LeetCode in Kotlin

3442. Maximum Difference Between Even and Odd Frequency I

Easy

You are given a string s consisting of lowercase English letters. Your task is to find the maximum difference between the frequency of two characters in the string such that:

• One of the characters has an even frequency in the string.
• The other character has an odd frequency in the string.

Return the maximum difference, calculated as the frequency of the character with an odd frequency minus the frequency of the character with an even frequency.

Example 1:

Input: s = “aaaaabbc”

Output: 3

Explanation:

• The character 'a' has an odd frequency of 5, and 'b' has an even frequency of 2.
• The maximum difference is 5 - 2 = 3.

Example 2:

Input: s = “abcabcab”

Output: 1

Explanation:

• The character 'a' has an odd frequency of 3, and 'c' has an even frequency of 2.
• The maximum difference is 3 - 2 = 1.

Constraints:

• 3 <= s.length <= 100
• s consists only of lowercase English letters.
• s contains at least one character with an odd frequency and one with an even frequency.

Solution

import kotlin.math.max
import kotlin.math.min

class Solution {
    fun maxDifference(s: String): Int {
        val freq = IntArray(26)
        var maxOdd = 0
        var minEven = 1000
        for (i in 0..<s.length) {
            freq[s[i].code - 'a'.code]++
        }
        for (j in freq) {
            if (j != 0 && j % 2 == 0) {
                minEven = min(minEven, j)
            } else {
                maxOdd = max(maxOdd, j)
            }
        }
        return maxOdd - minEven
    }
}

This site is open source. Improve this page.