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Just split one Stirng (a1b1a2b2a3b3.....) to two sets of odd and even (a1, a2, a3 ...) and (b1, b2, b3...). Sort them and put the two sorted together.
Count the unique strings as processed above.
Based on the theory that for a given sequence (a1, a2, a3 .... an)
ALL n! permutations can be get by exchange ax and ay (1<=x<=y<=n). This can be easily proven by Mathematical Induction.
code:
- lff0305 December 02, 2019