- Related Questions & Answers
- Count number of binary strings without consecutive 1's in C
- C/C++ Program to Count number of binary strings without consecutive 1’s?
- Program to Count number of binary strings without consecutive 1’s in C/C++?
- Python Program to Count number of binary strings without consecutive 1’
- Python program to find the length of the largest consecutive 1's in Binary Representation of a given string.
- Maximum length of consecutive 1’s in a binary string in Python using Map function
- 1's Complement vs 2's Complement
- 1's complement notation
- Count 1’s in a sorted binary array in C++
- MySQL's now() +1 day?
- C# program to check if there are K consecutive 1’s in a binary number
- Python program to check if there are K consecutive 1’s in a binary number?
- C# Program to Count the Number of 1's in the Entered Numbers
- Simpson's 1/3 Rule for definite integral
- Count the number of 1’s and 0’s in a binary array using STL in C++

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

In this problem, we have to find some binary numbers which have no consecutive 1s. In a 3-bit binary string, there are three binary numbers 011, 110, 111, who have consecutive 1s, and five numbers are there which have no consecutive 1s. So after applying this algorithm to 3-bit numbers, the answer will be 5.

If a[i] be the set of binary numbers, whose number of bits are I, and not containing any consecutive 1s, and b[i] is the set of binary number, where a number of bits are I, and containing consecutive 1s, then there are recurrence relations like:

a[i] := a[i - 1] + b[i - 1] b[i] := a[i - 1]

Input: This algorithm takes number of bits for a binary number. Let the input is 4. Output: It returns the number of binary strings which have no consecutive 1’s. Here the result is: 8. (There are 8 binary strings which has no consecutive 1’s)

countBinNums(n)

**Input:** n is the number of bits.**Output −** Count how many numbers are present which have no consecutive 1.

Begin define lists with strings ending with 0 and ending with 1 endWithZero[0] := 1 endWithOne[0] := 1 for i := 1 to n-1, do endWithZero[i] := endWithZero[i-1] + endWithOne[i-1] endWithOne[i] := endWithZero[i-1] done return endWithZero[n-1] + endWithOne[n-1] End

#include <iostream> using namespace std; int countBinNums(int n) { int endWithZero[n], endWithOne[n]; endWithZero[0] = endWithOne[0] = 1; for (int i = 1; i < n; i++) { endWithZero[i] = endWithZero[i-1] + endWithOne[i-1]; endWithOne[i] = endWithZero[i-1]; } return endWithZero[n-1] + endWithOne[n-1]; } int main() { int n; cout << "Enter number of bits: "; cin >> n; cout << "Number of binary numbers without consecitive 1's: "<<countBinNums(n) << endl; return 0; }

Enter number of bits: 4 Number of binary numbers without consecitive 1's: 8

Advertisements