LeetCode 578: Get Highest Answer Rate Question Solution in Python – A Step-by-Step Guide

What Is LeetCode 578: Get Highest Answer Rate Question?

In LeetCode 578: Get Highest Answer Rate Question, you’re given a list of tuples survey_log where each tuple (question_id, action) represents a survey action: "show" (question shown to a user) or "answer" (question answered by a user). Your task is to return the question_id with the highest answer rate—defined as the number of answers divided by the number of shows—among questions shown at least once. If there’s a tie, return the smallest question_id. For example, with survey_log = [(1, "show"), (1, "answer"), (2, "show"), (3, "show"), (3, "answer")], the result is 1 (100% rate). This problem builds on LeetCode 347: Top K Frequent Elements for frequency counting but focuses on ratio computation and selection.

Problem Statement

• Input: survey_log (List[Tuple[int, str]])—list of (question_id, action) where action is "show" or "answer".
• Output: Integer—question_id with highest answer rate; smallest ID if tied.
• Rules: Answer rate = answers / shows; consider only questions with ≥1 show; ties resolved by smallest ID.

Constraints

• 1 <= survey_log.length <= 10⁵
• 1 <= question_id <= 10⁵
• action is either "show" or "answer"

Examples

• Input: survey_log = [(1,"show"),(1,"answer"),(2,"show"),(3,"show"),(3,"answer")]
• Output: 1
• Rates: Q1: 1/1 = 1.0, Q2: 0/1 = 0.0, Q3: 1/1 = 1.0; Q1 wins (smallest ID).
• Input: survey_log = [(1,"show"),(2,"show"),(2,"answer")]
• Output: 2
• Rates: Q1: 0/1 = 0.0, Q2: 1/1 = 1.0; Q2 wins.
• Input: survey_log = [(1,"show")]
• Output: 1
• Rate: Q1: 0/1 = 0.0 (only shown question).

Understanding the Problem: Finding the Top Question

To solve LeetCode 578: Get Highest Answer Rate Question in Python, we need to calculate the answer rate for each question by counting "show" and "answer" actions, then identify the question with the highest rate, resolving ties by choosing the smallest question_id, handling up to 10⁵ log entries efficiently. A brute-force approach scanning the list repeatedly for each question (O(n²)) could work but would be slow. Instead, a dictionary-based approach tallies counts in one pass, making it fast and scalable. We’ll explore:

• Best Solution (Dictionary-Based Rate Calculation): O(n) time, O(k) space—fast and optimal (n = log length, k = unique questions).
• Alternative Solution (Brute-Force 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 Rate Calculation

Why Dictionary Wins

The dictionary-based rate calculation solution is the best for LeetCode 578 because it determines the highest answer rate question in O(n) time and O(k) space (k ≤ n, typically small) by using two dictionaries to track "show" and "answer" counts for each question in a single pass, then computing rates and selecting the winner efficiently. It’s like keeping a survey tally sheet, marking shows and answers as they come in, and spotlighting the most answered question—all in one quick, organized sweep!

How It Works

Think of this as a survey-rate champion:

• Step 1: Build Dictionaries:
• show_counts: Map question_id to number of "show" actions.
• answer_counts: Map question_id to number of "answer" actions.
• Step 2: Count Actions:
• Iterate survey_log, increment appropriate dictionary based on action.
• Step 3: Calculate Rates:
• For each question in show_counts:
• Rate = answer_counts[qid] / show_counts[qid] (default 0 if no answers).
• Track max rate and smallest ID for ties.
• Step 4: Return Result:
• question_id with highest rate.
• Why It Works:
• Single pass builds counts.
• Dictionary lookups enable O(1) rate computation.

It’s like a rate-calculating maestro!

Step-by-Step Example

Example: survey_log = [(1,"show"),(1,"answer"),(2,"show"),(3,"show"),(3,"answer")]

• Step 1: Initialize:
• show_counts = {}.
• answer_counts = {}.
• Step 2: Count actions:
• (1, "show"): show_counts = {1: 1}.
• (1, "answer"): answer_counts = {1: 1}.
• (2, "show"): show_counts = {1: 1, 2: 1}.
• (3, "show"): show_counts = {1: 1, 2: 1, 3: 1}.
• (3, "answer"): answer_counts = {1: 1, 3: 1}.
• Step 3: Calculate rates:
• Q1: 1/1 = 1.0.
• Q2: 0/1 = 0.0.
• Q3: 1/1 = 1.0.
• Max rate = 1.0, tied Q1 and Q3, pick smallest ID = 1.
• Result: 1.

Example: survey_log = [(1,"show"),(2,"show"),(2,"answer")]

• Step 2:
• show_counts = {1: 1, 2: 1}.
• answer_counts = {2: 1}.
• Step 3:
• Q1: 0/1 = 0.0.
• Q2: 1/1 = 1.0.
• Max rate = 1.0, Q2 wins.
• Result: 2.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codefrom typing import List, Tuple

class Solution:
    def getHighestAnswerRate(self, survey_log: List[Tuple[int, str]]) -> int:
        # Step 1: Initialize dictionaries for counts
        show_counts = {}
        answer_counts = {}

        # Step 2: Count shows and answers
        for qid, action in survey_log:
            if action == "show":
                show_counts[qid] = show_counts.get(qid, 0) + 1
            elif action == "answer":
                answer_counts[qid] = answer_counts.get(qid, 0) + 1

        # Step 3: Find question with highest answer rate
        max_rate = -1.0  # Use float for rate comparison
        winner_id = float('inf')  # Track smallest ID for ties

        for qid in show_counts:
            shows = show_counts[qid]
            answers = answer_counts.get(qid, 0)  # Default 0 if not answered
            rate = answers / shows
            if rate > max_rate or (rate == max_rate and qid < winner_id):
                max_rate = rate
                winner_id = qid

        # Step 4: Return winning question ID
        return winner_id

• Lines 6-7: Initialize dictionaries for show and answer counts.
• Lines 10-14: Iterate log, increment counts based on action.
• Lines 17-24: Calculate rates, track max rate and smallest ID.
• Line 27: Return winning question_id.
• Time Complexity: O(n)—single pass over log.
• Space Complexity: O(k)—dictionary size (k = unique questions).

It’s like a survey-rate calculator!

Alternative Solution: Brute-Force Traversal

Why an Alternative Approach?

The brute-force traversal approach identifies unique questions, then scans the list repeatedly to count shows and answers for each, running in O(n²) time and O(1) space (excluding minimal variables). It’s simple but inefficient for large logs, making it a good baseline for small datasets or understanding.

How It Works

Picture this as a survey-scanning seeker:

• Step 1: Get unique question IDs.
• Step 2: For each ID, count shows and answers by traversing list.
• Step 3: Compute rates, track max and smallest ID.
• Step 4: Return winner’s ID.

It’s like a frequency-scanning explorer!

Step-by-Step Example

Example: survey_log = [(1,"show"),(2,"show"),(2,"answer")]

• Step 1: Unique IDs: [1, 2].
• Step 2: Count:
• Q1: Shows = 1, Answers = 0.
• Q2: Shows = 1, Answers = 1.
• Step 3: Rates:
• Q1: 0/1 = 0.0.
• Q2: 1/1 = 1.0.
• Max = 1.0, ID = 2.
• Result: 2.

Code for Brute-Force Approach

Copy codefrom typing import List, Tuple

class Solution:
    def getHighestAnswerRate(self, survey_log: List[Tuple[int, str]]) -> int:
        # Step 1: Get unique question IDs
        unique_qids = set(qid for qid, _ in survey_log)

        # Step 2: Find highest rate
        max_rate = -1.0
        winner_id = float('inf')

        for qid in unique_qids:
            shows = 0
            answers = 0
            for log_qid, action in survey_log:
                if log_qid == qid:
                    if action == "show":
                        shows += 1
                    elif action == "answer":
                        answers += 1
            rate = answers / shows if shows > 0 else 0.0
            if rate > max_rate or (rate == max_rate and qid < winner_id):
                max_rate = rate
                winner_id = qid

        # Step 3: Return winner
        return winner_id

• Line 6: Extract unique IDs with set.
• Lines 9-22: For each ID, count shows/answers, compute rate.
• Line 25: Return winning ID.
• Time Complexity: O(n²)—nested loops.
• Space Complexity: O(1)—minimal extra space.

It’s a brute-force rate scanner!

Comparing the Two Solutions

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

Dictionary wins for efficiency!

Additional Examples and Edge Cases

• Single show: Lowest ID if no answers.
• All answers: Highest rate wins.
• Empty log: 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: Rate finesse—learn at Python Basics!
• Brute-Force: Scan simplicity.
• Surveys: Rates are fun.
• Python: Dict or loops, your pick!

Final Thoughts: Find That Top Question!

LeetCode 578: Get Highest Answer Rate Question in Python is a practical data challenge. Dictionary-based rate calculation 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 analyze? Head to Solve LeetCode 578 on LeetCode and pinpoint that top question today!