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

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