Minimum number of iterations - c++

Question

We are given an array with numbers from ranging from 1 to n (no duplicates) where n = size of the array.
We are allowed to do the following operation :
arr[i] = arr[arr[i]-1] , 0 <= i < n
Now, one iteration is considered when we perform above operation on the entire array.
Our task is to find the number of iterations after we encounter a previously encountered sequence.
Constraints :
a) Array has no duplicates
b) 1 <= arr[i] <= n , 0 <= i < n
c) 1 <= n <= 10^6
Ex 1:
n = 5
arr[] = {5, 4, 2, 1, 3}
After 1st iteration array becomes : {3, 1, 4, 5, 2}
After 2nd iteration array becomes : {4, 3, 5, 2, 1}
After 3rd iteration array becomes : {2, 5, 1, 3, 4}
After 4th iteration array becomes : {5, 4, 2, 1, 3}
In the 4th iteration, the sequence obtained is already seen before
So the expected output is 4.
This question was asked in one of job hiring tests, so I dont have any link to the question.
There were 2 sample test cases given out of which I remember one which is given above. I would really appreciate any help on this question
P.S.
I was able to code the brute force solution, where in I stored all the results in a Set and then kept advancing to the next permutation. But it gave TLE

Answer 1

First, note that an array of length n containing 1, 2, ..., n with no duplicates is a permutation.
Next, observe that arr[i] := arr[arr[i] - 1] is squaring the permutation.
That is, consider permutations as elements of the symmetric group S_n, where multiplication is composition of permutations.
Then the above operation is arr := arr * arr.
So, in terms of permutations and their composition, the question is as follows:
You are given a permutation p (= arr).
Consider permutations p, p^2, p^4, p^8, p^16, ...
What is the number of distinct elements among them?
Now, to solve it, consider the cycle notation of the permutation.
Every permutation is a product of disjoint cycles.
For example, 6 1 4 3 5 2 is the product of the following cycles: (1 6 2) (3 4) (5).
In other words, every application of this permutation:
moves elements at positions 1, 6, 2 along the cycle;
moves elements at positions 4, 3 along the cycle;
leaves element at position 5 in place.
So, when we consider p^k (take an identity permutation and apply the permutation p to it k times), we actually process three independent actions:
move elements at positions 1, 6, 2 along the cycle, k times;
move elements at positions 4, 3 along the cycle, k times;
leave element at position 5 in place, k times.
Now, take into account that, after d applications of a cycle of length d, it just returns all the respective elements to their initial places.
So, we can actually formulate p^k as:
move elements at positions 1, 6, 2 along the cycle, (k mod 3) times;
move elements at positions 4, 3 along the cycle, (k mod 2) times;
leave element at position 5 in place.
We can now prove (using Chinese Remainder Theorem, or just using general knowledge of group theory) that the permutations p, p^2, p^3, p^4, p^5, ... are all distinct up to p^m, where m is the least common multiple of all cycle lengths.
In our example with p = 6 1 4 3 5 2, we have p, p^2, p^3, p^4, p^5, and p^6 all distinct.
But p^6 is the identity permutation: moving six times along a cycle of length 2 or 3 results in the items at their initial places.
So p^7 is the same as p^1, p^8 is the same as p^2, and so on.
Our question however is harder: we want to know the number of distinct permutations not among p, p^2, p^3, p^4, p^5, ..., but among p, p^2, p^4, p^8, p^16, ...: p to the power of a power of two.
To do that, consider all cycle lengths c_1, c_2, ..., c_r in our permutation.
For each c_i, find the pre-period and period of 2^k mod c_i:
For example, c_1 = 3, and 2^k mod 3 look as 1, 2, 1, 2, 1, 2, ..., which is (1, 2) with pre-period 0 and period 2.
As another example, c_2 = 2, and 2^k mod 2 look as 1, 0, 0, 0, ..., which is 1, (0) with pre-period 1 and period 1.
In this problem, this part can be done naively, by just marking visited numbers mod c_i in some array.
By Chinese Remainder Theorem again, after all pre-periods are considered, the period of the whole system of cycles will be the least common multiple of all individual periods.
What remains is to consider pre-periods.
These can be processed with your naive solution anyway, as the lengths of pre-periods here is at most log_2 n.
The answer is the least common multiple of all individual periods, calculated as above, plus the length of the longest pre-period.