根据给定条件执行删除/旋转操作后数组A中元素的计数

给定两个大小分别为N 的二进制数组A[]和B[] ，任务是找出数组A[]中在执行以下操作后将剩余的元素数，直到没有元素可以被删除：

1. 如果数组A[]和B[]的起始元素相等，则删除这两个元素。
2. 否则，在删除数组A[]后，将数组A[]的起始字符附加到数组的末尾。

例子：

Input: A[] = {1, 1, 0, 1}, B[] = {1, 0, 1, 1}, N = 4 
Output: 0 
Explanation:
The operations are performed as follows:

1. A[0]( =1) = B[0]( =1): Delete the elements. Thereafter, the arrays are modified to {1, 0, 1} and {0, 1, 1} respectively.
2. A[0](=1) != B[0](= 0): Shift the A[0] to the end of the array A[]. Thereafter, the arrays are modified to { 0, 1, 1} and {0, 1, 1} respectively.
3. A[0]( =0) = B[0]( =0): Delete the elements. Thereafter, the arrays are modified to {1, 1} and {1, 1} respectively.
4. A[0]( =1) = B[0]( =1): Delete the elements. Thereafter, the arrays are modified to {1} and {1} respectively.
5. A[0]( =1) = B[0]( =1): Delete the elements. Thereafter, both arrays became empty.

Therefore, no elements are left in the array A[].

Input: A[] = {1, 0, 1, 1, 1, 1}, B[] = {1, 1, 0, 1, 0, 1}, N = 6 
Output: 2 

方法：给定的问题可以通过去除常见的 0 和 1 来解决，然后计算两个数组中 0 和 1 的唯一数量。考虑以下观察结果：

1. 只要数组A[] 中剩余的元素等于B[]的第一个元素，就可以删除这些元素。
2. 还可以观察到， A[]的元素顺序可以很容易地改变。
3. 因此，这个想法是保留A[] 中剩余的0和1的数量，如果在B[] 中遇到一个元素，使得相同的元素不再存在于A[] 中，则不再存在可以进行操作。

请按照以下步骤解决问题：

• 遍历数组A[]并计算变量中0和1的总数并将它们存储在变量中，分别说为零和一。
• 初始化一个变量说count as 0 存储执行的删除总数。
• 使用变量i遍历数组B[]并执行以下操作：
  • 如果B[i]等于0且zero>0 ，则将count的值增加1并将零减少1 。
  • 否则，如果B [i]等于1且一个> 0，则通过1和递减一个递增1计数的值。
  • 否则，退出循环，因为无法进一步执行更多操作。
• 最后，完成上述步骤后，打印N与count的差值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to calculate minimum size
// of the array A[] after performing
// the given operations
int minimumSizeAfterDeletion(int A[], int B[], int N)
{
    // Stores the count of 0s and 1s
    int zero = 0, one = 0;
 
    // Stores the total deletions performed
    int count = 0;
 
    // Traverse the array A[]
    for (int i = 0; i < N; i++) {
        if (A[i] == 0) {
            zero++;
        }
        else {
            one++;
        }
    }
 
    // Traverse array B[]
    for (int i = 0; i < N; i++) {
 
        // If the B[i] is 0 and zero is
        // greater than 0
        if (B[i] == 0 && zero > 0) {
            // Increment count by 1
            count++;
            // Decrement zero by 1
            zero--;
        }
 
        // Else if the B[i] is 1 and one is
        // greater than 0
        else if (B[i] == 1 && one > 0) {
            // Increment count by 1
            count++;
            // Decrement one by 1
            one--;
        }
 
        // Otherwise
        else {
            break;
        }
    }
 
    // Return the answer
    return N - count;
}
 
// Driver Code
int main()
{
 
    // Given Input
    int A[] = { 1, 0, 1, 1, 1, 1 };
    int B[] = { 1, 1, 0, 1, 0, 1 };
    int N = 6;
 
    // Function Call
    cout << minimumSizeAfterDeletion(A, B, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
     
// Function to calculate minimum size
// of the array A[] after performing
// the given operations
static int minimumSizeAfterDeletion(int A[], int B[],
                                    int N)
{
     
    // Stores the count of 0s and 1s
    int zero = 0, one = 0;
 
    // Stores the total deletions performed
    int count = 0;
 
    // Traverse the array A[]
    for(int i = 0; i < N; i++)
    {
        if (A[i] == 0)
        {
            zero++;
        }
        else
        {
            one++;
        }
    }
 
    // Traverse array B[]
    for(int i = 0; i < N; i++)
    {
         
        // If the B[i] is 0 and zero is
        // greater than 0
        if (B[i] == 0 && zero > 0)
        {
             
            // Increment count by 1
            count++;
             
            // Decrement zero by 1
            zero--;
        }
 
        // Else if the B[i] is 1 and one is
        // greater than 0
        else if (B[i] == 1 && one > 0)
        {
             
            // Increment count by 1
            count++;
             
            // Decrement one by 1
            one--;
        }
 
        // Otherwise
        else
        {
            break;
        }
    }
 
    // Return the answer
    return N - count;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Input
    int A[] = { 1, 0, 1, 1, 1, 1 };
    int B[] = { 1, 1, 0, 1, 0, 1 };
    int N = 6;
     
    // Function Call
    minimumSizeAfterDeletion(A, B, N);
    System.out.println(minimumSizeAfterDeletion(A, B, N));
}
}
 
// This code is contributed by Potta Lokesh

Python3

# Python3 program for the above approach
 
# Function to calculate minimum size
# of the array A[] after performing
# the given operations
def minimumSizeAfterDeletion(A, B, N):
     
    # Stores the count of 0s and 1s
    zero = 0
    one = 0
     
    # Stores the total deletions performed
    count = 0
     
    # Traverse the array A[]
    for i in range(N):
        if A[i] == 0:
            zero += 1
        else:
            one += 1
     
    # Traverse array B[]       
    for i in range(N):
         
        # If the B[i] is 0 and zero is
        # greater than 0
        if B[i] == 0 and zero > 0:
             
            # Increment count by 1
            count += 1
             
            # Decrement zero by 1
            zero -= 1
         
        # Else if the B[i] is 1 and one is
        # greater than 0
        elif B[i] == 1 and one > 0:
             
            # Increment count by 1
            count += 1
             
            # Decrement one by 1
            one -= 1
             
        # Otherwise
        else:
            break
     
    # Return the answer   
    return N - count
 
# Driver code
 
# Given input
A = [ 1, 0, 1, 1, 1, 1 ]
B = [ 1, 1, 0, 1, 0, 1 ]
N = 6
 
# Function call
print(minimumSizeAfterDeletion(A, B, N))
 
# This code is contributed by Parth Manchanda

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to calculate minimum size
// of the array A[] after performing
// the given operations
static int minimumSizeAfterDeletion(int []A, int []B, int N)
{
    // Stores the count of 0s and 1s
    int zero = 0, one = 0;
 
    // Stores the total deletions performed
    int count = 0;
 
    // Traverse the array A[]
    for (int i = 0; i < N; i++) {
        if (A[i] == 0) {
            zero++;
        }
        else {
            one++;
        }
    }
 
    // Traverse array B[]
    for (int i = 0; i < N; i++) {
 
        // If the B[i] is 0 and zero is
        // greater than 0
        if (B[i] == 0 && zero > 0) {
            // Increment count by 1
            count++;
            // Decrement zero by 1
            zero--;
        }
 
        // Else if the B[i] is 1 and one is
        // greater than 0
        else if (B[i] == 1 && one > 0) {
            // Increment count by 1
            count++;
            // Decrement one by 1
            one--;
        }
 
        // Otherwise
        else {
            break;
        }
    }
 
    // Return the answer
    return N - count;
}
 
// Driver Code
public static void Main()
{
 
    // Given Input
    int []A = { 1, 0, 1, 1, 1, 1 };
    int []B = { 1, 1, 0, 1, 0, 1 };
    int N = 6;
 
    // Function Call
    Console.Write(minimumSizeAfterDeletion(A, B, N));
}
}
 
// This code is contributed by ipg2016107.

Javascript

// Javascript program for the above approach
// Function to calculate minimum size
// of the array A[] after performing
// the given operations
function minimumSizeAfterDeletion(A, B, N)
{
     
    // Stores the count of 0s and 1s
    var zero = 0, one = 0;
 
    // Stores the total deletions performed
    var count = 0;
 
    // Traverse the array A[]
    for(var i = 0; i < N; i++)
    {
        if (A[i] == 0)
        {
            zero++;
        }
        else
        {
            one++;
        }
    }
 
    // Traverse array B[]
    for(var i = 0; i < N; i++)
    {
         
        // If the B[i] is 0 and zero is
        // greater than 0
        if (B[i] == 0 && zero > 0)
        {
             
            // Increment count by 1
            count++;
             
            // Decrement zero by 1
            zero--;
        }
 
        // Else if the B[i] is 1 and one is
        // greater than 0
        else if (B[i] == 1 && one > 0)
        {
             
            // Increment count by 1
            count++;
             
            // Decrement one by 1
            one--;
        }
 
        // Otherwise
        else
        {
            break;
        }
    }
 
    // Return the answer
    return N - count;
}
 
// Driver Code
    // Given Input
    var A = [ 1, 0, 1, 1, 1, 1 ];
    var B = [ 1, 1, 0, 1, 0, 1 ];
    var N = 6;
     
    // Function Call
    minimumSizeAfterDeletion(A, B, N);
    document.write(minimumSizeAfterDeletion(A, B, N));
 
// This code is contributed by shivanisinghss2110

输出

2

时间复杂度： O(N)
辅助空间： O(1)

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