将二进制字符串拆分为0和1相等数量的子字符串

给定长度为N的二进制字符串str ，任务是找到最大数量的子字符串str，可以将其划分为所有子字符串都平衡的状态，即它们具有相等的0和1数。如果无法拆分满足条件的str ，则打印-1 。

例子：

Input: str = “0100110101”
Output: 4
The required substrings are “01”, “0011”, “01” and “01”.

Input: str = “0111100010”
Output: 3

方法：初始化计数= 0和字符遍历字符串的字符和跟踪0，至今1的个数，只要0和1的数量相等增量计数。如果原始字符串的0和1s计数不相等，则打印-1，否则在遍历整个字符串之后打印count的值。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

// 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

# 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#

// 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

输出：

4

时间复杂度： O(N)其中N是字符串的长度
空间复杂度： O(1)