For each word in the dictionary, we check if it is a subsequence of s using Two Pointers. We update the result based on the length and lexicographical order.

1. res = "".
2. For each word in dictionary:
   • i = 0 (for word), j = 0 (for s).
   • While i < len(word) and j < len(s):
     • If word[i] == s[j], i += 1.
     • j += 1.
   • If i == len(word) (subsequence found):
     • If len(word) > len(res) or (len(word) == len(res) and word < res):
       • res = word.
3. Return res.

• Time Complexity: O(N * M) where N is dictionary size and M is avg length of s.
• Space Complexity: O(1).

def find_longest_word(s, dictionary):
    res = ""
    for word in dictionary:
        i = 0
        for char in s:
            if i < len(word) and word[i] == char:
                i += 1
        
        if i == len(word):
            if len(word) > len(res) or (len(word) == len(res) and word < res):
                res = word
                
    return res