LeetCode-in-Java

3527. Find the Most Common Response

Medium

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

Solution

import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class Solution {
    private boolean compareStrings(String str1, String str2) {
        int n = str1.length();
        int m = str2.length();
        int i = 0;
        int j = 0;
        while (i < n && j < m) {
            if (str1.charAt(i) < str2.charAt(j)) {
                return true;
            } else if (str1.charAt(i) > str2.charAt(j)) {
                return false;
            }
            i++;
            j++;
        }
        return n < m;
    }

    public String findCommonResponse(List<List<String>> responses) {
        int n = responses.size();
        Map<String, int[]> mp = new HashMap<>();
        String ans = responses.get(0).get(0);
        int maxFreq = 0;
        for (int row = 0; row < n; row++) {
            int m = responses.get(row).size();
            for (int col = 0; col < m; col++) {
                String resp = responses.get(row).get(col);
                int[] arr = mp.getOrDefault(resp, new int[] {0, -1});
                if (arr[1] != row) {
                    arr[0]++;
                    arr[1] = row;
                    mp.put(resp, arr);
                }
                if (arr[0] > maxFreq
                        || !ans.equals(resp) && arr[0] == maxFreq && compareStrings(resp, ans)) {
                    ans = resp;
                    maxFreq = arr[0];
                }
            }
        }
        return ans;
    }
}

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