[Python3] Two pointer fast solution
Setting up two pointers i, j for s and t, respectively.\n2. if s[i] == t[j], both i and j + 1\n3. if s[i] != t[j], seaching in i using i0 and look for t[j] and 
1. Setting up two pointers i, j for s and t, respectively.\n2. if s[i] == t[j], both i and j + 1\n3. if s[i] != t[j], seaching in i using i0 and look for t[j] and make sure there is nothing smaller than t[j] before find that, if can not find, return False\n4. Once find a s[i0] == t[j], continue searching if t[j+1] == t[j] and s[i0 + 1] == s[i0] untile s[i1] != s[i0].\n5. remove s[i0:i1] from s and continue the searching\n\n```\nclass Solution:\n    def isTransformable(self, s: str, t: str) -> bool:\n        i = 0\n        j = 0\n        \n        while i < len(s) and j < len(t):\n            if s[i] == t[j]:\n                i += 1\n                j += 1\n            else:\n                i0 = i\n                while i0 < len(s) and s[i0] > t[j]:\n                    i0 += 1\n                if s[i0] == t[j]:\n                    i1 = i0\n                    while i1 < len(s) and j < len(t) and s[i1] == t[j] == s[i0]:\n                        i1 += 1\n                        j += 1\n                    s= s[:i0] + s[i1:]\n                else:                \n                    return False\n        return True
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