如果连续三元组在给定的二进制数组中可被 3 整除，则通过删除中间元素查找 0 是否被删除更多或 1

给定一个大小为N的1和0 的二进制数组a[] 。如果a[i-1]+a[i]+a[i+1]可被3整除，则任务是删除一个元素。如果删除的1数量多于0则打印1 ，否则打印0 。

例子：

Input: a[] = { 1, 1, 1, 0, 1, 0, 0}
Output: 1
Explanation: Remove the second 1 from the left since that is the only ‘1’ for which (a[i]+a[i-1]+a[i+1]) %3==0. So print 1.

Input: a[] = { 1, 1}
Output: 0 
Explanation: No removal possible, so print 0.

方法：这个想法是基于这样的观察，如果两个邻居都等于当前元素，因为如果(a[i]=a[i-1]=a[i+1])那么总和总是可以被3整除。假设A存储删除的1 的数量， B存储删除的0 的数量。如果第 i 个元素等于邻居，则如果a[i] =1 ，则增加A ，否则增加B 。如果A的计数大于B ，则打印1 ，否则打印2=0 。请按照以下步骤解决问题：

• 将变量A和B初始化为0以存储移除的1和0的个数。
• 使用变量i迭代范围[0, N)并执行以下步骤：
  • 如果a[i-1]、a[i]和a[i+1]相等，则如果a[i]等于1，则A的值加1 ，否则B加1。
• 如果A大于B，则打印1 ，否则打印0。

下面是上述方法的实现。

C++

#include 
using namespace std;
 
// Function to if more number of
// 1's are removed or 0's
void solution(vector a)
{
 
    // Stores count of 1's removes
    int A = 0;
 
    // Stores count of 0's removes
    int B = 0;
 
    // Traverse the array
    for (int i = 1; i < a.size() - 1; i++) {
 
        // Check the divisibility
        if (a[i] == a[i - 1] && a[i] == a[i + 1]) {
 
            // Check for 1 or 0
            if (a[i] == 1)
                A++;
            else
                B++;
        }
    }
 
    // Print the result
    if (A > B)
        cout << ("1");
    else
        cout << ("0");
}
 
// Driver Code
int main()
{
 
    vector a = { 1, 1, 1, 0, 1, 0, 0 };
    solution(a);
    return 0;
}
 
// This code is contributed by lokeshpotta20.

Java

// Java program for above approach
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Function to if more number of
    // 1's are removed or 0's
    public static void solution(int[] a)
    {
 
        // Stores count of 1's removes
        int A = 0;
 
        // Stores count of 0's removes
        int B = 0;
 
        // Traverse the array
        for (int i = 1; i < a.length - 1; i++) {
 
            // Check the divisibility
            if (a[i] == a[i - 1]
                && a[i] == a[i + 1]) {
 
                // Check for 1 or 0
                if (a[i] == 1)
                    A++;
                else
                    B++;
            }
        }
 
        // Print the result
        if (A > B)
            System.out.println("1 ");
        else
            System.out.println("0 ");
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int a[] = { 1, 1, 1, 0, 1, 0, 0 };
        solution(a);
    }
}

Python3

# Python3 program for above approach
 
 
# Function to if more number of
# 1's are removed or 0's
def solution(a) :
 
    # Stores count of 1's removes
    A = 0;
 
    # Stores count of 0's removes
    B = 0;
 
    # Traverse the array
    for i in range(1 , len(a)- 1) :
 
        # Check the divisibility
        if (a[i] == a[i - 1] and a[i] == a[i + 1]) :
 
            # Check for 1 or 0
            if (a[i] == 1) :
                A += 1;
            else :
                B += 1;
 
    # Print the result
    if (A > B) :
        print("1");
    else :
        print("0");
 
# Driver Code
if __name__ == "__main__" :
 
 
    a = [ 1, 1, 1, 0, 1, 0, 0 ];
    solution(a);
 
    # This code is contributed by AnkThon

C#

// C# program for above approach
using System;
public class GFG {
 
    // Function to if more number of
    // 1's are removed or 0's
    public static void solution(int[] a)
    {
 
        // Stores count of 1's removes
        int A = 0;
 
        // Stores count of 0's removes
        int B = 0;
 
        // Traverse the array
        for (int i = 1; i < a.Length - 1; i++) {
 
            // Check the divisibility
            if (a[i] == a[i - 1]
                && a[i] == a[i + 1]) {
 
                // Check for 1 or 0
                if (a[i] == 1)
                    A++;
                else
                    B++;
            }
        }
 
        // Print the result
        if (A > B)
            Console.WriteLine("1 ");
        else
            Console.WriteLine("0 ");
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int []a = { 1, 1, 1, 0, 1, 0, 0 };
        solution(a);
    }
}
 
// This code is contributed by AnkThon

Javascript

 
 
// Function to if more number of
// 1's are removed or 0's
function solution(a) {
 
  // Stores count of 1's removes
  let A = 0;
 
  // Stores count of 0's removes
  let B = 0;
 
  // Traverse the array
  for (let i = 1; i < a.length - 1; i++) {
 
    // Check the divisibility
    if (a[i] == a[i - 1] && a[i] == a[i + 1]) {
 
      // Check for 1 or 0
      if (a[i] == 1)
        A++;
      else
        B++;
    }
  }
 
  // Print the result
  if (A > B)
    document.write(("1"));
  else
    document.write(("0"));
}
 
// Driver Code
 
 
let a = [1, 1, 1, 0, 1, 0, 0];
solution(a);
 
// This code is contributed by Saurabh Jaiswal.

输出

1 

时间复杂度： O(N)
辅助空间： O(1)