Divida la string binaria en substrings con el mismo número de 0 y 1

Dada una string binaria str de longitud N , la tarea es encontrar el recuento máximo de substrings consecutivas en las que str se puede dividir de manera que todas las substrings estén balanceadas, es decir, tengan el mismo número de 0 y 1 . Si no es posible dividir str cumpliendo las condiciones, imprima -1 .
Ejemplo: 
 

Entrada: str = “0100110101” 
Salida: 4 
Las substrings requeridas son “01”, “0011”, “01” y “01”.
Entrada: str = «0111100010» 
Salida: 3 

Entrada: string = «0000000000» 

Salida: -1
 

Enfoque: Inicialice el conteo = 0 y recorra la string carácter por carácter y realice un seguimiento del número de 0 y 1 hasta el momento, cada vez que el conteo de 0 y 1 sea igual, incremente el conteo. Como en la pregunta dada, si no es posible dividir la string, en ese momento, el conteo de 0 no debe ser igual al conteo de 1, luego devuelva -1 , de lo contrario, imprima el valor de conteo después del recorrido de la string completa.
A continuación se muestra la implementación del enfoque anterior:
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count
// of maximum substrings str
// can be divided into
int maxSubStr(string str, int n)
{
 
    // To store the count of 0s and 1s
    int count0 = 0, count1 = 0;
 
    // To store the count of maximum
    // substrings str can be divided into
    int cnt = 0;
    for (int i = 0; i < n; i++) {
        if (str[i] == '0') {
            count0++;
        }
        else {
            count1++;
        }
        if (count0 == count1) {
            cnt++;
        }
    }
 
    // It is not possible to
    // split the string
    if (count0!=count1) {
        return -1;
    }
 
    return cnt;
}
 
// Driver code
int main()
{
    string str = "0100110101";
    int n = str.length();
 
    cout << maxSubStr(str, n);
 
    return 0;
}

// Java implementation of the above approach
class GFG
{
 
// Function to return the count
// of maximum substrings str
// can be divided into
static int maxSubStr(String str, int n)
{
 
    // To store the count of 0s and 1s
    int count0 = 0, count1 = 0;
 
    // To store the count of maximum
    // substrings str can be divided into
    int cnt = 0;
    for (int i = 0; i < n; i++)
    {
        if (str.charAt(i) == '0')
        {
            count0++;
        }
        else
        {
            count1++;
        }
        if (count0 == count1)
        {
            cnt++;
        }
    }
 
    // It is not possible to
    // split the string
    if (count0 != count1)
    {
        return -1;
    }
    return cnt;
}
 
// Driver code
public static void main(String []args)
{
    String str = "0100110101";
    int n = str.length();
 
    System.out.println(maxSubStr(str, n));
}
}
 
// This code is contributed by PrinciRaj1992

# Python3 implementation of the approach
 
# Function to return the count
# of maximum substrings str
# can be divided into
def maxSubStr(str, n):
     
    # To store the count of 0s and 1s
    count0 = 0
    count1 = 0
     
    # To store the count of maximum
    # substrings str can be divided into
    cnt = 0
     
    for i in range(n):
        if str[i] == '0':
            count0 += 1
        else:
            count1 += 1
             
        if count0 == count1:
            cnt += 1
 
# It is not possible to
    # split the string
    if count0 != count1:
        return -1
             
    return cnt
 
# Driver code
str = "0100110101"
n = len(str)
print(maxSubStr(str, n))

// C# implementation of the above approach
using System;
 
class GFG
{
 
// Function to return the count
// of maximum substrings str
// can be divided into
static int maxSubStr(String str, int n)
{
 
    // To store the count of 0s and 1s
    int count0 = 0, count1 = 0;
 
    // To store the count of maximum
    // substrings str can be divided into
    int cnt = 0;
    for (int i = 0; i < n; i++)
    {
        if (str[i] == '0')
        {
            count0++;
        }
        else
        {
            count1++;
        }
        if (count0 == count1)
        {
            cnt++;
        }
    }
 
    // It is not possible to
    // split the string
    if (count0 != count1)
    {
        return -1;
    }
    return cnt;
}
 
// Driver code
public static void Main(String []args)
{
    String str = "0100110101";
    int n = str.Length;
 
    Console.WriteLine(maxSubStr(str, n));
}
}
 
// This code is contributed by PrinciRaj1992

<script>
 
// JavaScript implementation of the approach
 
// Function to return the count
// of maximum substrings str
// can be divided into
function maxSubStr(str, n)
{
 
    // To store the count of 0s and 1s
    var count0 = 0, count1 = 0;
 
    // To store the count of maximum
    // substrings str can be divided into
    var cnt = 0;
    for (var i = 0; i < n; i++) {
        if (str[i] == '0') {
            count0++;
        }
        else {
            count1++;
        }
        if (count0 == count1) {
            cnt++;
        }
    }
 
    // It is not possible to
    // split the string
    if (count0 != count1) {
        return -1;
    }
 
    return cnt;
}
 
// Driver code
var str = "0100110101";
var n = str.length;
document.write( maxSubStr(str, n));
 
</script>

4

Complejidad temporal: O(N) donde N es la longitud de la string 
Complejidad espacial: O(1)