LeetCode #1238 — MEDIUM

Circular Permutation in Binary Representation

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Backtracking · Bit Manipulation

Example 1

2
3

Example 2

3
2

Step 02

Core Insight

What unlocks the optimal approach

Use gray code to generate a n-bit sequence.
Rotate the sequence such that its first element is start.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1238: Circular Permutation in Binary Representation
class Solution {
    public List<Integer> circularPermutation(int n, int start) {
        int[] g = new int[1 << n];
        int j = 0;
        for (int i = 0; i < 1 << n; ++i) {
            g[i] = i ^ (i >> 1);
            if (g[i] == start) {
                j = i;
            }
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = j; i < j + (1 << n); ++i) {
            ans.add(g[i % (1 << n)]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1238: Circular Permutation in Binary Representation
func circularPermutation(n int, start int) []int {
	g := make([]int, 1<<n)
	j := 0
	for i := range g {
		g[i] = i ^ (i >> 1)
		if g[i] == start {
			j = i
		}
	}
	return append(g[j:], g[:j]...)
}

# Accepted solution for LeetCode #1238: Circular Permutation in Binary Representation
class Solution:
    def circularPermutation(self, n: int, start: int) -> List[int]:
        g = [i ^ (i >> 1) for i in range(1 << n)]
        j = g.index(start)
        return g[j:] + g[:j]

// Accepted solution for LeetCode #1238: Circular Permutation in Binary Representation
struct Solution;

impl Solution {
    fn circular_permutation(n: i32, start: i32) -> Vec<i32> {
        let mut res = vec![];
        for i in 0..1 << n {
            res.push(start ^ (i ^ i >> 1));
        }
        res
    }
}

#[test]
fn test() {
    let n = 2;
    let start = 3;
    let res = vec![3, 2, 0, 1];
    assert_eq!(Solution::circular_permutation(n, start), res);
    let n = 3;
    let start = 2;
    let res = vec![2, 3, 1, 0, 4, 5, 7, 6];
    assert_eq!(Solution::circular_permutation(n, start), res);
}

// Accepted solution for LeetCode #1238: Circular Permutation in Binary Representation
function circularPermutation(n: number, start: number): number[] {
    const ans: number[] = [];
    for (let i = 0; i < 1 << n; ++i) {
        ans.push(i ^ (i >> 1) ^ start);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(2^n)

Space

O(2^n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.