Problem Description

The problem asks to find a circular permutation of integers from 0 to 2n - 1 such that consecutive numbers differ by only one bit in their binary representation, and the last and first numbers also differ by one bit. The permutation must start with a given start integer.

Solution Approach

The core idea revolves around Gray codes. A Gray code is a binary numeral system where two successive values differ in only one bit. We can leverage the property of Gray codes to construct the required permutation.

There are two main approaches:

Approach 1: Generate Gray Codes, then rearrange

1. Generate Gray Codes: Generate all Gray codes for n bits. A simple way is using the formula gray(i) = i ^ (i >> 1).
2. Find Start Index: Locate the index of start within the generated Gray codes.
3. Rearrange: Split the Gray code array at the start index, and concatenate the two parts to form the circular permutation.

Approach 2: Optimized Gray Code Generation

This approach directly generates the desired permutation without the need for rearrangement. Since gray(0) = 0, we can calculate gray(i) ^ start for each i from 0 to 2n - 1. This directly generates the permutation starting with start.

Code Implementation (with explanations)

Approach 1 (Generate, then rearrange):

Python:

def circularPermutation(n: int, start: int) -> List[int]:
    gray_codes = [i ^ (i >> 1) for i in range(1 << n)] # Generate Gray codes
    start_index = gray_codes.index(start) #Find index of start
    return gray_codes[start_index:] + gray_codes[:start_index] #Rearrange

Java:

import java.util.*;
class Solution {
    public List<Integer> circularPermutation(int n, int start) {
        List<Integer> grayCodes = new ArrayList<>();
        for (int i = 0; i < (1 << n); ++i) {
            grayCodes.add(i ^ (i >> 1)); 
        }
        int startIndex = grayCodes.indexOf(start);
        List<Integer> result = new ArrayList<>(grayCodes.subList(startIndex, grayCodes.size()));
        result.addAll(grayCodes.subList(0, startIndex));
        return result;
    }
}

Approach 2 (Optimized Generation):

Python:

def circularPermutation(n: int, start: int) -> List[int]:
    return [i ^ (i >> 1) ^ start for i in range(1 << n)]

Java:

import java.util.*;
class Solution {
    public List<Integer> circularPermutation(int n, int start) {
        List<Integer> result = new ArrayList<>();
        for (int i = 0; i < (1 << n); ++i) {
            result.add(i ^ (i >> 1) ^ start);
        }
        return result;
    }
}

Time and Space Complexity Analysis

Approach 1:

• Time Complexity: O(2n). Generating Gray codes takes O(2n) time. Finding the index and rearranging also take O(2n) in the worst case.
• Space Complexity: O(2n) to store the Gray codes array.

Approach 2:

• Time Complexity: O(2n). The loop iterates 2n times.
• Space Complexity: O(2n) to store the result list. However, if we disregard the space for the output list, the space complexity is O(1).

Approach 2 is slightly more efficient as it avoids the extra step of finding the index and rearranging the array. Both approaches have the same dominant time complexity.