LeetCode in Kotlin

3042. Count Prefix and Suffix Pairs I

Easy

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 <= 50
• 1 <= words[i].length <= 10
• words[i] consists only of lowercase English letters.

Solution

class Solution {
    fun countPrefixSuffixPairs(words: Array<String>): Int {
        var count = 0
        for (i in words.indices) {
            for (j in i + 1 until words.size) {
                if (isPrefixAndSuffix(words[i], words[j])) {
                    count++
                }
            }
        }
        return count
    }

    private fun isPrefixAndSuffix(str1: String, str2: String): Boolean {
        val m = str1.length
        val n = str2.length
        if (m > n) {
            return false
        }
        for (i in 0 until m) {
            if (str1[i] != str2[i]) {
                return false
            }
        }
        for (i in 0 until m) {
            if (str1[i] != str2[n - m + i]) {
                return false
            }
        }
        return true
    }
}

This site is open source. Improve this page.