将以下内容翻译为中文：最小化删除0子串以从循环二进制字符串中删除所有0的出现

Input –  str = "0011001"

Input –  str = ‘0000000’

Output – 1

Input –  str = ‘00110010’

#include <bits/stdc++.h>
using namespace std;
int countRemovels(string str, int N){
   // to store the count of 0s
   int cnt = 0;
   bool isOne = false;
   
   // Iterate over the string
   for (int i = 0; i < N; i++){
   
      // If the current character is 0, increment the count of 0s
      if (str[i] == '0'){
         cnt++;
         
         // traverse the string until a 1 is found
         while (str[i] == '0'){
            i++;
         }
      }
      else{
      
         // If the current character is 1, then set isOne as true
         isOne = true;
      }
   }
   
   // If string contains only 0s, then return 1.
   if (!isOne)
      return 1;
      
   // If the first and last character is 0, then decrement the count, as the string is circular.
   if (str[0] == '0' && str[N - 1] == '0'){
      cnt--;
   }
   
   // return cnt
   return cnt;
}
int main(){
   string str = "0011001";
   int N = str.size();
   cout << "The total number of minimum substrings of consecutive zeros required to remove is - " << countRemovels(str, N);
   return 0;
}

The total number of minimum substrings of consecutive zeros required to remove is - 2<font face="sans-serif"><span style="font-size: 16px; background-color: rgb(255, 255, 255);">.</span></font></p><p>

#include <bits/stdc++.h>
using namespace std;
int countRemovels(string str, int N){
   // to store the count of 0s
   int cnt = 0;
   
   // to check if there is at least one 1
   bool isOne = false;
   
   // traverse the string
   for (int i = 0; i < N - 1; i++) {
   
      // if the current character is 0, the next is 1, then increment count by 1
      if (str[i] == '0' && str[i + 1] == '1'){
         cnt++;
      }
      else{
      
         // if the current character is 1, then set isOne to true
         isOne = true;
      }
   }
   
   // for circular string, if the last character is 0 and the first is 1, then increment count by 1
   if (str[N - 1] == '0' && str[0] == '1'){
      cnt++;
   }
   
   // if there is no 1 in the string, then return 1
   if (!isOne){
      return 1;
   }
   return cnt; // return cnt
}
int main(){
   string str = "0011001";
   int N = str.size();
   cout << "The total number of minimum substrings of consecutive zeros required to remove is - " << countRemovels(str, N);
   return 0;
}

The total number of minimum substrings of consecutive zeros required to remove is - 2