LeetCode in Kotlin

2423. Remove Letter To Equalize Frequency

Easy

You are given a 0-indexed string word, consisting of lowercase English letters. You need to select one index and remove the letter at that index from word so that the frequency of every letter present in word is equal.

Return true if it is possible to remove one letter so that the frequency of all letters in word are equal, and false otherwise.

Note:

• The frequency of a letter x is the number of times it occurs in the string.
• You must remove exactly one letter and cannot chose to do nothing.

Example 1:

Input: word = “abcc”

Output: true

Explanation: Select index 3 and delete it: word becomes “abc” and each character has a frequency of 1.

Example 2:

Input: word = “aazz”

Output: false

Explanation: We must delete a character, so either the frequency of “a” is 1 and the frequency of “z” is 2, or vice versa. It is impossible to make all present letters have equal frequency.

Constraints:

• 2 <= word.length <= 100
• word consists of lowercase English letters only.

Solution

class Solution {
    private fun isValidWord(arr: IntArray): Boolean {
        var temp = 0
        for (j in 0..25) {
            if (arr[j] == 0) {
                continue
            }
            if (temp == 0) {
                temp = arr[j]
                continue
            }
            if (arr[j] != temp) {
                return false
            }
        }
        return true
    }

    fun equalFrequency(word: String): Boolean {
        var ans = false
        // frequency array
        val arr = IntArray(26)
        for (element in word) {
            arr[element.code - 'a'.code] += 1
        }
        for (i in 0..25) {
            if (arr[i] != 0) {
                arr[i] -= 1
                if (isValidWord(arr)) {
                    ans = true
                    break
                }
                arr[i] += 1
            }
        }
        return ans
    }
}

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