Skip to content
Related Articles

Related Articles

Minimum number of alternate subsequences required to be removed to empty a Binary String

View Discussion
Improve Article
Save Article
  • Last Updated : 22 Sep, 2021

Given a binary string S consisting of N characters, the task is to print the minimum number of operations required to remove all the characters from the given string S by removing a single character or removing any subsequence of alternate characters in each operation.

Examples:

Input: S = “010101”
Output: 1
Explanation:
Below are the operations performed:
Operation 1: Consider the subsequence S[0, 5] i.e., “010101” as it contains alternating characters. Therefore, removing this modifies the string to “”.
Hence, the total number of operation required is 1.

Input: S = “00011”
Output: 3

Approach: The given problem can be solved by iterating over the string once and keep track of the maximum number of remaining 0s and 1s. Follow the steps below to solve the problem:

  • Initialize a variable, say ans to store the maximum number of 0s and 1s which are still left to be removed, a variable cnt0 to count the number of 0s, and a variable cnt1 to count the number of 1s.
  • Traverse the given string S from the beginning and perform the following steps:
    • If the current character is 0, then increment the value of cnt0 by 1 and decrement the value of cnt1 by 1 if it is greater than 0.
    • If the current character is 1, increment the value of cnt1 by 1 and decrement the value of cnt0 by 1 if it is greater than 0.
    • Update the value of ans to the maximum of ans, cnt1, and cnt0.
  • After completing the above steps, print the value of ans as the minimum number of operations required to remove all the characters.

Below is the implementation of the approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum number
// of operations to empty a binary string
int minOpsToEmptyString(string s)
{
    // Stores the resultant number of
    // operations
    int ans = INT_MIN;
 
    // Stores the number of 0s
    int cn0 = 0;
 
    // Stores the number of 1s
    int cn1 = 0;
 
    // Traverse the given string
    for (int i = 0;
         i < s.length(); i++) {
 
        if (s[i] == '0') {
 
            // To balance 0 with 1
            // if possible
            if (cn1 > 0)
                cn1--;
 
            // Increment the value
            // of cn0 by 1
            cn0++;
        }
        else {
 
            // To balance 1 with 0
            // if possible
            if (cn0 > 0)
                cn0--;
 
            // Increment the value
            // of cn1
            cn1++;
        }
 
        // Update the maximum number
        // of unused 0s and 1s
        ans = max({ ans, cn0, cn1 });
    }
 
    // Print the resultant count
    cout << ans;
}
 
// Driver Code
int main()
{
    string S = "010101";
    minOpsToEmptyString(S);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the minimum number
// of operations to empty a binary string
static void minOpsToEmptyString(String s)
{
     
    // Stores the resultant number of
    // operations
    int ans = Integer.MIN_VALUE;
  
    // Stores the number of 0s
    int cn0 = 0;
  
    // Stores the number of 1s
    int cn1 = 0;
  
    // Traverse the given string
    for(int i = 0; i < s.length(); i++)
    {
        if (s.charAt(i) == '0')
        {
             
            // To balance 0 with 1
            // if possible
            if (cn1 > 0)
                cn1--;
  
            // Increment the value
            // of cn0 by 1
            cn0++;
        }
        else
        {
             
            // To balance 1 with 0
            // if possible
            if (cn0 > 0)
                cn0--;
  
            // Increment the value
            // of cn1
            cn1++;
        }
  
        // Update the maximum number
        // of unused 0s and 1s
        ans = Math.max(ans, Math.max(cn0, cn1));
    }
  
    // Print the resultant count
    System.out.print(ans);
}
  
// Driver Code
public static void main(String[] args)
{
    String S = "010101";
    minOpsToEmptyString(S);
}
}
 
// This code is contributed by sanjoy_62

Python3




# Python3 program for the above approach
 
# Function to find the minimum number
# of operations to empty a binary string
def minOpsToEmptyString(s):
   
    # Stores the resultant number of
    # operations
    ans = -10**9
 
    # Stores the number of 0s
    cn0 = 0
 
    # Stores the number of 1s
    cn1 = 0
 
    # Traverse the given string
    for i in range(len(s)):
        if (s[i] == '0'):
 
            # To balance 0 with 1
            # if possible
            if (cn1 > 0):
                cn1 -= 1
 
            # Increment the value
            # of cn0 by 1
            cn0 += 1
        else:
 
            # To balance 1 with 0
            # if possible
            if (cn0 > 0):
                cn0 -= 1
 
            # Increment the value
            # of cn1
            cn1 += 1
 
        # Update the maximum number
        # of unused 0s and 1s
        ans = max([ans, cn0, cn1])
 
    # Print resultant count
    print (ans)
 
# Driver Code
if __name__ == '__main__':
    S = "010101"
    minOpsToEmptyString(S)
 
# This code is contributed by mohit kumar 29.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the minimum number
// of operations to empty a binary string
static void minOpsToEmptyString(string s)
{
     
    // Stores the resultant number of
    // operations
    int ans = 0;
  
    // Stores the number of 0s
    int cn0 = 0;
  
    // Stores the number of 1s
    int cn1 = 0;
  
    // Traverse the given string
    for(int i = 0; i < s.Length; i++)
    {
        if (s[i] == '0')
        {
             
            // To balance 0 with 1
            // if possible
            if (cn1 > 0)
                cn1--;
  
            // Increment the value
            // of cn0 by 1
            cn0++;
        }
        else
        {
             
            // To balance 1 with 0
            // if possible
            if (cn0 > 0)
                cn0--;
  
            // Increment the value
            // of cn1
            cn1++;
        }
  
        // Update the maximum number
        // of unused 0s and 1s
        ans = Math.Max(ans, Math.Max(cn0, cn1));
    }
  
    // Print the resultant count
    Console.Write(ans);
}
 
// Driver Code
public static void Main()
{
    string S = "010101";
    minOpsToEmptyString(S);
}
}
 
// This code is contributed by avijitmondal1998.

Javascript




<script>
// JavaScript program for the above approach
// Function to find the minimum number
// of operations to empty a binary string
function minOpsToEmptyString(s)
{
     
    // Stores the resultant number of
    // operations
    var ans = Number.MIN_VALUE;
  
    // Stores the number of 0s
    var cn0 = 0;
  
    // Stores the number of 1s
    var cn1 = 0;
  
    // Traverse the given string
    for(var i = 0; i < s.length; i++)
    {
        if (s.charAt(i) == '0')
        {
             
            // To balance 0 with 1
            // if possible
            if (cn1 > 0)
                cn1--;
  
            // Increment the value
            // of cn0 by 1
            cn0++;
        }
        else
        {
             
            // To balance 1 with 0
            // if possible
            if (cn0 > 0)
                cn0--;
  
            // Increment the value
            // of cn1
            cn1++;
        }
  
        // Update the maximum number
        // of unused 0s and 1s
        ans = Math.max(ans, Math.max(cn0, cn1));
    }
  
    // Print the resultant count
    document.write(ans);
}
  
// Driver Code
var S = "010101";
minOpsToEmptyString(S);
  
//This code is contributed by shivanisinghss2110
  
</script>

Output: 

1

 

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


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!