The task asks to find the most common response across multiple survey days, ensuring duplicates within each day are removed. If there is a tie in frequency, return the lexicographically smallest response. Hash tables and array scanning are key to solving this efficiently.
Problem Statement
You are given a 2D array responses, where each responses[i] represents survey responses for the ith day. The task is to determine the most frequent unique response across all days after removing duplicates within each individual day's response array.
If there is a tie for the most frequent response, return the lexicographically smallest one. Your solution should be efficient enough to handle up to 1,000 days and 1,000 responses per day, making array scanning and hash lookup important to solving the problem.
Examples
Example 1
Input: responses = [["good","ok","good","ok"],["ok","bad","good","ok","ok"],["good"],["bad"]]
Output: "good"
Example 2
Input: responses = [["good","ok","good"],["ok","bad"],["bad","notsure"],["great","good"]]
Output: "bad"
Constraints
- 1 <= responses.length <= 1000
- 1 <= responses[i].length <= 1000
- 1 <= responses[i][j].length <= 10
- responses[i][j] consists of only lowercase English letters
Solution Approach
Step 1: Remove Duplicates Within Each Day
The first step is to ensure that there are no duplicates within each day's response array. This can be achieved by converting each day's list of responses into a set, which automatically removes duplicates.
Step 2: Count Frequency of Each Response
After eliminating duplicates within each day, you can count the frequency of each unique response using a hash table (or HashMap). This will help efficiently track the occurrence of each response across all the days.
Step 3: Find the Most Common Response
After counting the frequencies, you need to find the response with the highest count. In case of a tie (same frequency), the lexicographically smaller response should be returned. This can be done using simple comparison logic while iterating through the frequency map.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity is determined by the need to process each response within each day, resulting in O(N * M) where N is the number of days and M is the average number of responses per day. The space complexity is O(N * M) to store the responses and their counts in the hash map.
What Interviewers Usually Probe
- Can the candidate efficiently handle large input sizes?
- Does the candidate apply hash maps effectively for counting frequencies?
- How well does the candidate handle tie-breaking between lexicographically equal counts?
Common Pitfalls or Variants
Common pitfalls
- Not removing duplicates within each day before counting frequencies can lead to incorrect results.
- Failure to account for ties in frequency, which may lead to returning the wrong response.
- Inefficient handling of large input sizes, especially with array scanning or hash lookup.
Follow-up variants
- Handle responses with varying case sensitivity (e.g., 'good' vs 'Good').
- Find the second most common response instead of the first.
- Return the least common response instead of the most common.
How GhostInterview Helps
- GhostInterview helps break down the problem into clear steps, focusing on array scanning and hash map usage for efficiency.
- It assists in identifying key patterns for optimizing time and space complexity during interviews.
- GhostInterview suggests common pitfalls and helps candidates avoid issues with large input handling and lexicographical ties.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the main pattern used in 'Find the Most Common Response'?
The primary pattern is array scanning combined with hash map lookup to efficiently count occurrences and handle ties.
How do I handle ties in the frequency of responses?
If multiple responses have the same frequency, return the lexicographically smallest one.
What is the time complexity of this problem?
The time complexity is O(N * M), where N is the number of days and M is the number of responses per day.
Why do I need to remove duplicates within each day?
Duplicates within each day would distort the frequency count and lead to incorrect results. Removing them ensures accuracy.
Can I solve this problem without using hash maps?
While it's possible, hash maps provide an efficient way to count frequencies and are optimal for this task, especially with large inputs.
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