Minimum Number of Operations to Reinitialize a Permutation

You are given an even integer n. You start with a permutation perm of size n where perm[i] = i. In one operation, you create a new array arr where: for even i, arr[i] = perm[i / 2], and for odd i, arr[i] = perm[n / 2 + (i - 1) / 2]. Then you assign perm = arr. Return the minimum non-zero number of operations needed to get back to the initial permutation.

This is LeetCode 1806: Minimum Number of Operations to Reinitialize a Permutation.

Why this problem matters

At first glance this looks like a simulation problem: apply the transformation, check if you are back to the start, repeat. And that works. But there is a much deeper idea hiding underneath. The operation defines a permutation on indices, and finding how many times you need to apply it is the same as finding the length of a cycle in that permutation. This concept, called the order of a permutation in group theory, appears in many math-heavy interview problems. Recognizing it lets you skip brute-force simulation of the full array and solve the problem in O(n) time with O(1) space.

The key insight

You do not need to simulate the entire array. Look at what the operation does to indices 0 and n-1: they never move. Now look at any index i between 1 and n-2. After one operation, index i moves to position 2 * i % (n - 1). That is the whole rule for the "middle" elements.

So the question becomes: starting from position 1, how many times do you need to double (mod n-1) before you land back on 1? In number theory terms, you are looking for the multiplicative order of 2 modulo n-1. You just follow position 1 through its orbit until it returns home.

The solution

def reinitialize_permutation(n: int) -> int:
    if n == 2:
        return 1
    ops = 0
    pos = 1
    while True:
        # Apply the mapping: even positions go to pos*2, odd go to 2*pos - (n-1)
        if pos < n // 2:
            pos = pos * 2
        else:
            pos = (pos - n // 2) * 2 + 1
        ops += 1
        if pos == 1:
            return ops

Here is an equivalent and cleaner formulation using modular arithmetic:

def reinitialize_permutation(n: int) -> int:
    ops = 0
    val = 1
    while True:
        val = val * 2 % (n - 1)
        ops += 1
        if val == 1:
            return ops

The first version simulates where index 1 travels using the exact rule from the problem. If the current position is in the first half of the array (an even destination), multiply by 2. Otherwise, use the odd-destination formula. The second version collapses both cases into val * 2 % (n - 1), which is valid for every index from 1 to n-2. Both return the same answer.

Visual walkthrough

Let's trace all four operations for n = 6 until the array returns to its starting configuration.

After four applications of the rule, every element is back where it started. Notice how 0 and 5 never moved. The four middle elements (indices 1 through 4) cycled through the same set of positions and all returned home at exactly the same step. This is not a coincidence. They all belong to the same permutation cycle, so they all have the same cycle length.

Complexity analysis

Time: O(n) in the worst case for the single-element approach. The cycle length is at most n-1, so you perform at most n-1 doublings.

Space: O(1) since you only track one integer position.

The building blocks

1. Permutation cycle detection

The core technique here is following a single element through repeated application of a permutation until it returns to its starting point. This pattern works for any permutation, not just this specific one.

pos = start
count = 0
while True:
    pos = permute(pos)
    count += 1
    if pos == start:
        return count

You pick any element, apply the mapping, and count iterations. The moment you see the starting value again, you have the cycle length for that element's orbit.

2. Multiplicative order

The cycle length for this problem equals the multiplicative order of 2 modulo n-1. That is the smallest positive integer k such that 2^k is congruent to 1 mod (n-1). You can compute it directly:

val = 2
ops = 1
while val % (n - 1) != 1:
    val = val * 2 % (n - 1)
    ops += 1

This is mathematically identical to tracking index 1 through the permutation. Every doubling mod (n-1) corresponds to one application of the operation.

Edge cases

• n = 2: The only non-fixed element is at index 1 in a 2-element array, so one operation returns it. Answer: 1.
• n = 4: The cycle length is 2. After two operations the array is back to the identity.
• Large even n: The cycle length can vary, but it is always bounded by n-1.
• Powers of 2: These tend to produce specific cycle patterns. For example, n = 8 gives a cycle length of 3, while n = 16 gives 4.

From understanding to recall

The jump from "simulate the full array every step" to "just follow one element" is the kind of optimization insight that shows up repeatedly in permutation problems. Once you see that the operation is really just multiplication by 2 in a modular world, the whole problem collapses to a simple loop. Spaced repetition helps make this cycle-detection reflex automatic, so the next time you see a problem asking "how many times until we return to the start," you reach for the single-element orbit trick without hesitation.

Related posts

• Next Permutation - Another permutation manipulation problem requiring understanding of element positions
• Permutations - Builds the foundation of how permutations work
• Rotate Array - A simpler cyclic operation on arrays that shares the "when does it return to start" flavor

If you want to build lasting fluency with permutation cycles, simulation problems, and modular arithmetic tricks, CodeBricks gives you a spaced repetition system designed for exactly these patterns. You practice a problem, move on, and revisit it at the optimal moment before you forget.