Last updated on August 4th, 2024 at 11:49 pm
Here, We see Gray Code LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.
List of all LeetCode Solution
Topics
Backtracking
Companies
Amazon
Level of Question
Medium
Gray Code LeetCode Solution
Table of Contents
Problem Statement
An n-bit gray code sequence is a sequence of 2n
integers where:
- Every integer is in the inclusive range
[0, 2n - 1]
, - The first integer is
0
, - An integer appears no more than once in the sequence,
- The binary representation of every pair of adjacent integers differs by exactly one bit, and
- The binary representation of the first and last integers differs by exactly one bit.
Given an integer n
, return any valid n-bit gray code sequence.
Example 1: Input: n = 2 Output: [0,1,3,2] Explanation: The binary representation of [0,1,3,2] is [00,01,11,10]. - 00 and 01 differ by one bit - 01 and 11 differ by one bit - 11 and 10 differ by one bit - 10 and 00 differ by one bit [0,2,3,1] is also a valid gray code sequence, whose binary representation is [00,10,11,01]. - 00 and 10 differ by one bit - 10 and 11 differ by one bit - 11 and 01 differ by one bit - 01 and 00 differ by one bit Example 2: Input: n = 1 Output: [0,1]
1. Gray Code Leetcode Solution C++
class Solution { public: vector<int> grayCode(int n) { vector<int> ans(1<<n); for (int i=0; i<(1<<n); i++) ans[i] = i^(i>>1); return ans; } };
2. Gray Code Leetcode Solution Java
class Solution { public List<Integer> grayCode(int n) { List<Integer> res = new ArrayList(); res.add(0); for(int i=1;i<=n;i++){ int count = res.size()-1; int add = (int)Math.pow(2,i-1); while(count>=0) res.add(add+res.get(count--)); } return res; } }
3. Gray Code Leetcode Solution JavaScript
var grayCode = function(n) { const codeCount = 1 << n; let result = []; for(let i = 0 ; i < codeCount ; i++){ code = i ^ ( i >> 1); result.push( code ); } return result };
4. Gray Code Leetcode Solution Python
class Solution(object): def grayCode(self, n): res = [0] for i in range(1, 2**n): res.append(res[-1] ^ (i & -i)) return res