Min flips of continuous characters to make all characters same in a string

Given a string consisting only of 1’s and 0’s. In one flip we can change any continuous sequence of this string. Find this minimum number of flips so the string consist of same characters only.
Examples:

Input : 00011110001110
Output : 2
We need to convert 1's sequence
so string consist of all 0's.
Input : 010101100011
Output : 4

Method 1 (Change in value encountered)
We need to find the min flips in string so all characters are equal. All we have to find numbers of sequence which consisting of 0’s or 1’s only. Then number of flips required will be half of this number as we can change all 0’s or all 1’s.

C++

// CPP program to find min flips in binary
// string to make all characters equal
#include <bits/stdc++.h>
using namespace std;

// To find min number of flips in binary string
int findFlips(char str[], int n)
{
    char last = ' '; int res = 0;

    for (int i = 0; i < n; i++) {

        // If last character is not equal
        // to str[i] increase res
        if (last != str[i])
            res++;
        last = str[i];
    }

    // To return min flips
    return res / 2;
}

// Driver program to check findFlips()
int main()
{
    char str[] = "00011110001110";
    int n = strlen(str);

    cout << findFlips(str, n);

    return 0;
}

Java

// Java program to find min flips in binary
// string to make all characters equal
public class minFlips {

    // To find min number of flips in binary string
    static int findFlips(String str, int n)
    {
        char last = ' '; int res = 0;

        for (int i = 0; i < n; i++) {

            // If last character is not equal
            // to str[i] increase res
            if (last != str.charAt(i))
                res++;
            last = str.charAt(i);
        }

        // To return min flips
        return res / 2;
    }

    // Driver program to check findFlips()
    public static void main(String[] args)
    {
        String str = "00011110001110";
        int n = str.length();

        System.out.println(findFlips(str, n));
    }
}

Python 3

# Python 3 program to find min flips in
# binary string to make all characters equal

# To find min number of flips in
# binary string
def findFlips(str, n):

    last = ' '
    res = 0

    for i in range( n) :

        # If last character is not equal
        # to str[i] increase res
        if (last != str[i]):
            res += 1
        last = str[i]

    # To return min flips
    return res // 2

# Driver Code
if __name__ == "__main__":

    str = "00011110001110"
    n = len(str)

    print(findFlips(str, n))

# This code is contributed by ita_c

C#

// C# program to find min flips in
// binary string to make all
// characters equal
using System;

public class GFG {

    // To find min number of flips
    // in binary string
    static int findFlips(String str, int n)
    {
        char last = ' '; int res = 0;

        for (int i = 0; i < n; i++) {

            // If last character is not
            // equal to str[i] increase
            // res
            if (last != str[i])
                res++;
            last = str[i];
        }

        // To return min flips
        return res / 2;
    }

    // Driver program to check findFlips()
    public static void Main()
    {
        String str = "00011110001110";
        int n = str.Length;

        Console.Write(findFlips(str, n));
    }
}

// This code is contributed by nitin mittal

Javascript

<script>

// JavaScript program to find min flips in binary
// string to make all characters equal


    // To find min number of flips in binary string
    function findFlips( str , n) {
        var last = ' ';
        var res = 0;

        for (i = 0; i < n; i++) {

            // If last character is not equal
            // to str[i] increase res
            if (last != str.charAt(i))
                res++;
            last = str.charAt(i);
        }

        // To return min flips
        return parseInt(res / 2);
    }

    // Driver program to check findFlips()

        var str = "00011110001110";
        var n = str.length;

        document.write(findFlips(str, n));


// This code contributed by aashish1995

</script>

PHP

<?php
// PHP program to find min flips in binary
// string to make all characters equal

// To find min number of
// flips in binary string
function findFlips($str, $n)
{
    $last = ' ';
    $res = 0;

    for ($i = 0; $i < $n; $i++)
    {

        // If last character is not equal
        // to str[i] increase res
        if ($last != $str[$i])
            $res++;
        $last = $str[$i];
    }

    // To return min flips
    return intval($res / 2);
}

    // Driver Code
    $str = "00011110001110";
    $n = strlen($str);

    echo findFlips($str, $n);

// This code is contributed by Sam007
?>

Output

2

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

Method 2(Count continuous 0 and 1 )

We can count the number of continuous 0 and continuous 1.
Since we have to take minimum, we use min function to take value with less continuous number. And output it.

Procedure:- Take two variable to count continuous 0 and 1. If 0 is encountered increment countZero and skip 0’s in continuation. Do same with 1 and its continuation. In the end the variables with less value is sent as output

C++

#include <bits/stdc++.h>
#include <iostream>

using namespace std;

int main()
{
    string str = "010101100011";
    int n = str.size();

    int countZero = 0;
    int countOne = 0;
    for (int i = 0; i < n; i++) {
        if (str[i] == '0') {
            countZero++;
            while ((i + 1) != n && str[i + 1] == '0') {
                i++;
                // cout<<"Skip 0";
            }
        }
        else {
            countOne++;
            while ((i + 1) != n && str[i + 1] == '1') {
                i++;
                // cout<<"Skip 1";
            }
        }
    }
    int mini = min(countOne, countZero);
    cout << mini << endl;

    return 0;
}

Java

public class Main {
    public static void main(String[] args) {
        String str = "010101100011";
        int n = str.length();

        int countZero = 0;
        int countOne = 0;

        for (int i = 0; i < n; i++) {
            if (str.charAt(i) == '0') {
                countZero++;
                while ((i + 1) != n && str.charAt(i + 1) == '0') {
                    i++;
                    // System.out.println("Skip 0");
                }
            } else {
                countOne++;
                while ((i + 1) != n && str.charAt(i + 1) == '1') {
                    i++;
                    // System.out.println("Skip 1");
                }
            }
        }

        int mini = Math.min(countOne, countZero);
        System.out.println(mini);
    }
}

C#

using System;

class Program
{
    static void Main()
    {
        string str = "010101100011";
        int n = str.Length;

        int countZero = 0;
        int countOne = 0;

        for (int i = 0; i < n; i++)
        {
            if (str[i] == '0')
            {
                countZero++; // Increment the count of '0's
                while ((i + 1) != n && str[i + 1] == '0')
                {
                    i++; // Skip consecutive '0's
                    // Console.WriteLine("Skip 0");
                }
            }
            else
            {
                countOne++; // Increment the count of '1's
                while ((i + 1) != n && str[i + 1] == '1')
                {
                    i++; // Skip consecutive '1's
                    // Console.WriteLine("Skip 1");
                }
            }
        }

        int mini = Math.Min(countOne, countZero); // Find the minimum count of '0's and '1's
        Console.WriteLine(mini); // Output the result
    }
}

Javascript

function countConsecutive(str) {
    const n = str.length;

    let countZero = 0;
    let countOne = 0;

    for (let i = 0; i < n; i++) {
        if (str[i] === '0') {
            countZero++;
            while (i + 1 < n && str[i + 1] === '0') {
                i++;
                // console.log("Skip 0");
            }
        } else {
            countOne++;
            while (i + 1 < n && str[i +1] === '1') {
                i++;
                // console.log("Skip 1");
            }
        }
    }

    const mini = Math.min(countOne, countZero);
    console.log(mini);
}

const str = "010101100011";
countConsecutive(str);

Output

4

Time Complexity: O(n) 

Auxiliary Space: O(1)

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