package LeetCode_745
/**
* 745. Prefix and Suffix Search
* https://leetcode.com/problems/prefix-and-suffix-search/description/
*
Given many words, words[i] has weight i.
Design a class WordFilter that supports one function, WordFilter.f(String prefix, String suffix).
It will return the word with given prefix and suffix with maximum weight. If no word exists, return -1.
Examples:
Input:
WordFilter(["apple"])
WordFilter.f("a", "e") // returns 0
WordFilter.f("b", "") // returns -1
Note:
words has length in range [1, 15000].
For each test case, up to words.length queries WordFilter.f may be made.
words[i] has length in range [1, 10].
prefix, suffix have lengths in range [0, 10].
words[i] and prefix, suffix queries consist of lowercase letters only.
* */
/*
* solution 1: HashMap, Time complexity:O(n*len^3 + n*len), Space complexity:O(n*len^3)
* n is number of words, len is the max length of word
* */
class WordFilter(words: Array<String>) {
//key is string form: perfix_suffix, value is Index
val map = HashMap<String, Int>()
init {
buildMap(words)
}
/*
* build map by word,
* for example: apple can generate the key: a_e, ap_le, app_ple, appl_pple, apple_apple
* */
private fun buildMap(words: Array<String>) {
var index = 0
for (word in words) {
val len = word.length
val perfixs = Array<String>(len + 1, { "" })
val suffixs = Array<String>(len + 1, { "" })
for (i in 0 until len) {
perfixs[i + 1] = perfixs[i] + word[i]
suffixs[i + 1] = word[len - i - 1] + suffixs[i]
}
//put perfixs and suffixs input map
for (perfix in perfixs) {
for (suffix in suffixs) {
//HashMap can replace and value if the same key
map.put(perfix + "_" + suffix, index)
}
}
index++
}
}
fun f(prefix: String, suffix: String): Int {
val key = prefix + "_" + suffix
if (map.containsKey(key)) {
return map.get(key)!!
}
return -1
}
}
/**
* Your WordFilter object will be instantiated and called as such:
* var obj = WordFilter(words)
* var param_1 = obj.f(prefix,suffix)
*/
