Minimize removal of substring of 0s to remove all occurrences of 0s from a circular Binary String

In this problem, we require to remove all zeros from the given binary string. Also, we require to remove pair of consecutive zeros at once and count the total number of pairs of zeros removal.

We can solve the problem by counting the number of pairs of consecutive zeros in the given string. In this tutorial, we will learn two different solutions to solve the problem.

Problem statement − We have given circular binary string str of length N. We need to find the minimum number of consecutive zeros required to remove all zeros from the string.

Sample Examples

Input –  str = "0011001"

Output – 2

Explanation

We can remove str[0] and str[1] together. After that, we can remove str[4] and str[5]. So, we need to remove 2 pairs of consecutive zeros.

Input –  str = ‘0000000’

Output – 1

Explanation

We can remove all zeros at once.

Input –  str = ‘00110010’

Output – 2

Explanation

We can remove str[0], str[1], and str[7] together as the binary string is circular. Next, we can remove str[5] and str[6] together.

Approach 1

In this approach, we will find the total number of pairs of consecutive zeros in the given string, which will answer the given problem.

Algorithm

• Step 1 − Initialize the ‘cnt’ variable with zero.

• Step 2 − Initialize the ‘isOne’ variable with a false value to keep track of the ones in the given string.

• Step 3 − Iterate through the string using the loop. In the loop, if the current character is ‘0’, increase the value of ‘cnt’ by 1.

• Step 4 − Use the while loop to make iterations until we keep finding ‘0’ as the next character and increment the value of ‘I’ by 1.

• Step 5 − If the current character is ‘1’, change the value of the ‘isOne’ variable to true, representing string contains at least one ‘1’.

• Step 6 − Once the iteration of the loop completes, If the value of ‘isOne’ is false, it means the string contains only zeros. Return 1 in such cases.

• Step 7 − If the first and last characters are ‘0’, decrease the value of the ‘cnt’ by 1 as the string is circular.

• Step 8 − Return the value of ‘cnt’.

Example

#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;
}

Output

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

Space complexity − O(1)

Approach 2

In this approach, we will count the minimum removal of substrings of zeros required to remove all zeros by counting the different adjacent elements.

Algorithm

• Step 1 − Define the ‘cnt’ and ‘isOne’ variables, and initialize with 0 and false, respectively.

• Step 2 − Use for loop to make N-1 iterations, where N is the length of the string.

• Step 3 − In the loop, check if the current character is ‘0,’ and the next character is ‘1’, increase the value of ‘cnt’ by 1. Otherwise, change the value of the ‘isOne’ variable to true.

• Step 4 − If the last character is ‘0’ and the first character is ‘1’, increase the value of ‘cnt’ by 1.

• Step 5 − If the value of the ‘isOne’ is false, return 1.

• Step 6 − Return the value of the ‘cnt’ variable.

Example

#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;
}

Output

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

Conclusion

We have seen two different solutions for the given problem. In the first approach, we count the total number of consecutive pairs of zeros, and in the second approach, we count the total number of mismatching adjacent characters.