Python Fast Solution
\nclass Solution:\n    def isTransformable(self, s: str, t: str) -> bool:\n        # If "s" is Already sorted in desc and has same characters as in "t" return T
```\nclass Solution:\n    def isTransformable(self, s: str, t: str) -> bool:\n        # If "s" is Already sorted in desc and has same characters as in "t" return True\n        # If "s" is Already sorted in asc and "s" not equal to "t" return False\n        \n        sorted_s = sorted(s)\n        sorted_t = sorted(t)\n        if sorted_s == sorted_t :\n            if \'\'.join(sorted_s[::-1]) == s:\n                return True\n            elif \'\'.join(sorted_s) == s and s != t:\n                return False\n        #Try Removing all repeating characters as 1111111 or 99999  by 1,9, ... etc from both strings\n        # if the length of remaining string remain unchanged continue with the string else reset the strings to the orignal value\n        # though its a overhead for most cases: last few tescases could be handled with following logic\n        tl = [t[0]]\n        sl = [s[0]]\n        \n        for i in range(1,len(t)):\n            if tl[-1] != t[i]:\n                tl.append(t[i])\n                \n        for i in range(1,len(s)):\n            if sl[-1] != s[i]:\n                sl.append(s[i])\n        if len(tl) != len(sl):\n            tl = list(t)\n            sl = list(s)\n        i = 0\n        # Then we simply check if char in "t" can be positioned in the given position in "s"         \n\t\t# we check for char in t if matches than increament the pointer in i\n        # we will only compare s[i:] and t[i:] as we increament the pointer by 1\n        # Return False if t[i] is not a part of s[i:]\n       \n        while i < len(tl):\n            # Characted Matched so we continue \n            if tl[i] == sl[i]:\n                i += 1\n                continue\n            else:\n                if tl[i] not in sl[i:]:\n                    return False\n                j = i + sl[i:].index(tl[i])\n                if j == i:\n                    i += 1\n                    continue\n                else:\n                    curr = sl[j]\n                    prev = sl[j-1]\n                    # Since sl[j-1] < sl[j]: ALready Sorted and does not match with t return False\n                    if prev < curr:\n                        return False\n                    else:\n                        sl[j-1] = curr\n                        sl[j] = prev\n        if sl == tl:\n            return True\n        else:\n            return False\n                \n\n```
['Python']