LeetCode in Kotlin

3805. Count Caesar Cipher Pairs

Medium

You are given an array words of n strings. Each string has length m and contains only lowercase English letters.

Two strings s and t are similar if we can apply the following operation any number of times (possibly zero times) so that s and t become equal.

• Choose either s or t.
• Replace every letter in the chosen string with the next letter in the alphabet cyclically. The next letter after 'z' is 'a'.

Count the number of pairs of indices (i, j) such that:

• i < j
• words[i] and words[j] are similar.

Return an integer denoting the number of such pairs.

Example 1:

Input: words = [“fusion”,”layout”]

Output: 1

Explanation:

words[0] = "fusion" and words[1] = "layout" are similar because we can apply the operation to "fusion" 6 times. The string "fusion" changes as follows.

• "fusion"
• "gvtjpo"
• "hwukqp"
• "ixvlrq"
• "jywmsr"
• "kzxnts"
• "layout"

Example 2:

Input: words = [“ab”,”aa”,”za”,”aa”]

Output: 2

Explanation:

words[0] = "ab" and words[2] = "za" are similar. words[1] = "aa" and words[3] = "aa" are similar.

Constraints:

• 1 <= n == words.length <= 105
• 1 <= m == words[i].length <= 105
• 1 <= n * m <= 105
• words[i] consists only of lowercase English letters.

Solution

class Solution {
    fun countPairs(words: Array<String>): Long {
        val a = words
        val b: MutableMap<String?, Int?> = HashMap<String?, Int?>()
        var c: Long = 0
        for (d in a) {
            val e = d.toCharArray()
            val f = e[0].code
            for (g in e.indices) {
                e[g] = ((e[g].code - f + 26) % 26).toChar()
            }
            val h = String(e)
            val i: Int = b.getOrDefault(h, 0)!!
            c += i.toLong()
            b.put(h, i + 1)
        }
        return c
    }
}

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