minimum is  always -1, 1, or 2 ?
1) S = aaaa...aa   => -1        (S = [a]*n)
2) S = aaabaaa   on center different symbol ==> -1 (S = [a]*m + 'b' +[a]*m)
3) S = ababababa...ababa  ==> -1   (S = [ab]*m + 'a' there S is palindrome)
4) S - is not palindrome,  ==> 1
5) S - is palindrome, there at least two different symbols => 2
   Proof 5:
let a = S[1], b = S[i] , and all S[1] = S[2] = ..=S[i-1] = a,  There S is palindrome.
    S = aaaabxy....baaaa
   Take T1 = aaab  T2 = xy...baaaa
   T1 - is not palindrome

    5.1)  T2 also is not palindrome, minimum cut is 2.
    5.2) T2 also palindrome
         5.2.1) i > 2, there two or more 'a' in T1.
           T1 = aaabx   T2 = y....baaaa
          There T1 and T2 both are not palndrome, (think about it).
         5.2.2. i = 2.  T1 = ab  But T2 palindrome, so S = abababababa...aba, this can't be allowed, because S != [ab]*m + 'a'  already checked.

My brute force solution pass the tests, the run id is 6945666. Maybe you can add more strong tests.