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## Problem

You are given a string s and an array of strings words.

You should add a closed pair of bold tag and to wrap the substrings in s that exist in words.

• If two such substrings overlap, you should wrap them together with only one pair of closed bold-tag.
• If two substrings wrapped by bold tags are consecutive, you should combine them.

Return s after adding the bold tags.

Example 1:

Input: s = "abcxyz123", words = ["abc","123"]
Output: "<b>abc</b>xyz<b>123</b>"
Explanation: The two strings of words are substrings of s as following: "abcxyz123".
We add <b> before each substring and </b> after each substring.


Example 2:

Input: s = "aaabbb", words = ["aa","b"]
Output: "<b>aaabbb</b>"
Explanation:
"aa" appears as a substring two times: "aaabbb" and "aaabbb".
"b" appears as a substring three times: "aaabbb", "aaabbb", and "aaabbb".
We add <b> before each substring and </b> after each substring: "<b>a<b>a</b>a</b><b>b</b><b>b</b><b>b</b>".
Since the first two <b>'s overlap, we merge them: "<b>aaa</b><b>b</b><b>b</b><b>b</b>".
Since now the four <b>'s are consecuutive, we merge them: "<b>aaabbb</b>".


Constraints:

• 1 <= s.length <= 1000
• 0 <= words.length <= 100
• 1 <= words[i].length <= 1000
• s and words[i] consist of English letters and digits.
• All the values of words are unique.

Note: This question is the same as 758: https://leetcode.com/problems/bold-words-in-string/

## Code

class Solution {
public String addBoldTag(String s, String[] words) {
boolean[] bold = new boolean[s.length()];
int boldEnd = 0;

for (int i = 0; i < s.length(); i++) {
for (String word : words) {
if (s.startsWith(word, i)) {
boldEnd = Math.max(boldEnd, i + word.length());
}
}

bold[i] = boldEnd > i;
}

StringBuilder res = new StringBuilder();

for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);

if (!bold[i]) {
res.append(c);
continue;
}

// first bold
if(i == 0 || !bold[i - 1]) {
res.append("<b>");
}
res.append(c);
// last bold
if(i == s.length() - 1 || !bold[i + 1]) {
res.append("</b>");
}
}

return res.toString();
}
}