LeetCode in Kotlin

3045. Count Prefix and Suffix Pairs II

Hard

You are given a 0-indexed string array words.

Let’s define a boolean function isPrefixAndSuffix that takes two strings, str1 and str2:

• isPrefixAndSuffix(str1, str2) returns true if str1 is both a prefix and a suffix of str2, and false otherwise.

For example, isPrefixAndSuffix("aba", "ababa") is true because "aba" is a prefix of "ababa" and also a suffix, but isPrefixAndSuffix("abc", "abcd") is false.

Return an integer denoting the number of index pairs (i_,_ j) such that i < j, and isPrefixAndSuffix(words[i], words[j]) is true.

Example 1:

Input: words = [“a”,”aba”,”ababa”,”aa”]

Output: 4

Explanation: In this example, the counted index pairs are:

i = 0 and j = 1 because isPrefixAndSuffix(“a”, “aba”) is true.

i = 0 and j = 2 because isPrefixAndSuffix(“a”, “ababa”) is true.

i = 0 and j = 3 because isPrefixAndSuffix(“a”, “aa”) is true.

i = 1 and j = 2 because isPrefixAndSuffix(“aba”, “ababa”) is true.

Therefore, the answer is 4.

Example 2:

Input: words = [“pa”,”papa”,”ma”,”mama”]

Output: 2

Explanation: In this example, the counted index pairs are:

i = 0 and j = 1 because isPrefixAndSuffix(“pa”, “papa”) is true.

i = 2 and j = 3 because isPrefixAndSuffix(“ma”, “mama”) is true.

Therefore, the answer is 2.

Example 3:

Input: words = [“abab”,”ab”]

Output: 0

Explanation: In this example, the only valid index pair is i = 0 and j = 1, and isPrefixAndSuffix(“abab”, “ab”) is false.

Therefore, the answer is 0.

Constraints:

• 1 <= words.length <= 105
• 1 <= words[i].length <= 105
• words[i] consists only of lowercase English letters.
• The sum of the lengths of all words[i] does not exceed 5 * 105.

Solution

class Solution {
    fun countPrefixSuffixPairs(words: Array<String>): Long {
        var ans: Long = 0
        val visited = BooleanArray(words.size)
        for (i in words.indices) {
            val p = words[i]
            if (!visited[i]) {
                var found = 1
                for (j in i + 1 until words.size) {
                    val s = words[j]
                    if (s.length >= p.length && s.startsWith(p) && s.endsWith(p)) {
                        ans += found.toLong()
                    }
                    if (p == s) {
                        found++
                        visited[j] = true
                    }
                }
            }
        }
        return ans
    }
}

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