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# Runtime: 271 ms (Top 15.83%) | Memory: 15.4 MB (Top 41.73%)
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class Solution:
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    def checkPalindromeFormation(self, a: str, b: str) -> bool:
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        def pal(x):
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            return x == x[::-1]
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        if pal(a) or pal(b):
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            return True
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        # either grow from inside to outside, or vice versa
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        ina = len(a) - 1
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        inb = 0
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        outa = 0
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        outb = len(b) - 1
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        while a[ina] == b[inb]:
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            ina -= 1
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            inb += 1
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            if ina <= inb:
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                return True  # short circuit found break point
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        # jump into each string now!?
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        # is a or b a palindrome in this portion from inb to ina
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        if pal(a[inb : ina + 1]) or pal(b[inb : ina + 1]):
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            return True  # either one is breakpoint, so check remainder is palindrome
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        while a[outa] == b[outb]:
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            outa += 1
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            outb -= 1
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            if outa >= outb:
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                return True
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        if pal(a[outa : outb + 1]) or pal(b[outa : outb + 1]):
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            return True  # either one is breakpoint, so check remainder
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        return False