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Longest Word in Dictionary through Deleting

Updated: Mar 25, 2021

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



Example 1:

Input:
s = "abpcplea", d = ["ale","apple","monkey","plea"]

Output: 
"apple"



Example 2:

Input:
s = "abpcplea", d = ["a","b","c"]

Output: 
"a"


Note:

  1. All the strings in the input will only contain lower-case letters.

  2. The size of the dictionary won't exceed 1,000.

  3. The length of all the strings in the input won't exceed 1,000.


Solution:


Approach 1: Sorting and two pointers


class Solution {
    public String findLongestWord(String s, List<String> d) {
        String res="";
        Collections.sort(d, new Comparator < String > () {
            public int compare(String s1, String s2) {
                return s2.length() != s1.length() ? s2.length() - s1.length() : s1.compareTo(s2);
            }
        });
        for(String word:d)
        {
            if(word.equals(s)) return word;
            int i=0,j=0;
                for(;i<s.length();i++)
                {
                    if(s.charAt(i)==word.charAt(j)) j++;
                    if(j==word.length()) return word;
                }
        }
       return "";
    }
}


Approach 2: No sorting


class Solution {
    public String findLongestWord(String s, List<String> d) {
    
        String res="";
        for (String word: d)
            if ( (word.length()>res.length() || word.length()==res.length()&& (word.compareTo(res)<0)) && isSubseq(word, s))               
                res=word;
        return res;
    }
    public boolean isSubseq(String a, String b)
    {
        int j = -1;
        for(int i = 0; i < a.length(); i++)
        {
            j = b.indexOf(a.charAt(i), j + 1);
            if(j == -1) {
                return false;
            }
        }
        return true;
    }
}
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