通过删除三元组将给定的二进制数组简化为单个元素

给定二进制数组arr [] cof size N ，任务是通过以下两个操作将数组简化为单个元素：

• 连续的0或1的三元组保持不变。
• 可以将由两个0和一个1组成的连续数组元素的三元组转换成更频繁的元素。

例子：

Input: arr[] = {0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1} 
Output: No 
Explanation: 
Following are the operations performed on the array: 
{0, 1, 1} -> 1 modifies the array to {1, 1, 1, 0, 0, 1, 1, 1, 1} 
{1, 0, 0} -> 0 modifies the array to {1, 1, 0, 1, 1, 1, 1} 
{1, 0, 1} -> 1 modifies the array to {1, 1, 1, 1, 1} 
Since, all the remaining elements are 1, they remain unchanged. 
Therefore, the array cannot be reduced to a single element.

Input: arr[] = {1, 0, 0, 0, 1, 1, 1} 
Output: Yes 
Explanation: 
Following are the operations performed on the array: 
{1, 0, 0} -> 0 {0, 0, 1, 1, 1} 
{0, 0, 1} -> 0 {0, 1, 1} 
{0, 1, 1} -> 1 {1}

方法：
请按照以下步骤解决问题：

• 计算0和1的频率。
• 计算它们各自计数的绝对差。
• 如果差异为1，则只能将数组缩小为1。因此，请打印“是”。
• 否则，打印编号。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
  
// Function to check if it is possible to
// reduce the array to a single element
void solve(int arr[], int n)
{
    // Stores frequency of 0's
    int countzeroes = 0;
  
    // Stores frequency of 1's
    int countones = 0;
  
    for (int i = 0; i < n; i++) {
  
        if (arr[i] == 0)
            countzeroes++;
        else
            countones++;
    }
  
    // Condition for array to be reduced
    if (abs(countzeroes - countones) == 1)
        cout << "Yes";
  
    // Otherwise
    else
        cout << "No";
}
  
// Driver Code
int main()
{
    int arr[] = { 0, 1, 0, 0, 1, 1, 1 };
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    solve(arr, n);
  
    return 0;
}

Java

// Java program to implement
// the above approach
class GFG{
  
// Function to check if it is possible to
// reduce the array to a single element
static void solve(int arr[], int n)
{
      
    // Stores frequency of 0's
    int countzeroes = 0;
  
    // Stores frequency of 1's
    int countones = 0;
  
    for(int i = 0; i < n; i++) 
    {
        if (arr[i] == 0)
            countzeroes++;
        else
            countones++;
    }
  
    // Condition for array to be reduced
    if (Math.abs(countzeroes - countones) == 1)
        System.out.print("Yes");
  
    // Otherwise
    else
        System.out.print("No");
}
  
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 0, 1, 0, 0, 1, 1, 1 };
  
    int n = arr.length;
  
    solve(arr, n);
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 program to implement
# the above approach
  
# Function to check if it is possible to
# reduce the array to a single element
def solve(arr, n):
  
    # Stores frequency of 0's
    countzeroes = 0;
  
    # Stores frequency of 1's
    countones = 0;
  
    for i in range(n):
        if (arr[i] == 0):
            countzeroes += 1;
        else:
            countones += 1;
      
    # Condition for array to be reduced
    if (abs(countzeroes - countones) == 1):
        print("Yes");
  
    # Otherwise
    else:
        print("No");
  
# Driver Code
if __name__ == '__main__':
      
    arr = [ 0, 1, 0, 0, 1, 1, 1 ];
  
    n = len(arr);
  
    solve(arr, n);
  
# This code is contributed by Amit Katiyar

C#

// C# program to implement
// the above approach
using System;
  
class GFG{
  
// Function to check if it is possible to
// reduce the array to a single element
static void solve(int []arr, int n)
{
      
    // Stores frequency of 0's
    int countzeroes = 0;
  
    // Stores frequency of 1's
    int countones = 0;
  
    for(int i = 0; i < n; i++) 
    {
        if (arr[i] == 0)
            countzeroes++;
        else
            countones++;
    }
  
    // Condition for array to be reduced
    if (Math.Abs(countzeroes - countones) == 1)
        Console.Write("Yes");
  
    // Otherwise
    else
        Console.Write("No");
}
  
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 0, 1, 0, 0, 1, 1, 1 };
  
    int n = arr.Length;
  
    solve(arr, n);
}
}
  
// This code is contributed by 29AjayKumar

输出：

Yes

时间复杂度： O(N)
辅助空间： O(1)