要求最小反转，以便两个相邻元素都不相同

给定大小为N的二进制数组arr [] 。任务是找到所需的最小反转数，以使两个相邻元素都不相同。一次反转后，元素可以从0更改为1或从1更改为0 。

例子：

Input: arr[] = {1, 1, 1}
Output: 1
Change arr[1] from 1 to 0 and
the array becomes {1, 0, 1}.

Input: arr[] = {1, 0, 0, 1, 0, 0, 1, 0}
Output: 3

方法：只有两种方法可以使数组为{1、0、1、0、1、0、1，…}或{0、1、0、1、0、1、0，…}。假设ans_a和ans_b分别是获取这些数组所需的更改计数。现在，最终答案将是min(ans_a，ans_b) 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the minimum
// inversions required so that no
// two adjacent elements are same
int min_changes(int a[], int n)
{
    // To store the inversions required
    // to make the array {1, 0, 1, 0, 1, 0, 1, ...}
    // and {0, 1, 0, 1, 0, 1, 0, ...} respectively
    int ans_a = 0, ans_b = 0;
  
    // Find all the changes required
    for (int i = 0; i < n; i++) {
        if (i % 2 == 0) {
            if (a[i] == 0)
                ans_a++;
            else
                ans_b++;
        }
        else {
            if (a[i] == 0)
                ans_b++;
            else
                ans_a++;
        }
    }
  
    // Return the required answer
    return min(ans_a, ans_b);
}
  
// Driver code
int main()
{
    int a[] = { 1, 0, 0, 1, 0, 0, 1, 0 };
    int n = sizeof(a) / sizeof(a[0]);
  
    cout << min_changes(a, n);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
  
// Function to return the minimum
// inversions required so that no
// two adjacent elements are same
static int min_changes(int a[], int n)
{
    // To store the inversions required
    // to make the array {1, 0, 1, 0, 1, 0, 1, ...}
    // and {0, 1, 0, 1, 0, 1, 0, ...} respectively
    int ans_a = 0, ans_b = 0;
  
    // Find all the changes required
    for (int i = 0; i < n; i++)
    {
        if (i % 2 == 0) 
        {
            if (a[i] == 0)
                ans_a++;
            else
                ans_b++;
        }
        else 
        {
            if (a[i] == 0)
                ans_b++;
            else
                ans_a++;
        }
    }
  
    // Return the required answer
    return Math.min(ans_a, ans_b);
}
  
// Driver code
public static void main(String[] args)
{
    int a[] = { 1, 0, 0, 1, 0, 0, 1, 0 };
    int n = a.length;
  
    System.out.println(min_changes(a, n));
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
  
# Function to return the minimum
# inversions required so that no
# two adjacent elements are same
def min_changes(a, n):
  
    # To store the inversions required
    # to make the array {1, 0, 1, 0, 1, 0, 1, ...}
    # and {0, 1, 0, 1, 0, 1, 0, ...} respectively
    ans_a = 0;
    ans_b = 0;
  
    # Find all the changes required
    for i in range(n):
        if (i % 2 == 0):
            if (a[i] == 0):
                ans_a += 1;
            else:
                ans_b += 1;
  
        else:
            if (a[i] == 0):
                ans_b += 1;
            else:
                ans_a += 1;
  
    # Return the required answer
    return min(ans_a, ans_b);
  
# Driver code
if __name__ == '__main__':
  
    a = [ 1, 0, 0, 1, 0, 0, 1, 0 ];
    n = len(a);
  
    print(min_changes(a, n));
  
# This code is contributed by Rajput-Ji

C#

// C# implementation of the approach
using System;
  
class GFG
{
  
// Function to return the minimum
// inversions required so that no
// two adjacent elements are same
static int min_changes(int []a, int n)
{
    // To store the inversions required
    // to make the array {1, 0, 1, 0, 1, 0, 1, ...}
    // and {0, 1, 0, 1, 0, 1, 0, ...} respectively
    int ans_a = 0, ans_b = 0;
  
    // Find all the changes required
    for (int i = 0; i < n; i++)
    {
        if (i % 2 == 0) 
        {
            if (a[i] == 0)
                ans_a++;
            else
                ans_b++;
        }
        else
        {
            if (a[i] == 0)
                ans_b++;
            else
                ans_a++;
        }
    }
  
    // Return the required answer
    return Math.Min(ans_a, ans_b);
}
  
// Driver code
public static void Main(String[] args)
{
    int []a = { 1, 0, 0, 1, 0, 0, 1, 0 };
    int n = a.Length;
  
    Console.WriteLine(min_changes(a, n));
}
}
  
// This code is contributed by Rajput-Ji

输出：

3