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

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; ` `} ` |

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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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