Given a string s and a list of strings dict, you need to add a closed pair of bold tag <b> and </b> to wrap
the substrings in s that exist in dict. If two such substrings overlap, you need to wrap them together by
only one pair of closed bold tag. Also, if two substrings wrapped by bold tags are consecutive, you need to
combine them.
Example 1:
Input:
s = "abcxyz123"
dict = ["abc","123"]
Output:
"<b>abc</b>xyz<b>123</b>"
Example 2:
Input:
s = "aaabbcc"
dict = ["aaa","aab","bc"]
Output:
"<b>aaabbc</b>c"
Note:
The given dict won't contain duplicates, and its length won't exceed 100.
All the strings in input have length in range [1, 1000].
public String addBoldTag(String input, String[] dict) {
HashSet<String> set = new HashSet<>();
int max = 0;
for(String s : dict) {
set.add(s);
max = Math.max(max, s.length());
}
ArrayList<int[]> arr = new ArrayList<>();
for(int i = 0; i < input.length(); i++) {
for(int j = Math.min(i + max, input.length()); j > i; j--) {
String candidate = input.substring(i, j);
if(set.contains(candidate)) {
if(arr.size() > 0 && arr.get(arr.size()-1)[1] >= i-1) {
arr.get(arr.size()-1)[1] = Math.max(j-1, arr.get(arr.size()-1)[1]);
} else {
arr.add(new int[]{i, j-1});
}
break;
}
}
}
StringBuilder sb = new StringBuilder();
int j = 0;
for(int i = 0; i < input.length(); i++) {
if(j < arr.size() && i == arr.get(j)[0]) sb.append("<b>");
sb.append(input.charAt(i));
if(j < arr.size() && i == arr.get(j)[1]) {
sb.append("</b>");
j++;
}
}
return sb.toString();
}