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
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]
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;
}
};
Code language: PHP (php)
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;
}
}
Code language: PHP (php)
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
};
Code language: JavaScript (javascript)
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