将数组分成3组，这样X3可被X2整除，X2可被X1整除

给定一个包含N个元素的数组A ( N可被3整除)，任务是将数字分成3组，让该组具有3个元素X1，X2和X3，该组应满足以下条件：

• X1，X2和X3成对区分
• X3可被X2整除
• X2可被X1整除

如果将阵列分为N / 3，则打印-1。这样的组是不可能的。
注意：数组的元素将在1到6(含)范围内。
例子：

Input : N = 6, A[] = {2, 2, 1, 1, 4, 6}
Output : 1 2 4
         1 2 6
Explanation: 
Group 1: Pairs = {(1,2), (2,4), (1,4)}
All pairs are distinct, 
4 is divisible by 2 and 2 by 1.
Group 2: Pairs = {(1,2), (2,6), (1,6)}
All pairs are distinct, 
6 is divisible by 2 and 2 by 1.

Input : N = 6, A[] = {1, 1, 1, 6, 6, 3}
Output : -1

方法：
由于数组的值在1到6之间，因此只能进行以下类型的分组：

• 1 2 4
• 1 2 6
• 1 3 6

通过计算每个元素的频率开始。由于1在所有组中都是通用的，因此它必须精确地出现N / 3次。 4只能放入始终包含2的第一种组中。因此2的计数应大于4的计数。其余2只能放入第二种组中。现在，剩下的数字必须放在第三类中。如果在任何时候该计数都小于要求，则答案将为-1。
下面是上述方法的实现：

C++

// C++ program to split array in groups of 3
 
#include 
using namespace std;
 
// Function to print the groups after
// splitting array in groups of 3
void printGroups(int n, int a[])
{
    int ct[7] = { 0 }, grps = n / 3, i;
 
    // Count occurrence of each element
    for (i = 0; i < n; i++)
        ct[a[i]]++;
 
    // Check if it is possible to form the groups
    if (ct[1] != grps || (ct[4] + ct[6]) != grps
              || (ct[2] + ct[3]) != grps || ct[4] > ct[2])
    {
        cout << -1;
        return;
    }
 
    // Print groups that end at 4
    for (i = 0; i < ct[4]; i++)
        cout << "1 2 4\n";
 
    // Print groups that end at 6, with 2
    // in the middle
    for (i = 0; i < ct[2] - ct[4]; i++)
        cout << "1 2 6\n";
 
    // Print groups that have a 3 in the middle
    for (i = 0; i < ct[3]; i++)
        cout << "1 3 6\n";
}
 
// Driver Code
int main()
{
    int n = 6;
    int a[n] = { 2, 2, 1, 1, 4, 6 };
 
    printGroups(n, a);
 
    return 0;
}

Java

// Java program to split array in groups of 3
class GFG
{
 
    // Function to print the groups after
    // splitting array in groups of 3
    static void printGroups(int n, int a[])
    {
        int ct[] = new int[7], grps = n / 3, i;
 
        // Count occurrence of each element
        for (i = 0; i < n; i++)
        {
            ct[a[i]]++;
        }
 
        // Check if it is possible to form the groups
        if (ct[1] != grps || (ct[4] + ct[6]) != grps
            || (ct[2] + ct[3]) != grps || ct[4] > ct[2])
        {
            System.out.print(-1);
            return;
        }
 
        // Print groups that end at 4
        for (i = 0; i < ct[4]; i++)
        {
            System.out.print("1 2 4\n");
        }
 
        // Print groups that end at 6, with 2
        // in the middle
        for (i = 0; i < ct[2] - ct[4]; i++)
        {
            System.out.print("1 2 6\n");
        }
         
        // Print groups that have a 3 in the middle
        for (i = 0; i < ct[3]; i++)
        {
            System.out.print("1 3 6\n");
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 6;
        int a[] = {2, 2, 1, 1, 4, 6};
 
        printGroups(n, a);
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 program to split array in
# groups of 3
 
# Function to print the groups after
# splitting array in groups of 3
def printGroups(n, a):
 
    ct = [0 for i in range(7)]
    grps = n // 3
    i = 0
 
    # Count occurrence of each element
    for i in range(n):
        ct[a[i]] += 1
 
    # Check if it is possible to
    # form the groups
    if (ct[1] != grps or (ct[4] + ct[6]) != grps or
       (ct[2] + ct[3]) != grps or ct[4] > ct[2]):
        print(-1)
        return
 
    # Print groups that end at 4
    for i in range(ct[4]):
        print("1 2 4")
 
    # Print groups that end at 6, with 2
    # in the middle
    for i in range(ct[2] - ct[4]):
        print("1 2 6")
 
    # Print groups that have a 3 in the middle
    for i in range(ct[3]):
        print("1 3 6")
 
# Driver Code
n = 6
a = [2, 2, 1, 1, 4, 6 ]
 
printGroups(n, a)
 
# This code is contributed
# by Mohit Kumar

C#

// C# program to split array in groups of 3
using System;
 
class GFG
{
 
    // Function to print the groups after
    // splitting array in groups of 3
    static void printGroups(int n, int []a)
    {
        int []ct = new int[7];
        int grps = n / 3, i;
 
        // Count occurrence of each element
        for (i = 0; i < n; i++)
        {
            ct[a[i]]++;
        }
 
        // Check if it is possible to form the groups
        if (ct[1] != grps || (ct[4] + ct[6]) != grps ||
           (ct[2] + ct[3]) != grps || ct[4] > ct[2])
        {
            Console.Write(-1);
            return;
        }
 
        // Print groups that end at 4
        for (i = 0; i < ct[4]; i++)
        {
            Console.Write("1 2 4\n");
        }
 
        // Print groups that end at 6, with 2
        // in the middle
        for (i = 0; i < ct[2] - ct[4]; i++)
        {
            Console.Write("1 2 6\n");
        }
         
        // Print groups that have a 3 in the middle
        for (i = 0; i < ct[3]; i++)
        {
            Console.Write("1 3 6\n");
        }
    }
 
    // Driver Code
    public static void Main()
    {
        int n = 6;
        int []a = {2, 2, 1, 1, 4, 6};
 
        printGroups(n, a);
    }
}
 
// This code is contributed
// by Akanksha Rai

PHP

 $ct[2])
    {
        echo -1;
        return;
    }
 
    // Print groups that end at 4
    for ($i = 0; $i < $ct[4]; $i++)
        echo "1 2 4\n";
 
    // Print groups that end at 6, with 2
    // in the middle
    for ($i = 0; $i < $ct[2] - $ct[4]; $i++)
        echo "1 2 6\n";
 
    // Print groups that have a 3 in the middle
    for ($i = 0; $i < $ct[3]; $i++)
        echo "1 3 6\n";
}
 
// Driver Code
$n = 6;
$a = array(2, 2, 1, 1, 4, 6);
 
printGroups($n, $a);
 
// This code is contributed
// by Akanksha Rai
?>

Javascript

// Javascript program to split array in groups of 3
 
// Function to print the groups after
// splitting array in groups of 3
function printGroups(n, a)
{
    let ct = new Array(7).fill(0), grps = parseInt(n / 3), i;
 
    // Count occurrence of each element
    for (i = 0; i < n; i++)
        ct[a[i]]++;
 
    // Check if it is possible to form the groups
    if (ct[1] != grps || (ct[4] + ct[6]) != grps
              || (ct[2] + ct[3]) != grps || ct[4] > ct[2])
    {
        document.write(-1);
        return;
    }
 
    // Print groups that end at 4
    for (i = 0; i < ct[4]; i++)
        document.write("1 2 4<br>");
 
    // Print groups that end at 6, with 2
    // in the middle
    for (i = 0; i < ct[2] - ct[4]; i++)
        document.write("1 2 6<br>");
 
    // Print groups that have a 3 in the middle
    for (i = 0; i < ct[3]; i++)
        document.write("1 3 6<br>");
}
 
// Driver Code
    let n = 6;
    let a = [ 2, 2, 1, 1, 4, 6 ];
 
    printGroups(n, a);
 

输出：

1 2 4
1 2 6

时间复杂度： O(N)