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

  • Last Updated : 21 May, 2021

Given circular binary string S of size N, the task is to count the minimum number of consecutive 0s required to be removed such that the string contains only 1s.

A circular string is a string whose first and last characters are considered to be adjacent to each other.

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Examples:



Input: S = “11010001”
Output: 2
Explanation:
Remove the substring {S[2]}. Now, the string modifies to “1110001”.
Remove the substring {S[3], …, S[5]} of consecutive 0s. Now, the string modifies to “1111”.
Therefore, the minimum count of removals required is 2.

Input: S = “00110000”
Output: 1

Approach: The idea to solve the given problem is to traverse the given string and count the number of substrings having the same number of 0s, say C. Now if the first and the last characters of the string are ‘0’, then print the value of (C – 1) as the minimum number of removals required. Otherwise, print the value of C as the result.

Note: If the given string contains all 0s, then the minimum number of removals required is 1. Consider this case separately.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count minimum number of
// removal of consecutive 0s required to
// make binary string consists only of 1s
int minRemovals(string str, int N)
{
    // Stores the count of removals
    int ans = 0;
 
    bool X = false;
 
    // Traverse the string S
    for (int i = 0; i < N; i++) {
 
        // If the current character is '0'
        if (str[i] == '0') {
 
            ans++;
 
            // Traverse until consecutive
            // characters are only '0's
            while (str[i] == '0') {
                i++;
            }
        }
 
        else {
            X = true;
        }
    }
 
    // If the binary string only
    // contains 1s, then return 1
    if (!X)
        return 1;
 
    // If the first and the last
    // characters are 0
    if (str[0] == '0'
        and str[N - 1] == '0') {
        ans--;
    }
 
    // Return the resultant count
    return ans;
}
 
// Driver Code
int main()
{
    string S = "11010001";
    int N = S.size();
    cout << minRemovals(S, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to count minimum number of
// removal of consecutive 0s required to
// make binary string consists only of 1s
static int minRemovals(String str, int N)
{
     
    // Stores the count of removals
    int ans = 0;
 
    boolean X = false;
 
    // Traverse the string S
    for(int i = 0; i < N; i++)
    {
         
        // If the current character is '0'
        if (str.charAt(i) == '0')
        {
            ans++;
 
            // Traverse until consecutive
            // characters are only '0's
            while (i < N && str.charAt(i) == '0')
            {
                i++;
            }
        }
 
        else
        {
            X = true;
        }
    }
 
    // If the binary string only
    // contains 1s, then return 1
    if (!X)
        return 1;
 
    // If the first and the last
    // characters are 0
    if (str.charAt(0) == '0' &&
        str.charAt(N - 1) == '0')
    {
        ans--;
    }
 
    // Return the resultant count
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    String S = "11010001";
    int N = S.length();
 
    System.out.println(minRemovals(S, N));
}
}
 
// This code is contributed by Kingash

Python3




# Python3 program for the above approach
 
# Function to count minimum number of
# removal of consecutive 0s required to
# make binary string consists only of 1s
def minRemovals(str, N):
     
    # Stores the count of removals
    ans = 0
    X = False
     
    # Traverse the string S
    i = 0
     
    while i < N:
         
        # If the current character is '0'
        if (str[i] == '0'):
            ans += 1
             
            # Traverse until consecutive
            # characters are only '0's
            while (str[i] == '0'):
                i += 1
        else:
            X = True
             
        i += 1
         
    # If the binary string only
    # contains 1s, then return 1
    if (not X):
        return 1
         
    # If the first and the last
    # characters are 0
    if (str[0] == '0' and str[N - 1] == '0'):
        ans -= 1
         
    # Return the resultant count
    return ans
     
# Driver Code
S = "11010001"
N = len(S)
 
print(minRemovals(S, N))
 
# This code is contributed by rohan07

C#




// C# program for the above approach
using System;
class GFG {
 
  // Function to count minimum number of
  // removal of consecutive 0s required to
  // make binary string consists only of 1s
  static int minRemovals(string str, int N)
  {
     
    // Stores the count of removals
    int ans = 0;
 
    bool X = false;
 
    // Traverse the string S
    for (int i = 0; i < N; i++) {
 
      // If the current character is '0'
      if (str[i] == '0') {
 
        ans++;
 
        // Traverse until consecutive
        // characters are only '0's
        while (str[i] == '0') {
          i++;
        }
      }
 
      else {
        X = true;
      }
    }
 
    // If the binary string only
    // contains 1s, then return 1
    if (!X)
      return 1;
 
    // If the first and the last
    // characters are 0
    if (str[0] == '0' && str[N - 1] == '0') {
      ans--;
    }
 
    // Return the resultant count
    return ans;
  }
 
  // Driver Code
  public static void Main()
  {
    string S = "11010001";
    int N = S.Length;
    Console.Write(minRemovals(S, N));
  }
}
 
// This code is contributed by subham348.

Javascript




<script>
// js program for the above approach
 
// Function to count minimum number of
// removal of consecutive 0s required to
// make binary consists only of 1s
function minRemovals(str, N)
{
   
  // Stores the count of removals
  let ans = 0;
 
  let X = false;
 
  // Traverse the S
  for (i = 0; i < N; i++) {
 
    // If the current character is '0'
    if (str[i] == '0') {
 
      ans++;
 
      // Traverse until consecutive
      // characters are only '0's
      while (str[i] == '0') {
        i++;
      }
    }
 
    else {
      X = true;
    }
  }
 
  // If the binary only
  // contains 1s, then return 1
  if (!X)
    return 1;
 
  // If the first and the last
  // characters are 0
  if (str[0] == '0' && str[N - 1] == '0') {
    ans--;
  }
 
  // Return the resultant count
  return ans;
}
 
  // Driver Code
 
let S = "11010001";
let N = S.length;
document.write(minRemovals(S, N));
 
// This code is contributed by mohit kumar 29.
</script>
Output: 
2

 

Time Complexity: O(N)
Auxiliary Space: O(1)




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