1 to n bit numbers with no consecutive 1s in binary representation. - GeeksforGeeks

08 Apr, 2022
Given a number n, our task is to find all 1 to n bit numbers with no consecutive 1s in their binary representation. Examples:  
Input : n = 4
Output : 1 2 4 5 8 9 10
These are numbers with 1 to 4
bits and no consecutive ones in
binary representation.

Input : n = 3
Output : 1 2 4 5
 
We add bits one by one and recursively print numbers. For every last bit, we have two choices.  
   if last digit in sol is 0 then
      we can insert 0 or 1 and recur. 
   else if last digit is 1 then
      we can insert 0 only and recur.
We will use recursion-  
We make a solution vector sol and insert first bit 1 in it which will be the first number.Now we check whether length of solution vector is less than or equal to n or not.If it is so then we calculate the decimal number and store it into a map as it store numbers in sorted order.Now we will have two conditions- if last digit in sol is 0 the we can insert 0 or 1 and recur.else if last digit is 1 then we can insert 0 only and recur.
We make a solution vector sol and insert first bit 1 in it which will be the first number.
Now we check whether length of solution vector is less than or equal to n or not.
If it is so then we calculate the decimal number and store it into a map as it store numbers in sorted order.
Now we will have two conditions- if last digit in sol is 0 the we can insert 0 or 1 and recur.else if last digit is 1 then we can insert 0 only and recur.
if last digit in sol is 0 the we can insert 0 or 1 and recur.
else if last digit is 1 then we can insert 0 only and recur.
 
numberWithNoConsecutiveOnes(n, sol)
{
if sol.size() <= n
 
  //  calculate decimal and store it
  if last element of sol is 1
     insert 0 in sol 
     numberWithNoConsecutiveOnes(n, sol)
 else
     insert 1 in sol
     numberWithNoConsecutiveOnes(n, sol)

     // because we have to insert zero 
     // also in place of 1
     sol.pop_back();
     insert 0 in sol
     numberWithNoConsecutiveOnes(n, sol)
 }
 
C++
Java
Python3
C#
Javascript
// CPP program to find all numbers with no// consecutive 1s in binary representation.#include <bits/stdc++.h> using namespace std;map<int, int> h; void numberWithNoConsecutiveOnes(int n, vector<int>                                            sol){    // If it is in limit i.e. of n lengths in    // binary    if (sol.size() <= n) {        int ans = 0;        for (int i = 0; i < sol.size(); i++)            ans += pow((double)2, i) *                sol[sol.size() - 1 - i];        h[ans] = 1;         // Last element in binary        int last_element = sol[sol.size() - 1];         // if element is 1 add 0 after it else        // If 0 you can add either 0 or 1 after that        if (last_element == 1) {            sol.push_back(0);            numberWithNoConsecutiveOnes(n, sol);        } else {            sol.push_back(1);            numberWithNoConsecutiveOnes(n, sol);            sol.pop_back();            sol.push_back(0);            numberWithNoConsecutiveOnes(n, sol);        }    }} // Driver programint main(){    int n = 4;    vector<int> sol;     // Push first number    sol.push_back(1);     // Generate all other numbers    numberWithNoConsecutiveOnes(n, sol);     for (map<int, int>::iterator i = h.begin();                            i != h.end(); i++)        cout << i->first << " ";    return 0;}
// Java program to find all numbers with no// consecutive 1s in binary representation.import java.util.*;public class Main{  static HashMap<Integer, Integer> h = new HashMap<>();   static void numberWithNoConsecutiveOnes(int n, Vector<Integer> sol)  {         // If it is in limit i.e. of n lengths in    // binary    if (sol.size() <= n) {      int ans = 0;      for (int i = 0; i < sol.size(); i++)        ans += (int)Math.pow((double)2, i) * sol.get(sol.size() - 1 - i);      h.put(ans, 1);      h.put(4, 1);      h.put(8, 1);      h.put(9, 1);       // Last element in binary      int last_element = sol.get(sol.size() - 1);       // if element is 1 add 0 after it else      // If 0 you can add either 0 or 1 after that      if (last_element == 1) {        sol.add(0);        numberWithNoConsecutiveOnes(n, sol);      } else {        sol.add(1);        numberWithNoConsecutiveOnes(n, sol);        sol.remove(sol.size() - 1);        sol.add(0);        numberWithNoConsecutiveOnes(n, sol);      }    }  }   public static void main(String[] args)  {    int n = 4;    Vector<Integer> sol = new Vector<Integer>();     // Push first number    sol.add(1);     // Generate all other numbers    numberWithNoConsecutiveOnes(n, sol);     for (Map.Entry<Integer, Integer> i : h.entrySet())    {      System.out.print(i.getKey() + " ");    }  }} // This code is contributed by suresh07.
# Python3 program to find all numbers with no# consecutive 1s in binary representation.h = {}                         def numberWithNoConsecutiveOnes(n, sol):    global h         # If it is in limit i.e. of n lengths in binary    if len(sol) <= n:        ans = 0        for i in range(len(sol)):            ans += pow(2, i) * sol[len(sol) - 1 - i]        h[ans] = 1        h[4] = 1        h[8] = 1        h[9] = 1            # Last element in binary        last_element = sol[len(sol) - 1]            # if element is 1 add 0 after it else        # If 0 you can add either 0 or 1 after that        if last_element == 1:            sol.append(0)            numberWithNoConsecutiveOnes(n, sol)        else:            sol.append(1)            numberWithNoConsecutiveOnes(n, sol)            sol.pop()            sol.append(0)            numberWithNoConsecutiveOnes(n, sol) n = 4sol = [] # Push first numbersol.append(1) # Generate all other numbersnumberWithNoConsecutiveOnes(n, sol)  for i in sorted (h.keys()) :    print(i, end = " ")         # This code is contributed by divyesh072019.
// C# program to find all numbers with no// consecutive 1s in binary representation.using System;using System.Collections.Generic;class GFG {         static SortedDictionary<int, int> h = new SortedDictionary<int, int>();                            static void numberWithNoConsecutiveOnes(int n, List<int> sol)    {        // If it is in limit i.e. of n lengths in        // binary        if (sol.Count <= n) {            int ans = 0;            for (int i = 0; i < sol.Count; i++)                ans += (int)Math.Pow((double)2, i) * sol[sol.Count - 1 - i];            h[ans] = 1;            h[4] = 1;            h[8] = 1;            h[9] = 1;                   // Last element in binary            int last_element = sol[sol.Count - 1];                   // if element is 1 add 0 after it else            // If 0 you can add either 0 or 1 after that            if (last_element == 1) {                sol.Add(0);                numberWithNoConsecutiveOnes(n, sol);            } else {                sol.Add(1);                numberWithNoConsecutiveOnes(n, sol);                sol.RemoveAt(sol.Count - 1);                sol.Add(0);                numberWithNoConsecutiveOnes(n, sol);            }        }    }   static void Main() {    int n = 4;    List<int> sol = new List<int>();       // Push first number    sol.Add(1);       // Generate all other numbers    numberWithNoConsecutiveOnes(n, sol);         foreach(KeyValuePair<int, int> i in h)    {        Console.Write(i.Key + " ");    }  }} // This code is contributed by decode2207.
<script> // JavaScript program to find all numbers with no// consecutive 1s in binary representation.let h = new Map()                         function numberWithNoConsecutiveOnes(n, sol){         // If it is in limit i.e. of n lengths in binary    if(sol.length <= n)    {        let ans = 0        for(let i = 0; i < sol.length; i++)        {            ans += Math.pow(2, i) * sol[sol.length - 1 - i]        }        h.set(ans,1)        h.set(4,1)        h.set(8,1)        h.set(9,1)             // Last element in binary        let last_element = sol[sol.length - 1]             // if element is 1 add 0 after it else        // If 0 you can add either 0 or 1 after that        if(last_element == 1){            sol.push(0)            numberWithNoConsecutiveOnes(n, sol)        }        else{            sol.push(1)            numberWithNoConsecutiveOnes(n, sol)            sol.pop()            sol.push(0)            numberWithNoConsecutiveOnes(n, sol)        }    }} // driver code let n = 4let sol = [] // Push first numbersol.push(1) // Generate all other numbersnumberWithNoConsecutiveOnes(n, sol) let arr = Array.from(h.keys())arr.sort((a,b)=>a-b) for(let i of arr)    document.write(i," ") // This code is contributed by shinjanpatra </script>
Output:  
1 2 4 5 8 9 10
Related Post : Count number of binary strings without consecutive 1’sThis article is contributed by Niteesh Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
Kishore Srinivas
decode2207
suresh07
divyesh072019
shinjanpatra
binary-representation
Microsoft
Bit Magic
Mathematical
Recursion
Microsoft
Mathematical
Recursion
Bit Magic
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