Binary representation of next larger number with same number of 1's and 0's in C program?

Suppose we have a binary number that represents a number n. We need to find a binary number that is larger than n but smallest, and that also has the same number of 0s and 1s. So if the number is 1011 (11 in decimal), then the output will be 1101 (13 in decimal). This problem can be solved using the next permutation calculation. Let's look at the algorithm to get this idea.

Algorithm

nextBin(bin) −

Begin
   len := length of the bin
   for i in range len-2, down to 1, do
      if bin[i] is 0 and bin[i+1] = 1, then
         exchange the bin[i] and bin[i+1]
         break
      end if
   done
   if i = 0, then there is no change, return
   otherwise j:= i + 2, k := len – 1
   while j < k, do
      if bin[j] is 1 and bin[k] is 0, then
         exchange bin[j] and bin[k]
         increase j and k by 1
      else if bin[i] is 0, then
         break
      else
         increase j by 1
      end if
   done
   return bin
End

Example

’s Chinese translation is:

Example

#include <iostream>
using namespace std;
string nextBinary(string bin) {
   int len = bin.size();
   int i;
   for (int i=len-2; i>=1; i--) {
      if (bin[i] == &#39;0&#39; && bin[i+1] == &#39;1&#39;) {
         char ch = bin[i];
         bin[i] = bin[i+1];
         bin[i+1] = ch;
         break;
      }
   }
   if (i == 0)
   "No greater number is present";
   int j = i+2, k = len-1;
   while (j < k) {
      if (bin[j] == &#39;1&#39; && bin[k] == &#39;0&#39;) {
         char ch = bin[j];
         bin[j] = bin[k];
         bin[k] = ch;
         j++;
         k--;
      }
      else if (bin[i] == &#39;0&#39;)
         break;
      else
         j++;
   }
   return bin;
}
int main() {
   string bin = "1011";
   cout << "Binary value of next greater number = " << nextBinary(bin);
}

Output

Binary value of next greater number = 1101

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