LeetCode in Kotlin

318. Maximum Product of Word Lengths

Medium

Given a string array words, return the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. If no such two words exist, return 0.

Example 1:

Input: words = [“abcw”,”baz”,”foo”,”bar”,”xtfn”,”abcdef”]

Output: 16

Explanation: The two words can be “abcw”, “xtfn”.

Example 2:

Input: words = [“a”,”ab”,”abc”,”d”,”cd”,”bcd”,”abcd”]

Output: 4

Explanation: The two words can be “ab”, “cd”.

Example 3:

Input: words = [“a”,”aa”,”aaa”,”aaaa”]

Output: 0

Explanation: No such pair of words.

Constraints:

• 2 <= words.length <= 1000
• 1 <= words[i].length <= 1000
• words[i] consists only of lowercase English letters.

Solution

class Solution {
    fun maxProduct(words: Array<String>): Int {
        val n = words.size
        var res = 0
        val masks = IntArray(n)
        for (i in 0 until n) {
            masks[i] = getMask(words[i])
        }
        for (i in 0 until n) {
            for (j in i + 1 until n) {
                if (masks[i] and masks[j] == 0) {
                    res = Math.max(res, words[i].length * words[j].length)
                }
            }
        }
        return res
    }

    private fun getMask(s: String): Int {
        var mask = 0
        for (i in 0 until s.length) {
            val c = s[i]
            mask = mask or (1 shl c.code - 'a'.code)
        }
        return mask
    }
}

This site is open source. Improve this page.