3527. Find the Most Common Response LeetCode Solution

Last updated: May 5, 2025

In this guide, you will get 3527. Find the Most Common Response LeetCode Solution with the best time and space complexity. The solution to Find the Most Common Response problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Find the Most Common Response solution in C++
Find the Most Common Response solution in Java
Find the Most Common Response solution in Python
Additional Resources

3527. Find the Most Common Response LeetCode Solution image

Problem Statement of Find the Most Common Response

You are given a 2D string array responses where each responses[i] is an array of strings representing survey responses from the ith day.
Return the most common response across all days after removing duplicate responses within each responses[i]. If there is a tie, return the lexicographically smallest response.

Example 1:

Input: responses = [[“good”,”ok”,”good”,”ok”],[“ok”,”bad”,”good”,”ok”,”ok”],[“good”],[“bad”]]
Output: “good”
Explanation:

After removing duplicates within each list, responses = [[“good”, “ok”], [“ok”, “bad”, “good”], [“good”], [“bad”]].
“good” appears 3 times, “ok” appears 2 times, and “bad” appears 2 times.
Return “good” because it has the highest frequency.

Example 2:

Input: responses = [[“good”,”ok”,”good”],[“ok”,”bad”],[“bad”,”notsure”],[“great”,”good”]]
Output: “bad”
Explanation:

After removing duplicates within each list we have responses = [[“good”, “ok”], [“ok”, “bad”], [“bad”, “notsure”], [“great”, “good”]].
“bad”, “good”, and “ok” each occur 2 times.
The output is “bad” because it is the lexicographically smallest amongst the words with the highest frequency.

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

Complexity Analysis

Time Complexity: O(\Sigma |\texttt{responses[i]}|)
Space Complexity: O(\Sigma |\texttt{responses[i]}|)

3527. Find the Most Common Response LeetCode Solution in C++

class Solution {
 public:
  string findCommonResponse(vector<vector<string>>& responses) {
    string ans;
    unordered_map<string, int> count;
    int maxFreq = 0;

    for (const vector<string>& response : responses)
      for (const string& response :
           unordered_set<string>{response.begin(), response.end()})
        ++count[response];

    for (const pair<const string, int>& entry : count)
      maxFreq = max(maxFreq, entry.second);

    for (const auto& [response, freq] : count)
      if (freq == maxFreq && (ans.empty() || response < ans))
        ans = response;

    return ans;
  }
};
/* code provided by PROGIEZ */

3527. Find the Most Common Response LeetCode Solution in Java

class Solution {
  public String findCommonResponse(List<List<String>> responses) {
    String ans = "";
    Map<String, Integer> count = new HashMap<>();
    int maxFreq = 0;

    for (List<String> response : responses)
      for (final String r : new HashSet<>(response))
        count.merge(r, 1, Integer::sum);

    for (final int freq : count.values())
      maxFreq = Math.max(maxFreq, freq);

    for (Map.Entry<String, Integer> entry : count.entrySet()) {
      final String response = entry.getKey();
      final int freq = entry.getValue();
      if (freq == maxFreq && (ans.isEmpty() || response.compareTo(ans) < 0))
        ans = response;
    }

    return ans;
  }
}
// code provided by PROGIEZ

3527. Find the Most Common Response LeetCode Solution in Python

class Solution:
  def findCommonResponse(self, responses: list[list[str]]) -> str:
    count = collections.Counter()

    for response in responses:
      for response in set(response):
        count[response] += 1

    maxFreq = max(count.values())
    return min([response
                for response, count in count.items()
                if count == maxFreq])
 # code by PROGIEZ

Additional Resources

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