Check if all bits can be made same by flipping two consecutive bits

Given a binary string, the task is to find whether all the digits of the string can be made equal i.e either 0 or 1 by flipping two consecutive bits any number of times.

Examples:

Input: 01011
Output: YES
Explanation:
Flip 2nd and 3rd bit -> 00111, 
again flipping 1'st and 2'nd bit -> 11111

Input: 100011
Output: NO
Explanation:
No number of moves can ever 
equalize all elements of the array.

Approach:
On careful observation, toggling of i’th and j’th bit can be done by toggling from i’th bit like (i, i+1), (i+1, i+2) …. (j-1, j) here every bit is toggling twice (if bit is toggle twice then its come to its initial value) except i and j then ultimately i’th and j’th bits toggle. Therefore, it can be said that it is only not possible to make all digits in binary string equal when the count of both 1 and 0 is odd.

Below is the implementation of the above approach:

C++

// C++ program for the
// above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if
// Binary string can be
// made equal
string canMake(string& s)
{

    int o = 0, z = 0;

    // Counting occurence of
    // zero and one in binary
    // string
    for (int i = 0; i < s.size(); i++) {
        if (s[i] - '0' == 1)
            o++;
        else
            z++;
    }

    // From above observation
    if (o % 2 == 1 && z % 2 == 1)
        return "NO";
    else
        return "YES";
}

// Driver code
int main()
{

    string s = "01011";
    cout << canMake(s) << '\n';
    return 0;
}

Java

// Java program for the above approach
class GFG
{

    // Function to check if
    // Binary string can be
    // made equal
    static String canMake(String s)
    {

        int o = 0, z = 0;

        // Counting occurence of
        // zero and one in binary
        // string
        for (int i = 0; i < s.length(); i++)
        {
            if (s.charAt(i) - '0' == 1)
                o++;
            else
                z++;
        }

        // From above observation
        if (o % 2 == 1 && z % 2 == 1)
            return "NO";
        else
            return "YES";
    }

    // Driver code
    public static void main (String[] args)
    {

        String s = "01011";
        System.out.println(canMake(s)) ;

    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 program for the above approach

# Function to check if
# Binary string can be
# made equal
def canMake(s) :

    o = 0; z = 0;

    # Counting occurence of
    # zero and one in binary
    # string
    for i in range(len(s)) :
        if (ord(s[i]) - ord('0') == 1) :
            o += 1;
        else :
            z += 1;

    # From above observation
    if (o % 2 == 1 and z % 2 == 1) :
        return "NO";
    else :
        return "YES";

# Driver code
if __name__ == "__main__" :

    s = "01011";
    print(canMake(s));

# This code is contributed by AnkitRai01

C#

// C# program for the above approach
using System;

class GFG
{

    // Function to check if
    // Binary string can be
    // made equal
    static string canMake(string s)
    {

        int o = 0, z = 0;

        // Counting occurence of
        // zero and one in binary
        // string
        for (int i = 0; i < s.Length; i++)
        {
            if (s[i] - '0' == 1)
                o++;
            else
                z++;
        }

        // From above observation
        if (o % 2 == 1 && z % 2 == 1)
            return "NO";
        else
            return "YES";
    }

    // Driver code
    public static void Main()
    {
        string s = "01011";
        Console.WriteLine(canMake(s)) ;
    }
}

// This code is contributed by AnkitRai01

Output:

YES

Time Complexity: O(n), where n is the length of the given Binary number

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