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

2.6

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-

  1. We make a solution vector sol and insert first bit 1 in it which will be the first number.
  2. Now we check whether length of solution vector is less than or equal to n or not.
  3. If it is so then we calculate the decimal number and store it into a map as it store numbers in sorted order.
  4. 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.
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)
 }
// 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 program
int 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;
}

Output:

1 2 4 5 8 9 10

Related Post :
Count number of binary strings without consecutive 1’s

This article is contributed by Niteesh Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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