'''
Gist for Sergio ;) answering to a question in his interview

You are given a string s and an array of strings words.
You should add a closed pair of bold tag <b> and </b> 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.

Example 1:

Input: s = "abcxyz123", words = ["abc","123"]
Output: "<b>abc</b>xyz<b>123</b>"
Example 2:

Input: s = "aaabbcc", words = ["aaa","aab","bc"]
Output: "<b>aaabbc</b>c"

Return s after adding the bold tags.

It's hard to describe, so let's code:
'''

import re

def letsDoIt(s,chars):
    final = s
    final2 = final
    for word in chars:
        counter = -1
        target = ""
        final = final2
        open = -1
        for i in range(0,len(final)):

            target = final2[i:i+len(word)]

            if "<" not in target and ">" not in target:

                if counter >= 0:
                    counter += 1

                if counter == len(word) and open>0:
                    #needs a fix in one detected case, so this if will check if there is an open <b> more than </b>
                    if len(re.findall('<b>', final2)) > len(re.findall('</b>', final2)) and i+len(word)>final2.rfind("<b>")+len("<b>"):
                        counter = -1
                        final2 = final2[:i]+"</b>"+final2[i:]

                if target == word:
                    if counter == -1:
                        final2 = final2[:i]+"<b>"+final2[i:]
                        open = i
                    counter = 0

            else: # needed two cases: if we read from left to right, so never will find a '<b>', just in case of '</b>'
                if len(final2) > i+len(word)-1 and final2[i+len(word)-1] == '<':
                    temp = final2[i:]
                    temp = temp[:temp.find(">")+1+len(word)-1]
                    if word in temp.replace("</b>",""):
                        leftSize = len(temp.split("</b>")[0])
                        rightSize = len(word)-leftSize

                        temp = final2[:i]
                        temp2 = final2[i:i+leftSize]
                        temp3 = final2[i+leftSize+len("</b>"):i+leftSize+len("</b>")+rightSize]
                        temp3 += "</b>"+temp2[i+leftSize+len("</b>")+rightSize:]
                        temp4 = final2[i+len("</b>")+leftSize+rightSize:]

                        final2 = temp+temp2+temp3+temp4
                    elif word in temp[:temp.find("<b>")]+temp[temp.find("<b>")+len("<b>"):]:

                        leftSize = len(temp.split("<b>")[0])
                        rightSize = len(word)-leftSize

                        temp = final2[:i]+"<b>"
                        temp2 = final2[i:i+leftSize]
                        temp3 = final2[i+leftSize+len("<b>"):i+leftSize+len("<b>")+rightSize]
                        temp3 += temp2[i+leftSize+len("<b>")+rightSize:]
                        temp4 = final2[i+len("<b>")+leftSize+rightSize:]

                        final2 = temp+temp2+temp3+temp4

            #put it for and when len(word) < index
            '''
            if len(re.findall('<b>', final2))+1==len(re.findall('</b>', final2)) and final2.rfind("<b>") >= final2.rfind("</b>") and final2.rfind("<b>")+len(word)>=i:
                final2=final2[:final2.rfind("<b>")+len("<b>")+len(word)]+"</b>"+final2[final2.rfind("<b>")+len("<b>")+len(word):]
            '''
            #it's the same, done at different way
            if len(re.findall('<b>', final2)) > len(re.findall('</b>', final2)) and i+len(word)>final2.rfind("<b>")+len("<b>"):
                if i-open==len(word):
                    temp = final2[:final2.rfind("<b>")+len("<b>")+len(word)]
                    temp2 = final2[final2.rfind("<b>")+len("<b>")+len(word):]
                    final2 = temp+"</b>"+temp2


    #first replace is because requirements of the exercise say that
    #sorry for the second, time issues (just 15 minutes) in elif of special cases sometimes it writes two <b>, because I don't have time and simply just need a check if temp ends with <b> or not
    return final2.replace("</b><b>","").replace("<b><b>","")


if __name__ == "__main__":

    #s = "abcxyz123aaaaaaaaaa"
    #chars = ["abc","123"]

    s = "aaaabbcc"
    chars = ["aaa","bc","aab"]

    print(letsDoIt(s,chars))