Solutions

1. Mark Bold Characters

class Solution:
    def addBoldTag(self, s: str, words: List[str]) -> str:
        n = len(s)
        bold = [False] * n
        
        for word in words:
            start = s.find(word)
            while start != -1:
                for i in range(start, start + len(word)):
                    bold[i] = True
                    
                start = s.find(word, start + 1)

        open_tag = "<b>"
        close_tag = "</b>"
        ans = []
        
        for i in range(n):
            if bold[i] and (i == 0 or not bold[i - 1]):
                ans.append(open_tag)
                
            ans.append(s[i])
            
            if bold[i] and (i == n - 1 or not bold[i + 1]):
                ans.append(close_tag)
        
        return "".join(ans)

Time & Space Complexity

The time complexity may differ between languages. It is dependent on how the built-in method is implemented.
For this analysis, we will assume that we are using Java.

Time complexity: $O(m \cdot (n^2 \cdot k - n \cdot k^2))$
Space complexity: $O(n)$

Where $n$ is s.length, $m$ is words.length, and $k$ is the average length of the words.