Python O(n) Easy to understand one pass solution with detailed explanation 100% time and space
Create a map for each string and its corresponding abbrevation.\nIterate the string, check if we find a \'&\' if we do then we find the next \';\' check if the 
Create a map for each string and its corresponding abbrevation.\nIterate the string, check if we find a \'&\' if we do then we find the next \';\' check if the substring between \'&->;\' is in hashmap if so replace it and continue iterating\nOtherwise we add each character.\nTime complexity : O(n) \n```   \n def entityParser(self, text: str) -> str:\n        res,i = "",0\n        hmap = {\'&quot;\' : \'"\', \'&apos;\' : \'\\\'\', \'&amp;\' : \'&\', \'&gt;\' : \'>\',\'&lt;\' : \'<\', \'&frasl;\' : \'/\'}\n        while i < len(text):\n            if text[i] == \'&\':\n                j=text.find(\';\',i+1)\n                if j != -1 and text[i:j+1] in hmap:\n                    res+=hmap[text[i:j+1]]\n                    i=j+1\n                    continue\n            res+=text[i]\n            i+=1\n                \n        return res\n```
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