Deletions of “01” or “10” in binary string to make it free from “01” or “10”

Given a binary string str, the task is to find the count of deletion of the sub-string “01” or “10” from the string so that the given string is free from these sub-strings. Print the minimum number of deletions.
Examples:

Input: str = “11010” 
Output: 2 
The resultant string will be “1”
Input: str = “1000101” 
Output: 3 
Resultant string, str = “0” 

Approach: We are deleting “01” and “10” and the binary string contains only characters ‘0’ and ‘1’. Therefore, the minimum number of deletions will be equal to the minimum of the count of ‘0’ and ‘1’.

Algorithm:

Step 1: Start
Step 2: create a function that returns a value that takes one string and one int value as input.
Step 3: In the function initialize two integer values say “count0” and “count1” both to 0;
Step 4: Now traverse through the characters of the string
Step 5: If the current character is ‘0’, increment “count0”.
       If the current character is ‘1’, increment “count1”.
Step 6: Return the minimum value among both count0 and count1.
Step 7: End

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count of deletions
// of sub-strings "01" or "10"
int substrDeletion(string str, int len)
{

    // To store the count of 0s and 1s
    int count0 = 0, count1 = 0;

    for (int i = 0; i < len; i++) {
        if (str[i] == '0')
            count0++;
        else
            count1++;
    }

    return min(count0, count1);
}

// Driver code
int main()
{
    string str = "010";
    int len = str.length();
    cout << substrDeletion(str, len);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to return the count of deletions
// of sub-strings "01" or "10"
static int substrDeletion(String str, int len)
{

    // To store the count of 0s and 1s
    int count0 = 0, count1 = 0;

    for (int i = 0; i < len; i++)
    {
        if (str.charAt(i) == '0')
            count0++;
        else
            count1++;
    }

    return Math.min(count0, count1);
}

// Driver code
public static void main(String[] args)
{
    String str = "010";
    int len = str.length();
    System.out.println(substrDeletion(str, len));
}
}

// This code is contributed by Code_Mech.

Python3

# Python3 implementation of the approach

# Function to return the count of
# deletions of sub-strings "01" or "10"
def substrDeletion(string, length) :

    # To store the count of 0s and 1s
    count0 = 0;
    count1 = 0;

    for i in range(length) :
        if (string[i] == '0') :
            count0 += 1;
        else :
            count1 += 1;

    return min(count0, count1);

# Driver code
if __name__ == "__main__" :

    string = "010";
    length = len(string);

    print(substrDeletion(string, length));

# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the count of deletions
// of sub-strings "01" or "10"
static int substrDeletion(string str, int len)
{

    // To store the count of 0s and 1s
    int count0 = 0, count1 = 0;

    for (int i = 0; i < len; i++)
    {
        if (str[i] == '0')
            count0++;
        else
            count1++;
    }

    return Math.Min(count0, count1);
}

// Driver code
public static void Main()
{
    string str = "010";
    int len = str.Length;
    Console.Write(substrDeletion(str, len));
}
}

// This code is contributed by Ita_c.

Javascript

<script>
// Javascript implementation of the approach

// Function to return the count of deletions
// of sub-strings "01" or "10"
function substrDeletion(str,len)
{
    // To store the count of 0s and 1s
    let count0 = 0, count1 = 0;

    for (let i = 0; i < len; i++)
    {
        if (str[i] == '0')
            count0++;
        else
            count1++;
    }

    return Math.min(count0, count1);
}

// Driver code
let str = "010";
let len = str.length;
document.write(substrDeletion(str, len));



// This code is contributed by rag2127
</script>

Output

1

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.