Minimum removals to make a string concatenation of a substring of 0s followed by a substring of 1s

Given binary string str of length N​​​​, the task is to find the minimum number of characters required to be deleted from the given binary string to make a substring of 0s followed by a substring of 1s.

Examples:

Input: str = “00101101”
Output: 2
Explanation: Removing str[2] and str[6] or removing str[3] and str[6] modifies the given binary string to “000111” or “001111” respectively. The number of removals required in both the cases is 2, which is the minimum possible.

Input: str = “111000001111”
Output: 3

Brute Force Approach:

1. Initialize the minimum number of deletions required to be the length of the input string since we can delete all characters in the worst case to satisfy the condition.
2. Iterate over each character i in the input string.
3. Calculate the number of 1s before i and the number of 0s after i using the count() function in the STL.
4. Calculate the total number of deletions required to make a substring of 0s followed by a substring of 1s using i as the midpoint.
5. Update the minimum number of deletions required if the current substring requires fewer deletions.
6. Return the minimum number of deletions required.

Below is the code of above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count minimum removals
// required to make a given string
// concatenation of substring of 0s
// followed by substring of 1s
int minimumDeletions(string s)
{
    int n = s.length();
    int minDeletions = n;
    for(int i=0;i<n;i++)
    {
        int numOnesBefore = count(s.begin(), s.begin()+i, '1');
        int numZerosAfter = count(s.begin()+i, s.end(), '0');
        minDeletions = min(minDeletions, numOnesBefore + numZerosAfter);
    }
    return minDeletions;
}


// Driver Code
int main()
{
    string stri = "00101101";

    cout << minimumDeletions(stri) << endl;

    return 0;
}

Java

import java.util.*;

public class Main
{

  // Function to count minimum removals
  // required to make a given string
  // concatenation of substring of 0s
  // followed by substring of 1s
  public static int minimumDeletions(String s) {
    int n = s.length();
    int minDeletions = n;
    for(int i=0;i<n;i++) {
      int numOnesBefore = count(s.substring(0, i), '1');
      int numZerosAfter = count(s.substring(i), '0');
      minDeletions = Math.min(minDeletions, numOnesBefore + numZerosAfter);
    }
    return minDeletions;
  }

  // Utility function to count occurrences of a character in a string
  public static int count(String s, char c) {
    int count = 0;
    for (int i = 0; i < s.length(); i++) {
      if (s.charAt(i) == c) {
        count++;
      }
    }
    return count;
  }

  // Driver Code
  public static void main(String[] args) {
    String stri = "00101101";
    System.out.println(minimumDeletions(stri));
  }
}

Python

# Function to count minimum removals
# required to make a given string
# concatenation of substring of 0s
# followed by substring of 1s


def minimum_deletions(s):
    n = len(s)
    min_deletions = n

    for i in range(n):
        num_ones_before = s[:i].count('1')
        num_zeros_after = s[i:].count('0')
        min_deletions = min(min_deletions, num_ones_before + num_zeros_after)

    return min_deletions


# Driver Code
if __name__ == "__main__":
    stri = "00101101"
    print(minimum_deletions(stri))

C#

using System;

class Program
{
    // Function to count minimum removals
    // required to make a given string
    // concatenation of substring of 0s
    // followed by substring of 1s
    static int MinimumDeletions(string s)
    {
        int n = s.Length;
        int minDeletions = n;
        for (int i = 0; i < n; i++)
        {
            int numOnesBefore = CountOccurrences(s.Substring(0, i), '1');
            int numZerosAfter = CountOccurrences(s.Substring(i), '0');
            minDeletions = Math.Min(minDeletions, numOnesBefore + numZerosAfter);
        }
        return minDeletions;
    }

    // Helper function to count occurrences of a character in a string
    static int CountOccurrences(string input, char target)
    {
        int count = 0;
        foreach (char c in input)
        {
            if (c == target)
                count++;
        }
        return count;
    }

    // Driver Code
    static void Main()
    {
        string stri = "00101101";
        Console.WriteLine(MinimumDeletions(stri));
    }
}

Javascript

// Function to count minimum removals
// required to make a given string
// concatenation of substring of 0s
// followed by substring of 1s
function minimumDeletions(s) {
    const n = s.length;
    let minDeletions = n;
    for (let i = 0; i < n; i++) {
        const numOnesBefore = s.substring(0, i).split('1').length - 1;
        const numZerosAfter = s.substring(i).split('0').length - 1;
        minDeletions = Math.min(minDeletions, numOnesBefore + numZerosAfter);
    }
    return minDeletions;
}

// Driver Code
const stri = "00101101";
console.log(minimumDeletions(stri));

Output

2

Time Complexity:  O(N^2), where N is the length of the input string. The outer loop iterates over all possible positions in the string, and the count function inside the loop has a time complexity of O(N) for each call, giving an overall time complexity of O(N^2).

Auxiliary Space: O(1) because we are not using any additional data structures to store information about the input string.

Efficient Approach: The above approach can be optimized by having an auxiliary space that keeps the count of the number of 0s after 1s. Using this pre-computation, the time complexity can be improved by a factor of N. Below are the steps:

• Initialize a variable, say ans, to store the minimum number of characters required to be deleted.
• Initialize an array, say zeroCount[], to count the number of 0s present after a given index.
• Traverse the string str from the end over the range [N – 2, 0] and if the current character is 0, then update zeroCount[i] as (zeroCount[i + 1] + 1). Otherwise, update zeroCount[i] as zeroCount[i + 1].
• Initialize a variable, say oneCount, to count the number of 1s.
• Traverse the given string again. For every character found to be ‘1’, update ans as the minimum of ans and (oneCount + zeroCount[i]).
• After the above steps, if the value of ans remains same as its initialized value, then 0 characters are required to be deleted. Otherwise, ans is the required number of characters to be deleted.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
#include <iostream>

using namespace std;

// Function to count minimum removals
// required to make a given string
// concatenation of substring of 0s
// followed by substring of 1s
int minimumDeletions(string s)
{

    // Stores the length of the string
    int n = s.size();

    // Precompute the count of 0s
    int zeroCount[n];

    // Check for the last character
    zeroCount[n - 1] = (s[n - 1] == '0') ? 1 : 0;

    // Traverse and update zeroCount array
    for(int i = n - 2; i >= 0; i--)

        // If current character is 0,
        zeroCount[i] = (s[i] == '0') ?

                       // Update aCount[i] as
                       // aCount[i + 1] + 1
                       zeroCount[i + 1] + 1 :

                       // Update as aCount[i + 1]
                       zeroCount[i + 1];

    // Keeps track of deleted 1s
    int oneCount = 0;

    // Stores the count of removals
    int ans = INT_MAX;

    // Traverse the array
    for(int i = 0; i < n; i++)
    {

        // If current character is 1
        if (s[i] == '1')
        {

            // Update ans
            ans = min(ans,
                      oneCount +
                      zeroCount[i]);
            oneCount++;
        }
    }

    // If all 1s are deleted
    ans = min(ans, oneCount);

    // Return the minimum
    // number of deletions
    return (ans == INT_MAX) ? 0 : ans;
}

// Driver Code
int main()
{
    string stri = "00101101";

    cout << minimumDeletions(stri) << endl;

    return 0;
}

// This code is contributed by AnkThon

Java

// Java program for the above approach

import java.io.*;

class GFG {

    // Function to count minimum removals
    // required to make a given string
    // concatenation of substring of 0s
    // followed by substring of 1s
    public static int minimumDeletions(String s)
    {
        // Stores the length of the string
        int n = s.length();

        // Precompute the count of 0s
        int zeroCount[] = new int[n];

        // Check for the last character
        zeroCount[n - 1] = (s.charAt(n - 1)
                            == '0')
                               ? 1
                               : 0;

        // Traverse and update zeroCount array
        for (int i = n - 2; i >= 0; i--)

            // If current character is 0,
            zeroCount[i] = (s.charAt(i) == '0')

                               // Update aCount[i] as
                               // aCount[i + 1] + 1
                               ? zeroCount[i + 1] + 1

                               // Update as aCount[i + 1]
                               : zeroCount[i + 1];

        // Keeps track of deleted 1s
        int oneCount = 0;

        // Stores the count of removals
        int ans = Integer.MAX_VALUE;

        // Traverse the array
        for (int i = 0; i < n; i++) {

            // If current character is 1
            if (s.charAt(i) == '1') {

                // Update ans
                ans = Math.min(ans,
                               oneCount
                                   + zeroCount[i]);
                oneCount++;
            }
        }

        // If all 1s are deleted
        ans = Math.min(ans, oneCount);

        // Return the minimum
        // number of deletions
        return (ans == Integer.MAX_VALUE)
            ? 0
            : ans;
    }

    // Driver Code
    public static void main(String[] args)
    {
        String str = "00101101";
        System.out.println(
            minimumDeletions(str));
    }
}

Python3

# Python3 program for the above approach

# Function to count minimum removals
# required to make a given string
# concatenation of substring of 0s
# followed by substring of 1s
def minimumDeletions(s):

    # Stores the length of the string
    n = len(s)

    # Precompute the count of 0s
    zeroCount = [ 0 for i in range(n)]

    # Check for the last character
    zeroCount[n - 1] = 1 if s[n - 1] == '0' else 0

    # Traverse and update zeroCount array
    for i in range(n - 2, -1, -1):

        # If current character is 0,
        zeroCount[i] = zeroCount[i + 1] + 1 if (s[i] == '0') else zeroCount[i + 1]

    # Keeps track of deleted 1s
    oneCount = 0

    # Stores the count of removals
    ans = 10**9

    # Traverse the array
    for i in range(n):

        # If current character is 1
        if (s[i] == '1'):

            # Update ans
            ans = min(ans,oneCount + zeroCount[i])
            oneCount += 1

    # If all 1s are deleted
    ans = min(ans, oneCount)

    # Return the minimum
    # number of deletions
    return 0 if ans == 10**18 else ans

# Driver Code
if __name__ == '__main__':
    str = "00101101"
    print(minimumDeletions(str))

    # 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 count minimum removals
    // required to make a given string
    // concatenation of substring of 0s
    // followed by substring of 1s
    public static int minimumDeletions(String s)
    {
        // Stores the length of the string
        int n = s.Length;

        // Precompute the count of 0s
        int []zeroCount = new int[n];

        // Check for the last character
        zeroCount[n - 1] = (s[n - 1]
                            == '0')
                               ? 1
                               : 0;

        // Traverse and update zeroCount array
        for (int i = n - 2; i >= 0; i--)

            // If current character is 0,
            zeroCount[i] = (s[i] == '0')

                               // Update aCount[i] as
                               // aCount[i + 1] + 1
                               ? zeroCount[i + 1] + 1

                               // Update as aCount[i + 1]
                               : zeroCount[i + 1];

        // Keeps track of deleted 1s
        int oneCount = 0;

        // Stores the count of removals
        int ans = int.MaxValue;

        // Traverse the array
        for (int i = 0; i < n; i++) {

            // If current character is 1
            if (s[i] == '1') {

                // Update ans
                ans = Math.Min(ans,
                               oneCount
                                   + zeroCount[i]);
                oneCount++;
            }
        }

        // If all 1s are deleted
        ans = Math.Min(ans, oneCount);

        // Return the minimum
        // number of deletions
        return (ans == int.MaxValue)
            ? 0
            : ans;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        String str = "00101101";
        Console.WriteLine(
            minimumDeletions(str));
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript program for above approach

// Function to count minimum removals
    // required to make a given string
    // concatenation of substring of 0s
    // followed by substring of 1s
    function minimumDeletions(s)
    {
        // Stores the length of the string
        let n = s.length;

        // Precompute the count of 0s
        let zeroCount = [];

        // Check for the last character
        zeroCount[n - 1] = (s[n - 1]
                            == '0')
                               ? 1
                               : 0;

        // Traverse and update zeroCount array
        for (let i = n - 2; i >= 0; i--)

            // If current character is 0,
            zeroCount[i] = (s[i] == '0')

                           // Update aCount[i] as
                          // aCount[i + 1] + 1
                          ? zeroCount[i + 1] + 1

                         // Update as aCount[i + 1]
                           : zeroCount[i + 1];

        // Keeps track of deleted 1s
        let oneCount = 0;

        // Stores the count of removals
        let ans = Number.MAX_VALUE;

        // Traverse the array
        for (let i = 0; i < n; i++) {

            // If current character is 1
            if (s[i] == '1') {

                // Update ans
                ans = Math.min(ans,
                           oneCount
                           + zeroCount[i]);
                oneCount++;
            }
        }

        // If all 1s are deleted
        ans = Math.min(ans, oneCount);

        // Return the minimum
        // number of deletions
        return (ans == Number.MAX_VALUE)
            ? 0
            : ans;
    }

// Driver Code

    let str = "00101101";
        document.write(
            minimumDeletions(str));

</script>

Output

2

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