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Maximize the number of times a character can be removed from substring 01 from given Binary String

  • Difficulty Level : Hard
  • Last Updated : 13 Jul, 2021

Given a binary string S of size N, the task is to find the maximum number of operations that can be performed on S, by selecting any substring “01” and removing any character from it in a single move, reducing the length of the string by 1.

Examples:

Input: S = “001111”, N = 6
Output: 5
Explanation: 
One way to perform the operations is:

  1. Select the substring “01” over the range [1, 2] and erase the S[2] ( = ‘1’). The string modifies to “00111”.
  2. Select the substring “01” over the range [1, 2] and erase the S[2] ( = ‘1’). The string modifies to “0011”.
  3. Select the substring “01” over the range [1, 2] and erase the S[2] ( = ‘1’). The string modifies to “001”.
  4. Select the substring “01” over the range [1, 2] and erase the S[1] ( = ‘0’). The string modifies to “01”.
  5. Select the substring “01” over the range [0, 1] and erase the S[1] ( = ‘1’). The string modifies to “0”.
  6. Now no characters can be removed.

Therefore, the total number of operations performed is 5, which is the maximum possible.

Input: S=”0101″, N=4
Output: 3



 

Approach: The given problem can be solved based on the following observations: 

  1. Any 1s present in the prefix of S cannot be removed because there are no 0s before them.
  2. Any 0s present in the suffix of S cannot be removed because there are no 1s after them.
  3. All other characters are removable.
  4. If there are X removable characters, at most X-1 operations can be performed because, eventually, only a single character will remain which cannot be removed.

Follow the steps below to solve the problem:

  • Initialize two variables, say X and Y as 0, which store the number of 0s in the suffix and the number of 1s in the prefix respectively, which cannot be removed.
  • Iterate over the characters of the string S and perform the following steps:
    • If the current character is ‘1′, increment Y by 1.
    • Otherwise, stop traversing.
  • Iterate over the characters of the string S in reverse order and perform the following steps:
    • If the current character is 0, increment X by 1.
    • Otherwise, stop traversing.
  • If the sum of X and Y is equal to N, print 0 as there are no removable characters.
  • Otherwise, print N-(X+Y)-1 as the answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum moves
// that can be performed on a string
int maxOperations(string S, int N)
{
    // Stores 0s in suffix
    int X = 0;
    // Stores 1s in prefix
    int Y = 0;
 
    // Iterate over the characters of
    // the string
    for (int i = 0; i < N; i++) {
        if (S[i] == '0')
            break;
        Y++;
    }
    // Iterate until i is greater than
    // or equal to 0
    for (int i = N - 1; i >= 0; i--) {
        if (S[i] == '1')
            break;
        X++;
    }
 
    // If N is equal to x+y
    if (N == X + Y)
        return 0;
 
    // Return answer
    return N - (X + Y) - 1;
}
// Driver code
int main()
{
    // Input
    string S = "001111";
    int N = S.length();
 
    // Function call
    cout << maxOperations(S, N) << endl;
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to find the maximum moves
// that can be performed on a string
static int maxOperations(String S, int N)
{
     
    // Stores 0s in suffix
    int X = 0;
     
    // Stores 1s in prefix
    int Y = 0;
 
    // Iterate over the characters of
    // the string
    for(int i = 0; i < N; i++)
    {
        if (S.charAt(i) == '0')
            break;
             
        Y++;
    }
     
    // Iterate until i is greater than
    // or equal to 0
    for(int i = N - 1; i >= 0; i--)
    {
        if (S.charAt(i) == '1')
            break;
             
        X++;
    }
 
    // If N is equal to x+y
    if (N == X + Y)
        return 0;
 
    // Return answer
    return N - (X + Y) - 1;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Input
    String S = "001111";
    int N = S.length();
     
    // Function call
    System.out.println(maxOperations(S, N));
}
}
 
// This code is contributed by sanjoy_62

Python3




# Python3 program for the above approach
 
# Function to find the maximum moves
# that can be performed on a string
def maxOperations(S, N):
     
    # Stores 0s in suffix
    X = 0
     
    # Stores 1s in prefix
    Y = 0
 
    # Iterate over the characters of
    # the string
    for i in range(N):
        if (S[i] == '0'):
            break
         
        Y += 1
         
    # Iterate until i is greater than
    # or equal to 0
    i = N - 1
    while(i >= 0):
        if (S[i] == '1'):
            break
         
        X += 1
 
    # If N is equal to x+y
    if (N == X + Y):
        return 0
 
    # Return answer
    return N - (X + Y) - 1
 
# Driver code
if __name__ == '__main__':
     
    # Input
    S = "001111"
    N = len(S)
 
    # Function call
    print(maxOperations(S, N))
     
# This code is contributed by SURENDRA_GANGWAR

C#




// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the maximum moves
// that can be performed on a string
static int maxOperations(String S, int N)
{
     
    // Stores 0s in suffix
    int X = 0;
 
    // Stores 1s in prefix
    int Y = 0;
 
    // Iterate over the characters of
    // the string
    for(int i = 0; i < N; i++)
    {
        if (S[i] == '0')
            break;
 
        Y++;
    }
 
    // Iterate until i is greater than
    // or equal to 0
    for(int i = N - 1; i >= 0; i--)
    {
        if (S[i] == '1')
            break;
 
        X++;
    }
 
    // If N is equal to x+y
    if (N == X + Y)
        return 0;
 
    // Return answer
    return N - (X + Y) - 1;
}
 
// Driver code
static void Main()
{
     
    // Input
    String S = "001111";
    int N = S.Length;
 
    // Function call
    Console.WriteLine(maxOperations(S, N));
}
}
 
// This code is contributed by abhinavjain194

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to find the maximum moves
// that can be performed on a string
function maxOperations(S, N)
{
     
    // Stores 0s in suffix
    let X = 0;
 
    // Stores 1s in prefix
    let Y = 0;
 
    // Iterate over the characters of
    // the string
    for(let i = 0; i < N; i++)
    {
        if (S[i] == '0')
            break;
 
        Y++;
    }
 
    // Iterate until i is greater than
    // or equal to 0
    for(let i = N - 1; i >= 0; i--)
    {
        if (S[i] == '1')
            break;
 
        X++;
    }
 
    // If N is equal to x+y
    if (N == X + Y)
        return 0;
 
    // Return answer
    return N - (X + Y) - 1;
}
             
 
// Driver Code
 
    // Input
    let S = "001111";
    let N = S.length;
     
    // Function call
    document.write(maxOperations(S, N));
 
</script>
Output
5

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

 

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