package LeetCode_30
/**
* 30. Substring with Concatenation of All Words
* https://leetcode.com/problems/substring-with-concatenation-of-all-words/
*
* You are given a string s and an array of strings words of the same length.
* Return all starting indices of substring(s) in s that is a concatenation of each word in words exactly once,
* in any order, and without any intervening characters. You can return the answer in any order.
Example 1:
Input: s = "barfoothefoobarman", words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoo" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.
Example 2:
Input: s = "wordgoodgoodgoodbestword", words = ["word","good","best","word"]
Output: []
Example 3:
Input: s = "barfoofoobarthefoobarman", words = ["bar","foo","the"]
Output: [6,9,12]
Constraints:
1. 1 <= s.length <= 104
2. s consists of lower-case English letters.
3. 1 <= words.length <= 5000
4. 1 <= words[i].length <= 30
5. words[i] consists of lower-case English letters.
* */
class Solution {
/*
* solution: HashMap, Time complexity:O(n*l), Space complexity:O(n), l is length of s, n is count of word
* */
fun findSubstring(s: String, words: Array<String>): List<Int> {
if (s == null || s.isEmpty() || words == null || words.isEmpty()) {
return ArrayList()
}
val frequencyMap = HashMap<String, Int>()
for (word in words) {
frequencyMap.put(word, frequencyMap.getOrDefault(word, 0) + 1)
}
val result = ArrayList<Int>()
val wordLength = words[0].length
val totalWordCount = words.size
var i = 0
//while (i <= s.length - wordLength * totalWordCount) {
while (i < s.length) {
val seedWord = HashMap<String, Int>()
for (j in 0 until totalWordCount) {
//check each word one by one
val startIndex = i + j * wordLength
var endIndex = startIndex + wordLength
if (endIndex > s.lastIndex) {
endIndex = s.lastIndex+1
}
val curWord = s.substring(startIndex, endIndex)
if (!frequencyMap.contains(curWord)) {
break
}
seedWord.put(curWord, seedWord.getOrDefault(curWord, 0) + 1)
println(seedWord)
if (seedWord.get(curWord) ?: 0 > frequencyMap.getOrDefault(curWord, 0)) {
//handle case: capcapmap,cap's frequency in frequencyMap is 1, so just need the second 'cap'
break
}
//if j can go through the length of word
if (j + 1 == totalWordCount) {
result.add(i)
}
}
i++
}
return result
}
}
