通过重复将两个连续的 0 替换为单个 1 使给定的二进制字符串相等

//including the required libraries
#include <bits/stdc++.h>
using namespace std;

//convert consecutive 0's to 1
bool convertBinary(string s1, string s2){

   //fetching the lengths of both the strings
   int len1 = s1.length();
   int len2 = s2.length();
   string temp ="";

   //maintaining counters of both the strings
   int i = 0, j = 0;

   //iterating over both the strings simultaneously
   while (i < len1 && j < len2) {

      //if both the characters are equivalent in nature
      //skip to next character
      if (s1[i] == s2[j]) {
         temp+=s1[i];

         //incrementing both pointers
         i++;
         j++;
      }

      // if both characters differ
      else {

         // checking if '00' of s1 can be converted to '1' of s2
         if(s2[j]=='1' && s1[i]=='0'){

            //checking if i+1th index exists and is equivalent to 0
            if(i+1 < len1 && s1[i+1]=='0'){

               //conversion is possible
               //skip two 0's of s1 since converted to 1 in s2
               temp+='1';
               i += 2;
               j++;
            } else {
               return false;
            }
         }

         // If not possible to combine
         else {
            return false;
         }
      }
   }
   cout<<"Entered string2 "<<s2<<"\n";
   cout<<"Converted string1 "<<temp<<"\n";

   //check if both the strings are returned to end position
   if (i == len1 && j == len2)
      return true;
      return false;
}

// calling the conversion rate
int main(){
   string str1 = "100100";
   string str2 = "1111";

   //capturing result
   cout<<"Entered string1 "<<str1<<"\n";
   bool res = convertBinary(str1, str2);
   if (res)
      cout << "First string can be converted to second";
   else
      cout << "First string can't be converted to second";
   return 0;
}