将二进制字符串转换为全 0 或全 1 所需的最少操作

给定一个二进制字符串str ，任务是找到使字符串的所有字符相同所需的最小操作数，即结果字符串包含全 0 或全 1。在单个操作中，任何连续 0 的块都可以转换为相同长度的连续 1 块，反之亦然。
例子：

Input: str = “000111”
Output: 1
In a single operation, either change all 0s to 1s
or change all 1s to 0s.
Input: str = “0011101010”
Output: 3
All the 1s can be converted to 0s in 3 operations.

编程需要懂一点英语

方法：问题是将所有字符转换为一个字符。现在因为转换整个连续的字符组算作一个步骤。您可以计算由于它们之间存在其他字符而彼此分隔的不同组的数量。现在，步数将只是这两个数字中的最小值。因此，答案将是连续块 0 的计数或连续块 1 的计数中的最小值。
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
// minimum operations required
int minOperations(string str, int n)
{
    int count = 0;
    for (int i = 0; i < n - 1; i++) {
 
        // Increment count when consecutive
        // characters are different
        if (str[i] != str[i + 1])
            count++;
    }
 
    // Answer is rounding off the
    // (count / 2) to lower
    return (count + 1) / 2;
}
 
// Driver code
int main()
{
    string str = "000111";
    int n = str.length();
 
    cout << minOperations(str, n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Function to return the count of
// minimum operations required
static int minOperations(String str, int n)
{
    int count = 0;
    for (int i = 0; i < n - 1; i++)
    {
 
        // Increment count when consecutive
        // characters are different
        if (str.charAt(i) != str.charAt(i + 1))
            count++;
    }
 
    // Answer is rounding off the
    // (count / 2) to lower
    return (count + 1) / 2;
}
 
// Driver code
public static void main(String[] args)
{
    String str = "000111";
    int n = str.length();
 
    System.out.println(minOperations(str, n));
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 implementation of the approach
 
# Function to return the count of
# minimum operations required
def minOperations(str, n):
    count = 0
    for i in range(n - 1):
 
        # Increment count when consecutive
        # characters are different
        if (str[i] != str[i + 1]):
            count += 1
 
    # Answer is rounding off the
    # (count / 2) to lower
    return (count + 1) // 2
 
# Driver code
str = "000111"
n = len(str)
 
print(minOperations(str, n))
 
# This code is contributed
# by Mohit Kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the count of
// minimum operations required
static int minOperations(string str, int n)
{
    int count = 0;
    for (int i = 0; i < n - 1; i++)
    {
 
        // Increment count when consecutive
        // characters are different
        if (str[(i)] != str[(i + 1)])
            count++;
    }
 
    // Answer is rounding off the
    // (count / 2) to lower
    return (count + 1) / 2;
}
 
// Driver code
public static void Main()
{
    string str = "000111";
    int n = str.Length;
 
    Console.WriteLine(minOperations(str, n));
}
}
 
// This code is contributed by Code_Mech

Javascript

输出：