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.

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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