A backtracking approach to generate n bit Gray Codes Last Updated : 05 Feb, 2024 Comments Improve Suggest changes Like Article Like Report Given a number n, the task is to generate n bit Gray codes (generate bit patterns from 0 to 2^n-1 such that successive patterns differ by one bit) Examples: Input : 2 Output : 0 1 3 2Explanation : 00 - 001 - 111 - 310 - 2Input : 3 Output : 0 1 3 2 6 7 5 4 We have discussed an approach in Generate n-bit Gray CodesThis article provides a backtracking approach to the same problem. Idea is that for each bit out of n bit we have a choice either we can ignore it or we can invert the bit so this means our gray sequence goes upto 2 ^ n for n bits. So we make two recursive calls for either inverting the bit or leaving the bit as it is. C++ // CPP program to find the gray sequence of n bits. #include <iostream> #include <vector> using namespace std; /* we have 2 choices for each of the n bits either we can include i.e invert the bit or we can exclude the bit i.e we can leave the number as it is. */ void grayCodeUtil(vector<int>& res, int n, int& num) { // base case when we run out bits to process // we simply include it in gray code sequence. if (n == 0) { res.push_back(num); return; } // ignore the bit. grayCodeUtil(res, n - 1, num); // invert the bit. num = num ^ (1 << (n - 1)); grayCodeUtil(res, n - 1, num); } // returns the vector containing the gray // code sequence of n bits. vector<int> grayCodes(int n) { vector<int> res; // num is passed by reference to keep // track of current code. int num = 0; grayCodeUtil(res, n, num); return res; } // Driver function. int main() { int n = 3; vector<int> code = grayCodes(n); for (int i = 0; i < code.size(); i++) cout << code[i] << endl; return 0; } Java // JAVA program to find the gray sequence of n bits. import java.util.*; class GFG { static int num; /* we have 2 choices for each of the n bits either we can include i.e invert the bit or we can exclude the bit i.e we can leave the number as it is. */ static void grayCodeUtil(Vector<Integer> res, int n) { // base case when we run out bits to process // we simply include it in gray code sequence. if (n == 0) { res.add(num); return; } // ignore the bit. grayCodeUtil(res, n - 1); // invert the bit. num = num ^ (1 << (n - 1)); grayCodeUtil(res, n - 1); } // returns the vector containing the gray // code sequence of n bits. static Vector<Integer> grayCodes(int n) { Vector<Integer> res = new Vector<Integer>(); // num is passed by reference to keep // track of current code. num = 0; grayCodeUtil(res, n); return res; } // Driver function. public static void main(String[] args) { int n = 3; Vector<Integer> code = grayCodes(n); for (int i = 0; i < code.size(); i++) System.out.print(code.get(i) +"\n"); } } // This code is contributed by Rajput-Ji Python3 # Python3 program to find the # gray sequence of n bits. """ we have 2 choices for each of the n bits either we can include i.e invert the bit or we can exclude the bit i.e we can leave the number as it is. """ def grayCodeUtil(res, n, num): # base case when we run out bits to process # we simply include it in gray code sequence. if (n == 0): res.append(num[0]) return # ignore the bit. grayCodeUtil(res, n - 1, num) # invert the bit. num[0] = num[0] ^ (1 << (n - 1)) grayCodeUtil(res, n - 1, num) # returns the vector containing the gray # code sequence of n bits. def grayCodes(n): res = [] # num is passed by reference to keep # track of current code. num = [0] grayCodeUtil(res, n, num) return res # Driver Code n = 3 code = grayCodes(n) for i in range(len(code)): print(code[i]) # This code is contributed by SHUBHAMSINGH10 C# // C# program to find the gray sequence of n bits. using System; using System.Collections.Generic; class GFG { static int num; /* we have 2 choices for each of the n bits either we can include i.e invert the bit or we can exclude the bit i.e we can leave the number as it is. */ static void grayCodeUtil(List<int> res, int n) { // base case when we run out bits to process // we simply include it in gray code sequence. if (n == 0) { res.Add(num); return; } // ignore the bit. grayCodeUtil(res, n - 1); // invert the bit. num = num ^ (1 << (n - 1)); grayCodeUtil(res, n - 1); } // returns the vector containing the gray // code sequence of n bits. static List<int> grayCodes(int n) { List<int> res = new List<int>(); // num is passed by reference to keep // track of current code. num = 0; grayCodeUtil(res, n); return res; } // Driver function. public static void Main(String[] args) { int n = 3; List<int> code = grayCodes(n); for (int i = 0; i < code.Count; i++) Console.Write(code[i] +"\n"); } } // This code is contributed by 29AjayKumar JavaScript <script> // Javascript program to find the gray sequence of n bits. /* We have 2 choices for each of the n bits either we can include i.e invert the bit or we can exclude the bit i.e we can leave the number as it is. */ function grayCodeUtil(res, n, num) { // Base case when we run out bits to process // we simply include it in gray code sequence. if (n == 0) { res.push(num[0]); return; } // Ignore the bit. grayCodeUtil(res, n - 1, num); // Invert the bit. num[0] = num[0] ^ (1 << (n - 1)); grayCodeUtil(res, n - 1, num); } // Returns the vector containing the gray // code sequence of n bits. function grayCodes(n) { let res = []; // num is passed by reference to keep // track of current code. let num = [0]; grayCodeUtil(res, n, num); return res; } // Driver code let n = 3; let code = grayCodes(n); for(let i = 0; i < code.length; i++) document.write(code[i] + "<br>"); // This code is contributed by gfgking </script> Output: 01326754Time Complexity: O(2n)Auxiliary Space: O(n) Comment More infoAdvertise with us Next Article Permutations of given String F foreverrookie Follow Improve Article Tags : Misc Bit Magic Backtracking DSA gray-code +1 More Practice Tags : BacktrackingBit MagicMisc Similar Reads Backtracking Algorithm Backtracking algorithms are like problem-solving strategies that help explore different options to find the best solution. 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