LeetCode in Kotlin

30. Substring with Concatenation of All Words

Hard

You are given a string s and an array of strings words. All the strings of words are of the same length.

A concatenated substring in s is a substring that contains all the strings of any permutation of words concatenated.

• For example, if words = ["ab","cd","ef"], then "abcdef", "abefcd", "cdabef", "cdefab", "efabcd", and "efcdab" are all concatenated strings. "acdbef" is not a concatenated substring because it is not the concatenation of any permutation of words.

Return the starting indices of all the concatenated substrings in s. You can return the answer in any order.

Example 1:

Input: s = “barfoothefoobarman”, words = [“foo”,”bar”]

Output: [0,9]

Explanation:

Since words.length == 2 and words[i].length == 3, the concatenated substring has to be of length 6.

The substring starting at 0 is “barfoo”. It is the concatenation of [“bar”,”foo”] which is a permutation of words.

The substring starting at 9 is “foobar”. It is the concatenation of [“foo”,”bar”] which is a permutation of words.

The output order does not matter. Returning [9,0] is fine too.

Example 2:

Input: s = “wordgoodgoodgoodbestword”, words = [“word”,”good”,”best”,”word”]

Output: []

Explanation:

Since words.length == 4 and words[i].length == 4, the concatenated substring has to be of length 16.

There is no substring of length 16 is s that is equal to the concatenation of any permutation of words.

We return an empty array.

Example 3:

Input: s = “barfoofoobarthefoobarman”, words = [“bar”,”foo”,”the”]

Output: [6,9,12]

Explanation:

Since words.length == 3 and words[i].length == 3, the concatenated substring has to be of length 9.

The substring starting at 6 is “foobarthe”. It is the concatenation of [“foo”,”bar”,”the”] which is a permutation of words.

The substring starting at 9 is “barthefoo”. It is the concatenation of [“bar”,”the”,”foo”] which is a permutation of words.

The substring starting at 12 is “thefoobar”. It is the concatenation of [“the”,”foo”,”bar”] which is a permutation of words.

Constraints:

• 1 <= s.length <= 104
• 1 <= words.length <= 5000
• 1 <= words[i].length <= 30
• s and words[i] consist of lowercase English letters.

Solution

class Solution {
    fun findSubstring(s: String, words: Array<String>): List<Int> {
        val indices: MutableList<Int> = ArrayList()
        if (words.size == 0) {
            return indices
        }
        // Put each word into a HashMap and calculate word frequency
        val wordMap: MutableMap<String, Int> = HashMap()
        for (word in words) {
            wordMap[word] = wordMap.getOrDefault(word, 0) + 1
        }
        val wordLength = words[0].length
        val window = words.size * wordLength
        for (i in 0 until wordLength) {
            // move a word's length each time
            var j = i
            while (j + window <= s.length) {

                // get the subStr
                val subStr = s.substring(j, j + window)
                val map: MutableMap<String, Int> = HashMap()
                // start from the last word
                for (k in words.indices.reversed()) {
                    // get the word from subStr
                    val word = subStr.substring(k * wordLength, (k + 1) * wordLength)
                    val count = map.getOrDefault(word, 0) + 1
                    // if the num of the word is greater than wordMap's, move (k * wordLength) and
                    // break
                    if (count > wordMap.getOrDefault(word, 0)) {
                        j = j + k * wordLength
                        break
                    } else if (k == 0) {
                        indices.add(j)
                    } else {
                        map[word] = count
                    }
                }
                j = j + wordLength
            }
        }
        return indices
    }
}

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