Circular Permutation in Binary Representation - Video Solutions

Problem Overview: You need to generate a permutation of all 2^n integers represented with n bits such that adjacent numbers differ by exactly one bit. The sequence must also be circular, meaning the last and first numbers differ by one bit as well. The permutation must start with a given value start.

Approach 1: Gray Code Generation (Time: O(2^n), Space: O(2^n))

The key observation is that Gray code already produces a sequence where consecutive numbers differ by exactly one bit. The standard formula for the i-th Gray code is g = i ^ (i >> 1). This naturally generates a valid Hamiltonian cycle of the n‑bit hypercube, which satisfies the circular constraint. To make the sequence start with start, simply XOR every generated Gray code value with start. XOR preserves the one‑bit difference property because it applies the same transformation to all elements. Iterate from 0 to 2^n - 1, compute the Gray code, apply XOR with start, and append to the result list.

This approach leverages properties of bit manipulation and the mathematical structure of Gray codes. It avoids searching or validation because the sequence is guaranteed to satisfy the adjacency condition.

Approach 2: Backtracking with Validity Check (Time: O(2^n * n), Space: O(2^n))

Another way is to construct the permutation using backtracking. Treat each n‑bit number as a node in a graph where edges connect numbers whose XOR has exactly one set bit. Start from start, mark it as visited, and recursively try all numbers that differ by one bit and have not been used yet. Continue until the sequence contains all 2^n numbers. At the final step, verify that the last element differs from the starting element by one bit to satisfy the circular condition.

This approach explicitly explores the solution space using backtracking. Each step flips one bit of the current number to generate candidates, then checks if the candidate has already been used. While slower than the Gray code method, it demonstrates how the problem maps to a Hamiltonian path search on the n‑bit graph, which is a useful insight for problems involving bit states and math properties of binary representations.

Recommended for interviews: The Gray code approach is the expected solution. It runs in O(2^n) time with simple bit operations and directly guarantees the one‑bit adjacency property. Backtracking proves you understand the underlying graph structure, but interviewers typically prefer the Gray code insight because it is both optimal and elegant.