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 a, b, ..... y, z ^ ...... ^ <- moving first ^ ...... ^ <- moving last 

 a, b, ..... y, z ^ ...... ^ <- moving first ^ ...... ^ <- moving last 

We can combine both conditions into if and have just one loop:

 int result = 0; for (int i = 0; i < A.length; ++i) if (A[i] != A[A.length - 1] || A[0] != A[A.length - 1 - i]) { result = A.length - i - 1; break; }  

Edit: Let's proof that at least one of the indexes is either 0 or length - 1. Let's do it by the contradiction. Suppose we have a solution like

Edit: Let's proof that at least one of the indexes is either 0 or length - 1. Let's do it by contradiction. Suppose we have a solution like

Items to the left of c must be d, otherwise we can take a or b indexes and have an improved solution. Items to right of d must be c or we can once again push last to the right and have a better solution. So we have

Items to the left of c must be d, otherwise we can take a or b indexes and have an improved solution. Items to right of d must be c or we can once again push last index to the right and have a better solution. So we have