 2: For each candidate, count votes by traversing list.
• Step 3: Track max count and winner.
• Step 4: Return winner’s name.

It’s like a frequency-counting explorer!

Step-by-Step Example

Example: votes = ["Alice","Bob","Alice","Charlie"]

• Step 1: Unique candidates: ["Alice", "Bob", "Charlie"].
• Step 2: Count votes:
• "Alice": Scan → 2.
• "Bob": Scan → 1.
• "Charlie": Scan → 1.
• Step 3: Max = 2, winner = "Alice".
• Result: "Alice".

Code for Brute-Force Approach

Copy codeclass Solution:
    def findWinner(self, votes: List[str]) -> str:
        # Step 1: Get unique candidates (simplified by scanning)
        max_votes = 0
        winner = ""

        # Step 2: Check each candidate
        for i in range(len(votes)):
            candidate = votes[i]
            count = 0
            # Count votes for this candidate
            for vote in votes:
                if vote == candidate:
                    count += 1
            # Update winner if higher count
            if count > max_votes:
                max_votes = count
                winner = candidate

        # Step 3: Return winner
        return winner

• Lines 4-5: Initialize max and winner.
• Lines 8-16: For each candidate, count votes, update max.
• Line 19: Return winner.
• Time Complexity: O(n²)—nested loops.
• Space Complexity: O(1)—minimal extra space.

It’s a brute-force vote scanner!

Comparing the Two Solutions

• Dictionary (Best):
• Pros: O(n), O(k), fast.
• Cons: Extra space for dictionary.
• Brute-Force (Alternative):
• Pros: O(n²), O(1), space-efficient.
• Cons: Slower.

Dictionary wins for efficiency!

Additional Examples and Edge Cases

• ["Alice"]: "Alice".
• ["A","A","B"]: "A".
• Empty list: Undefined (assume handled).

Dictionary handles them all!

Complexity Recap

• Dictionary: Time O(n), Space O(k).
• Brute-Force: Time O(n²), Space O(1).

Dictionary’s the speed champ!

Key Takeaways

• Dictionary: Vote tallying—learn at Python Basics!
• Brute-Force: Scan simplicity.
• Elections: Votes are fun.
• Python: Dict or brute, your pick!

Final Thoughts: Crown That Winner!

LeetCode 574: Winning Candidate in Python is a delightful data challenge. Dictionary-based vote counting is your fast track, while brute-force offers a thorough dive. Want more? Try LeetCode 347: Top K Frequent Elements or LeetCode 692: Top K Frequent Words. Ready to tally? Head to Solve LeetCode 574 on LeetCode and declare that winner today!