C/C++ program: Count number of binary strings without consecutive 1's?

A binary number is a number that contains only two digits, that is, only 0 and 1. Each binary number is a stream of binary bits, which we think of as a binary string. For this string, we need to find the number of length N binary strings that do not contain consecutive 1's.

For example, for N=5, the binary string that satisfies the given condition is 00000 00001 00010 00100 00101 01000 01001 01010 10000 10001 10010 10100 10101.

One way is to generate all N-digit strings and print only the strings that satisfy the given condition. However, this approach is not efficient when it comes to large-scale operations.

Another way is to use recursion. At each step of the recursion, we append 0 and 1 to the partially formed number and recurse with one less number. The key is that only if the last digit of the partially formed number is 0, we append 1 and recurse. This way, there will be no consecutive 1's in the output string.

Input: n = 5
Output: Number of 5-digit binary strings without any consecutive 1's are 13

Example

#include <iostream>
#include <string>
using namespace std;
int countStrings(int n, int last_digit) {
   if (n == 0)
      return 0;
   if (n == 1) {
      if (last_digit)
         return 1;
      else
         return 2;
   }
   if (last_digit == 0)
      return countStrings(n - 1, 0) + countStrings(n - 1, 1);
   else
      return countStrings(n - 1, 0);
}
int main() {
   int n = 5;
   cout << "Number of " << n << "-digit binary strings without any "
   "consecutive 1&#39;s are " << countStrings(n, 0);
   return 0;
}

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