LeetCode in Kotlin

1238. Circular Permutation in Binary Representation

Medium

Given 2 integers n and start. Your task is return any permutation p of (0,1,2.....,2^n -1) such that :

• p[0] = start
• p[i] and p[i+1] differ by only one bit in their binary representation.
• p[0] and p[2^n -1] must also differ by only one bit in their binary representation.

Example 1:

Input: n = 2, start = 3

Output: [3,2,0,1]

Explanation: The binary representation of the permutation is (11,10,00,01). All the adjacent element differ by one bit. Another valid permutation is [3,1,0,2]

Example 2:

Input: n = 3, start = 2

Output: [2,6,7,5,4,0,1,3]

Explanation: The binary representation of the permutation is (010,110,111,101,100,000,001,011).

Constraints:

• 1 <= n <= 16
• 0 <= start < 2 ^ n

Solution

class Solution {
    fun circularPermutation(n: Int, start: Int): List<Int> {
        val l1: MutableList<Int> = ArrayList()
        for (i in 0 until (1 shl n)) {
            l1.add(start xor (i xor (i shr 1)))
        }
        return l1
    }
}

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