LeetCode in Kotlin

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

class Solution {
    private fun compareStrings(str1: String, str2: String): Boolean {
        val n = str1.length
        val m = str2.length
        var i = 0
        var j = 0
        while (i < n && j < m) {
            if (str1[i] < str2[j]) {
                return true
            } else if (str1[i] > str2[j]) {
                return false
            }
            i++
            j++
        }
        return n < m
    }

    fun findCommonResponse(responses: List<List<String>>): String {
        val n = responses.size
        val mp: MutableMap<String, IntArray> = HashMap()
        var ans = responses[0][0]
        var maxFreq = 0
        for (row in 0..<n) {
            val m = responses[row].size
            for (col in 0..<m) {
                val resp = responses[row][col]
                val arr = mp.getOrDefault(resp, intArrayOf(0, -1))
                if (arr[1] != row) {
                    arr[0]++
                    arr[1] = row
                    mp.put(resp, arr)
                }
                if (arr[0] > maxFreq ||
                    (ans != resp) && arr[0] == maxFreq && compareStrings(resp, ans)
                ) {
                    ans = resp
                    maxFreq = arr[0]
                }
            }
        }
        return ans
    }
}

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