# 1238. Circular Permutation in Binary Representation
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).
# Solution
Approach 1: generate gray codes.
# Code (Python)
Approach 1:
def circularPermutation(self, n: int, start: int) -> List[int]:
# idea: generate naively.
# n bits starts with 'start', example:
#000
#001
#011
#010
#110
#111
#101
#100
#1100
# basically for each extra digit, flip that digit and splice the reverse of previous results
result = [start, start ^ 1]
for i in range(1, n):
# tip: flip a bit by XORing with a mask 0..010..0
higher_bits = (result[-1] ^ (1 << i)) & ~ ((1 << i) - 1)
for num in result[::-1]:
result.append((num & ((1 << i) - 1)) | higher_bits)
return result
def circularPermutation(self, n: int, start: int) -> List[int]:
# idea: generate gray code, then shift them
codes = [i ^ (i >> 1) for i in range(1 << n)]
index = codes.index(start) # find the starting point
return codes[index: ] + codes[: index]
def circularPermutation(self, n: int, start: int) -> List[int]:
# somehow(?) shift the start
return [start ^ i ^ (i >> 1) for i in range(1 << n)]
Approach 2:
# Code (C++)
Approach 1:
Approach 2: