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    \$\begingroup\$ Rule 10 says this is not what he wants. Also, did you mean s[0]? s[1] will not work if s has less than 2 characters, while s[0] works as long as s is not empty. \$\endgroup\$ Commented Jan 13, 2014 at 5:05
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    \$\begingroup\$ I thought my solution sufficiently different from AaA. No idea why I used s[1] initially. \$\endgroup\$ Commented Jan 13, 2014 at 5:14