LeetCode-in-Java

524. Longest Word in Dictionary through Deleting

Medium

Given a string s and a string array dictionary, return the longest string in the dictionary that can be formed by deleting some of the given string characters. If there is more than one possible result, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string.

Example 1:

Input: s = “abpcplea”, dictionary = [“ale”,”apple”,”monkey”,”plea”]

Output: “apple”

Example 2:

Input: s = “abpcplea”, dictionary = [“a”,”b”,”c”]

Output: “a”

Constraints:

• 1 <= s.length <= 1000
• 1 <= dictionary.length <= 1000
• 1 <= dictionary[i].length <= 1000
• s and dictionary[i] consist of lowercase English letters.

Solution

import java.util.ArrayDeque;
import java.util.Deque;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class Solution {
    private static class Pair {
        private String word;
        private int index;

        public Pair(String word, int index) {
            this.index = index;
            this.word = word;
        }

        public String getWord() {
            return word;
        }

        public void setWord(String word) {
            this.word = word;
        }

        public int getIndex() {
            return index;
        }

        public void setIndex(int index) {
            this.index = index;
        }
    }

    public String findLongestWord(String s, List<String> dictionary) {
        Map<Character, Deque<Pair>> map = new HashMap<>();
        for (char c = 'a'; c <= 'z'; c++) {
            map.put(c, new ArrayDeque<>());
        }

        for (String word : dictionary) {
            map.get(word.charAt(0)).offerFirst(new Pair(word, 0));
        }
        int maxLen = 0;
        String res = "";
        for (int i = 0; i < s.length(); i++) {
            if (!map.get(s.charAt(i)).isEmpty()) {
                Deque<Pair> deque = map.get(s.charAt(i));
                int size = deque.size();
                for (int j = 0; j < size; j++) {
                    Pair pair = deque.pollLast();
                    assert pair != null;
                    if (pair.index == pair.word.length() - 1) {
                        if (maxLen < pair.word.length()) {
                            maxLen = pair.word.length();
                            res = pair.word;
                        } else if (maxLen == pair.word.length() && res.compareTo(pair.word) > 0) {
                            res = pair.word;
                        }
                    } else {
                        pair.index++;
                        map.get(pair.word.charAt(pair.index)).offerFirst(pair);
                    }
                }
            }
        }
        return res;
    }
}

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