检查数组的任何排列是否包含不被3整除的每个相邻对的和

给定一个由N个整数组成的数组arr [] ，任务是检查每个相邻元素对的和不能被3整除的数组元素是否存在任何排列。如果可能，请打印“是” 。否则，打印“否” 。

例子：

Input: arr[] = {1, 2, 3, 3}
Output: Yes
Explanation:
Since there exist at least 1 combination {3, 2, 3, 1} where sum of every adjacent pairs is not divisible by 3.

Input: arr[] = {3, 6, 1, 9}
Output: No

天真的方法：最简单的方法是生成给定数组的所有置换，并检查是否存在其中两个相邻元素之和不能被3整除的布置。如果发现是真的，则打印“是” 。否则，打印“否” 。

时间复杂度： O(N！)
辅助空间： O(1)

高效方法：为了优化上述方法，我们的想法是观察所有数组元素{{0，1，2}}的唯一可能余数。要将这三个数字分开，以使两个相邻元素的和不能被3整除，请按照以下步骤操作：

1. 计数所有的数字成具有余数为0，1和2三个部分。设计数分别为a ， b和c 。
2. 现在，将余数为0的数与余数为1或2的数进行排列，以使它们的和不会被3整除。以下是可以满足此条件的条件：
   • 如果a≥1且a≤b + c + 1
   • 如果a和b都等于0且c> 0
   • 如果a和c都等于0并且b> 0
3. 如果无法以上述方式排列所有数字，则不会发生排列，以使它们的相邻元素之和不能被3整除。因此，请打印“ No” 。
4. 如果发现步骤2中的条件为真，则打印“是” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to checks if any permutation
// of the array exists whose sum of
// adjacent pairs is not divisible by 3
void factorsOf3(int arr[], int N)
{
    int a = 0, b = 0, c = 0;
    for (int i = 0; i < N; i++) {
 
        // Count remainder 0
        if (arr[i] % 3 == 0)
            a++;
 
        // Count remainder 1
        else if (arr[i] % 3 == 1)
            b++;
 
        // Count remainder 2
        else if (arr[i] % 3 == 2)
            c++;
    }
 
    // Condition for valid arrangements
    if (a >= 1 && a <= b + c + 1)
        cout << "Yes" << endl;
    else if (a == 0 && b == 0 && c > 0)
        cout << "Yes" << endl;
    else if (a == 0 && c == 0 && b > 0)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 2, 3, 3 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    factorsOf3(arr, N);
 
    return 0;
}

Java

// Java program for
// the above approach
class GFG{
 
// Function to checks if any permutation
// of the array exists whose sum of
// adjacent pairs is not divisible by 3
static void factorsOf3(int arr[], int N)
{
  int a = 0, b = 0, c = 0;
  for (int i = 0; i < N; i++)
  {
    // Count remainder 0
    if (arr[i] % 3 == 0)
      a++;
 
    // Count remainder 1
    else if (arr[i] % 3 == 1)
      b++;
 
    // Count remainder 2
    else if (arr[i] % 3 == 2)
      c++;
  }
 
  // Condition for valid arrangements
  if (a >= 1 && a <= b + c + 1)
    System.out.print("Yes" + "\n");
  else if (a == 0 && b == 0 && c > 0)
    System.out.print("Yes" + "\n");
  else if (a == 0 && c == 0 && b > 0)
    System.out.print("Yes" + "\n");
  else
    System.out.print("No" + "\n");
}
 
// Driver Code
public static void main(String[] args)
{
  // Given array arr[]
  int arr[] = {1, 2, 3, 3};
 
  int N = arr.length;
 
  // Function Call
  factorsOf3(arr, N);
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach
 
# Function to checks if any permutation
# of the array exists whose sum of
# adjacent pairs is not divisible by 3
def factorsOf3(arr, N):
 
    a = 0
    b = 0
    c = 0
 
    for i in range(N):
 
        # Count remainder 0
        if (arr[i] % 3 == 0):
            a += 1
        # Count remainder 1
        elif (arr[i] % 3 == 1):
            b += 1
        # Count remainder 2
        elif (arr[i] % 3 == 2):
            c += 1
 
    # Condition for valid arrangements
    if (a >= 1 and a <= b + c + 1):
        print("Yes")
    elif (a == 0 and b == 0 and c > 0):
        print("Yes")
    elif (a == 0 and c == 0 and b > 0):
        print("Yes")
    else:
        print("No")
 
# Driver Code
 
# Given array arr[]
arr = [ 1, 2, 3, 3 ]
N = len(arr)
 
# Function call
factorsOf3(arr, N)
 
# This code is contributed by Shivam Singh

C#

// C# program for
// the above approach
using System;
class GFG{
 
// Function to checks if any
// permutation of the array
// exists whose sum of
// adjacent pairs is not
// divisible by 3
static void factorsOf3(int []arr,
                       int N)
{
  int a = 0, b = 0, c = 0;
  for (int i = 0; i < N; i++)
  {
    // Count remainder 0
    if (arr[i] % 3 == 0)
      a++;
 
    // Count remainder 1
    else if (arr[i] % 3 == 1)
      b++;
 
    // Count remainder 2
    else if (arr[i] % 3 == 2)
      c++;
  }
 
  // Condition for valid arrangements
  if (a >= 1 && a <= b + c + 1)
    Console.Write("Yes" + "\n");
  else if (a == 0 && b == 0 && c > 0)
    Console.Write("Yes" + "\n");
  else if (a == 0 && c == 0 && b > 0)
    Console.Write("Yes" + "\n");
  else
    Console.Write("No" + "\n");
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given array []arr
  int []arr = {1, 2, 3, 3};
 
  int N = arr.Length;
 
  // Function Call
  factorsOf3(arr, N);
}
}
 
// This code is contributed by shikhasingrajput

Javascript

 
// Javascript program for the above approach
 
// Function to checks if any permutation
// of the array exists whose sum of
// adjacent pairs is not divisible by 3
function factorsOf3(arr, N)
{
    let a = 0, b = 0, c = 0;
    for(let i = 0; i < N; i++)
    {
         
        // Count remainder 0
        if (arr[i] % 3 == 0)
            a++;
         
        // Count remainder 1
        else if (arr[i] % 3 == 1)
            b++;
         
        // Count remainder 2
        else if (arr[i] % 3 == 2)
            c++;
    }
     
    // Condition for valid arrangements
    if (a >= 1 && a <= b + c + 1)
        document.write("Yes" + "<br/>");
    else if (a == 0 && b == 0 && c > 0)
        document.write("Yes" + "<br/>");
    else if (a == 0 && c == 0 && b > 0)
        document.write("Yes" + "<br/>");
    else
        document.write("No" + "<br/>");
}
 
// Driver Code
 
// Given array arr[]
let arr = [ 1, 2, 3, 3 ];
 
let N = arr.length;
 
// Function Call
factorsOf3(arr, N);
 
// This code is contributed by chinmoy1997pal 
 

输出：

Yes

时间复杂度： O(N)
辅助空间： O(1)