• 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 1: Gray Code (DP) + Rotation

Gray code starts with 0, need to rotate after generating the list.

Time complexity: O(2^n)
Space complexity: O(2^n)

// Author: Huahua
class Solution {
public:
  vector<int> circularPermutation(int n, int start) {
    vector<vector<int>> dp(n + 1);
    dp[0] = {0};    
    for (int i = 1; i <= n; ++i) {
      dp[i] = dp[i - 1];
      for (int j = dp[i - 1].size() - 1; j >= 0; --j)
        dp[i].push_back(dp[i - 1][j] | (1 << (i - 1)));
    }
    for (auto it = begin(dp[n]); it != end(dp[n]); ++it)
      if (*it == start) {
        rotate(begin(dp[n]), it, end(dp[n]));
        break;
      }
    return dp[n];
  }
};

Solution 2: Gray code with a start

Time complexity: O(2^n)
Space complexity: O(1)

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

The post 花花酱 LeetCode 1238. Circular Permutation in Binary Representation appeared first on Huahua's Tech Road.