Description Link to heading

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 Link to heading

It’s the same as 89.gray-code, we just make res[i] = res[i] ^ start, since x ^ 0 = x holds for any x.

Code Link to heading 

class Solution {
  public:
    vector<int> circularPermutation(int n, int start) {
        vector<int> res(1 << n, 0);
        for (int i = 0; i < res.size(); i++) {
            res[i] = i ^ (i / 2) ^ start;
        }
        return res;
    }
};