LeetCode in Kotlin

1806. Minimum Number of Operations to Reinitialize a Permutation

Medium

You are given an even integer n. You initially have a permutation perm of size n where perm[i] == i (0-indexed).

In one operation, you will create a new array arr, and for each i:

• If i % 2 == 0, then arr[i] = perm[i / 2].
• If i % 2 == 1, then arr[i] = perm[n / 2 + (i - 1) / 2].

You will then assign arr to perm.

Return the minimum non-zero number of operations you need to perform on perm to return the permutation to its initial value.

Example 1:

Input: n = 2

Output: 1

Explanation: perm = [0,1] initially.

After the 1^st operation, perm = [0,1]

So it takes only 1 operation.

Example 2:

Input: n = 4

Output: 2

Explanation: perm = [0,1,2,3] initially.

After the 1^st operation, perm = [0,2,1,3]

After the 2^nd operation, perm = [0,1,2,3]

So it takes only 2 operations.

Example 3:

Input: n = 6

Output: 4

Constraints:

• 2 <= n <= 1000
• n is even.

Solution

class Solution {
    fun reinitializePermutation(n: Int): Int {
        val factor = n - 1
        if (factor < 2) {
            return 1
        }
        var powerOfTwo = 2
        var ops = 1
        while (powerOfTwo != 1) {
            powerOfTwo = (powerOfTwo shl 1) % factor
            ops++
        }
        return ops
    }
}

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