In the recently concluded Long challenge, in the problem “CLONEME”, I wrote an optimal brute force solution. The solution is as follows :
 I stored the prefix sum, sum of square and sum of cubes of elements in an array.
 Check if the 2 subarrays are permutation of each other or not. For this simply check if the sum, sum of squares, sum of cubes etc are exactly same are not.
 Now, we need to deal with one mismatch only. Now, I found the difference of sum and sum of sqaures of 2 subarrays. If only one element mismatches(let them be a and b, then we have the following equations :
S_1  S_2 = a  b and {S_1}^2  {S_2}^2 = a^2  b^2.
Then a + b = \frac{{S_1}^2  {S_2}^2}{S_1  S_2} i.e. {S_1}^2  {S_2}^2 should be divisble by S_1  S_2.
Also, if S_1 == S_2, this means atleast 2 elements differ as, there not permutation as assured by the prefix sums checked above.

Now, using difference in sum of cubes in the 2 subarrays, I confirm again if the 2 elements I found are valid out. i.e if the difference of the sum of cubes of 2 subarrays is exactly equal to the difference of the cube of the 2 numbers found.

If all the above methods fails, I simply do a brute force algorithm, i.e. construct the 2 subarrays, sort them and then check if they mismatch at one position only.
I had tried to construct test cases such that my solution TLE, but was unsuccessful. Can someone here would like to challenge my solution for the problem ?
Remember, that almost all the queries should be different as I should cache the previous queries answer which is generally used and such test cases are avoided by making problem. (i.e. if you find a particluar query for which my solution will execute the brute force solution, then do not give the same query again and again in the test case).
Happy coding