Formatted question description: https://leetcode.ca/all/1238.html

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

Related Topics:
Math

Solution 1. Gray Code

// OJ: https://leetcode.com/problems/circular-permutation-in-binary-representation/

// Time: O(2^N)
// Space: O(1)
// Ref: https://leetcode.com/problems/circular-permutation-in-binary-representation/discuss/414203/JavaC%2B%2BPython-4-line-Gray-Code
class Solution {
public:
    vector<int> circularPermutation(int n, int start) {
        vector<int> ans;
        for (int i = 0; i < (1 << n); ++i) ans.push_back(start ^ i ^ (i >> 1));
        return ans;
    }
};

Java

class Solution {
    public List<Integer> circularPermutation(int n, int start) {
        List<Integer> grayCodeList = new ArrayList<Integer>();
        grayCodeList.add(0);
        grayCodeList.add(1);
        for (int i = 2; i <= n; i++) {
            int power2 = (int) Math.pow(2, i - 1);
            int size = grayCodeList.size();
            for (int j = size - 1; j >= 0; j--) {
                int prevNum = grayCodeList.get(j);
                int curNum = prevNum + power2;
                grayCodeList.add(curNum);
            }
        }
        List<Integer> permutation = new ArrayList<Integer>();
        int index = grayCodeList.indexOf(start);
        int size = grayCodeList.size();
        for (int i = 0; i < size; i++) {
            permutation.add(grayCodeList.get(index));
            index = (index + 1) % size;
        }
        return permutation;
    }
}

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