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# 524. Longest Word in Dictionary through Deleting

### Description

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.

### Constraints

* All the strings in the input will only contain lower-case letters.
* The size of the dictionary won't exceed 1,000.
* The length of all the strings in the input won't exceed 1,000.

### Approach

### Links

* GeeksforGeeks
* [Leetcode](https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/)
* ProgramCreek
* YouTube

### **Examples**

{% tabs %}
{% tab title="Example 1" %}
**Input:** s = "abpcplea", d = \["ale","apple","monkey","plea"]

**Output:** "apple"
{% endtab %}

{% tab title="Example 2" %}
**Input:** s = "abpcplea", d = \["a","b","c"]

**Output:** "a"
{% endtab %}
{% endtabs %}

### **Solutions**

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : O(n*x). One iteration over all strings is required. 
 *    Here n refers to the number of strings in list d and x refers to 
 *    average string length.
 * Space complexity : O(x). result variable is used.
 */

class Solution {
    public String findLongestWord(String s, List<String> d) {
        String result = "";
        if(s == null || d == null) {
            return result;
        }
        for(String str: d) {
            if(isSubsequence(s, str)) {
                if(str.length() > result.length() || 
                   str.length() == result.length() && str.compareTo(result) < 0) {
                    result = str;
                }
            }
        }
        return result;
    }
    
    private boolean isSubsequence(String s1, String s2) {
        if(s2.length() > s1.length()) {
            return false;
        }
        int i = 0;
        for(int j = 0; j < s1.length() && i < s2.length(); j++) {
            if(s1.charAt(j) == s2.charAt(i)) {
                i++;
            }
        }
        return i == s2.length();
    }
}
```

{% endtab %}
{% endtabs %}

### **Follow up**

*