Verifique si todos los bits se pueden hacer iguales al voltear dos bits consecutivos

Dada una string binaria, la tarea es encontrar si todos los dígitos de la string pueden igualarse, es decir, 0 o 1, cambiando dos bits consecutivos cualquier número de veces.
Ejemplos: 
 

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.

Enfoque: 
con una observación cuidadosa, se puede alternar entre i-ésimo y j-ésimo bit alternando entre i-ésimo bit como (i, i+1), (i+1, i+2) …. (j-1, j) aquí cada bit se alterna dos veces (si el bit se alterna dos veces, entonces llega a su valor inicial) excepto i y j, luego, en última instancia, los bits i’th y j’th se alternan. Por lo tanto, se puede decir que solo no es posible hacer que todos los dígitos en una string binaria sean iguales cuando el conteo de 1 y 0 es impar.
A continuación se muestra la implementación del enfoque anterior: 
 

// 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 occurrence 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 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 occurrence 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 program for the above approach
 
# Function to check if
# Binary string can be
# made equal
def canMake(s) :
 
    o = 0; z = 0;
 
    # Counting occurrence 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# 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 occurrence 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

<script>
 
// javascript program for the above approach
 
// Function to check if
// Binary string can be
// made equal
function canMake(s)
{
 
    var o = 0, z = 0;
 
    // Counting occurrence of
    // zero and one in binary
    // string
    for (i = 0; i < s.length; i++)
    {
        if (s.charAt(i).charCodeAt(0) - '0'.charCodeAt(0) == 1)
            o++;
        else
            z++;
    }
 
    // From above observation
    if (o % 2 == 1 && z % 2 == 1)
        return "NO";
    else
        return "YES";
}
 
// Driver code
var s = "01011";
document.write(canMake(s)) ;
 
// This code is contributed by Rajput-Ji
 
</script>

YES

Complejidad de tiempo: O(n), donde n es la longitud del número binario dado