### anupomkar's blog

By anupomkar, history, 2 months ago,

Given a permutation array P (1-based indexing), find the total number of steps required to sort the same array using the permutation array.

Constraints: n = arr.length 2<=n<=10^5

Example: Say P =[2,5,4,3,1]

Copy the array from P to arr arr = [2,5,4,3,1] step1 -> arr = [5,1,3,4,2] {Here the elements are arranged according to the permutation array, (i.e, permutation array P is considered here as index array to arrange elements) Here, 2nd element (5) comes 1st place because in P, 2 is at 1st position 5th element (1) comes 2nd place because in P, 5 is at 2nd position and so on...}

step2 -> arr = [1,2,4,3,5] step3 -> arr = [2,5,4,3,1] step4 -> arr = [5,1,4,3,2] step5 -> arr = [1,2,3,4,5]

Hence, the total number of steps required = 5

Expecting O(n) solution according to given constraints.

• +4

 » 2 months ago, # |   +1 For a permutation $p$ with length $n$, consider a graph consisting of $n$ vertices numbered from $1$ to $n$ , with a directed edge $u \to v$ iff $p_u=v$. Since every vertex has exactly one incoming edge and one outgoing edge, the graph must be made up of several cycles.Then the problem can be rephrased as follows:There are $n$ chesses on the graph. Initially, chess $i(1 \le i \le n)$ is on the vertex $p_i$. Every turn, all chesses will simultaneously move forward along one edge. After how many turns, every chess $i$ is exactly on vertex $i$ (for the first time)?The answer to the problem is obviously the $\operatorname{lcm}$ of the lengths of the cycles in the graph minus one. ($\operatorname{lcm}$: the Lowest Common Multiple)By calculating $\gcd$ (Greatest Common Divisor) in $O(\log{n})$ time, we can eaily get an $O(n\log{n})$ solution. However, the answer to the problem may be too large, so that probably we need to get the answer modulo by some number. Instead of calculate $\operatorname{lcm}$ directly, we can precompute the prime factors of the lengths of the cycles, so that we can simply calculate $\max$ of the exponents of each prime factor. Finally, we just need to multiple the powers of the primes together, getting the answer modulo by some number. The total time complexity is $O(n)$, implementing properly.
•  » » 2 months ago, # ^ |   0 I totally forgot to consider the LCM, I kept on talking the product of all cycle lengths. How did you realise that LCM has to be calculated? And are there any similar problems you've seen before, if yes please mention them here?
•  » » » 2 months ago, # ^ |   +1 For a cycle of length $k$, the chess on it will go back to the same place every $k$ turns. So if every chess has to go back, the number of turns must be the multiple of the length of every cycle, and the smallest one is naturally the $\operatorname{lcm}$ of the lengths.I believe this problem is classic, especially the trick to divide a permutation into several cycles. However, I can't list specific problems out.
•  » » » » 2 months ago, # ^ |   0 Thanks! Understood the thing behind it.
•  » » » 7 weeks ago, # ^ |   +1 I think this problem is similar, https://codeforces.com/contest/1690/problem/F
•  » » 5 weeks ago, # ^ |   0 Also, you can find the steps needed to achieve any arbitrary array (or report that it is impossible to achieve) with the Chinese Remainder Theorem.
 » 7 weeks ago, # |   0 Did you get the same problem in gameskraft?.I got it in that one
•  » » 7 weeks ago, # ^ |   0 Yes! Its the OA question of Gameskraft
•  » » » 5 weeks ago, # ^ |   0 were there any other problems ?