打印所有可能的方法以将数组拆分为K个子集

给定大小为N且整数K的数组arr [] ，任务是打印所有可能的方法以将给定的数组拆分为K个子集。

例子：

Input: arr[] = { 1, 2, 3 }, K = 2
Output: { {{ 1, 2 }, { 3 }}, {{ 1, 3 }, { 2 }}, {{ 1 }, { 2, 3 }}}.

Input: arr[] = { 1, 2, 3, 4 }, K = 2
Output: { {{ 1, 2, 3 }, { 4 }}, {{ 1, 2, 4 }, { 3 }}, {{ 1, 2 }, { 3, 4 }}, {{ 1, 3, 4 }, { 2 }}, {{ 1, 3 }, { 2, 4 }}, {{ 1, 4 }, { 2, 3 }}, {{ 1 }, { 2 3, 4 }} }

方法：可以使用回溯来生成和打印所有子集来解决该问题。请按照以下步骤解决问题：

1. 使用以下递归关系遍历数组并将元素插入K个子集中的任何一个：

PartitionSub(i, K, N)

{

   for (j = 0; j < K; j++) {

      sub[j].push_back(arr[i])

      PartitionSub(i + 1, K, N)

      sub[j].pop_back()

   }

}

2. 如果K等于0或K> N ，则无法生成子集。
3. 如果插入到K个子集中的数组元素的数量等于N ，则打印该子集的元素。

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// arr: Store input array
// i: Stores current index of arr
// N: Stores length of arr
// K: Stores count of subsets
// nos: Stores count of feasible subsets formed
// v: Store K subsets of the given array
 
// Utility function to find all possibe
// ways to split array into K subsets
void PartitionSub(int arr[], int i,
                int N, int K, int nos,
                vector >& v)
{
 
    // If count of elements in K subsets
    // are greater than or equal to N
    if (i >= N) {
 
        // If count of subsets
        // formed is equal to K
        if (nos == K) {
 
            // Print K subsets by splitting
            // array into K subsets
            for (int x = 0; x < v.size(); x++) {
 
                cout << "{ ";
 
                // Print current subset
                for (int y = 0; y < v[x].size(); y++) {
                    cout << v[x][y];
 
                    // If current element is the last
                    // element of the subset
                    if (y == v[x].size() - 1) {
 
                        cout << " ";
                    }
 
                    // Otherwise
                    else {
 
                        cout << ", ";
                    }
                }
 
                if (x == v.size() - 1) {
 
                    cout << "}";
                }
                else {
 
                    cout << "}, ";
                }
            }
            cout << endl;
        }
 
        return;
    }
 
    for (int j = 0; j < K; j++) {
 
        // If any subset is occupied,
        // then push the element
        // in that first
        if (v[j].size() > 0) {
            v[j].push_back(arr[i]);
 
            // Recursively do the same
            // for remaining elements
            PartitionSub(arr, i + 1, N, K, nos, v);
 
            // Backtrack
            v[j].pop_back();
        }
 
        // Otherwise, push it in an empty
        // subset and increase the
        // subset count by 1
        else {
 
            v[j].push_back(arr[i]);
            PartitionSub(arr, i + 1, N, K, nos + 1, v);
            v[j].pop_back();
 
            // Break to avoid the case of going in
            // other empty subsets, if available,
            // and forming the same combination
            break;
        }
    }
}
 
// Function to to find all possibe ways to
// split array into K subsets
void partKSubsets(int arr[], int N, int K)
{
 
    // Stores K subset by splitting array
    // into K subsets
    vector > v(K);
 
    // Size of each subset must
    // be less than the number of elements
    if (K == 0 || K > N) {
 
        cout << "Not Possible" << endl;
    }
    else {
 
        cout << "The Subset Combinations are: " << endl;
        PartitionSub(arr, 0, N, K, 0, v);
    }
}
 
// Driver Code
int main()
{
 
    // Given array
    int arr[] = { 1, 2, 3, 4 };
 
    // Given K
    int K = 2;
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Prints all possible
    // splits into subsets
    partKSubsets(arr, N, K);
}

Java

// Java program for above approach
import java.util.*;
import java.lang.*;
class Gfg
{
 
  // arr: Store input array
  // i: Stores current index of arr
  // N: Stores length of arr
  // K: Stores count of subsets
  // nos: Stores count of feasible subsets formed
  // v: Store K subsets of the given array
 
  // Utility function to find all possibe
  // ways to split array into K subsets
  static void PartitionSub(int arr[], int i,
                           int N, int K, int nos,
                           ArrayList> v)
  {
 
    // If count of elements in K subsets
    // are greater than or equal to N
    if (i >= N)
    {
 
      // If count of subsets
      // formed is equal to K
      if (nos == K)
      {
 
        // Print K subsets by splitting
        // array into K subsets
        for (int x = 0; x < v.size(); x++)
        {
 
          System.out.print("{ ");
 
          // Print current subset
          for (int y = 0; y < v.get(x).size(); y++)
          {
            System.out.print(v.get(x).get(y));
 
            // If current element is the last
            // element of the subset
            if (y == v.get(x).size() - 1)
            {
              System.out.print(" ");
            }
 
            // Otherwise
            else
            {
              System.out.print(", ");
            }
          }
 
          if (x == v.size() - 1)
          {
            System.out.print("}");
          }
          else
          {
            System.out.print("}, ");
          }
        }
        System.out.println();;
      }
      return;
    }
 
    for (int j = 0; j < K; j++)
    {
 
      // If any subset is occupied,
      // then push the element
      // in that first
      if (v.get(j).size() > 0)
      {
        v.get(j).add(arr[i]);
 
        // Recursively do the same
        // for remaining elements
        PartitionSub(arr, i + 1, N, K, nos, v);
 
        // Backtrack
        v.get(j).remove(v.get(j).size()-1);
      }
 
      // Otherwise, push it in an empty
      // subset and increase the
      // subset count by 1
      else
      {
 
        v.get(j).add(arr[i]);
        PartitionSub(arr, i + 1, N, K, nos + 1, v);
        v.get(j).remove(v.get(j).size()-1);
 
        // Break to avoid the case of going in
        // other empty subsets, if available,
        // and forming the same combination
        break;
      }
    }
  }
 
  // Function to to find all possibe ways to
  // split array into K subsets
  static void partKSubsets(int arr[], int N, int K)
  {
 
    // Stores K subset by splitting array
    // into K subsets
    ArrayList> v = new ArrayList<>();
 
    for(int i = 0; i < K; i++)
      v.add(new ArrayList<>());
 
    // Size of each subset must
    // be less than the number of elements
    if (K == 0 || K > N)
    {
      System.out.println("Not Possible");
    }
    else
    {
      System.out.println("The Subset Combinations are: ");
      PartitionSub(arr, 0, N, K, 0, v);
    }
  }
 
  // Driver function
  public static void main (String[] args)
  {
 
    // Given array
    int arr[] = { 1, 2, 3, 4 };
 
    // Given K
    int K = 2;
 
    // Size of the array
    int N = arr.length;
 
    // Prints all possible
    // splits into subsets
    partKSubsets(arr, N, K);
  }
}
 
// This code is contributed by offbeat

Python3

# Python 3 program for the above approach
 
# arr: Store input array
# i: Stores current index of arr
# N: Stores length of arr
# K: Stores count of subsets
# nos: Stores count of feasible subsets formed
# v: Store K subsets of the given array
 
# Utility function to find all possibe
# ways to split array into K subsets
def PartitionSub(arr, i, N, K, nos, v):
   
    # If count of elements in K subsets
    # are greater than or equal to N
    if (i >= N):
       
        # If count of subsets
        # formed is equal to K
        if (nos == K):
           
            # Print K subsets by splitting
            # array into K subsets
            for x in range(len(v)):
                print("{ ", end = "")
 
                # Print current subset
                for y in range(len(v[x])):
                    print(v[x][y], end = "")
 
                    # If current element is the last
                    # element of the subset
                    if (y == len(v[x]) - 1):
                        print(" ", end = "")
 
                    # Otherwise
                    else:
                        print(", ", end = "")
 
                if (x == len(v) - 1):
                    print("}", end = "")
                 
                else:
                    print("}, ", end = "")
            print("\n", end = "")
        return
    for j in range(K):
       
        # If any subset is occupied,
        # then push the element
        # in that first
        if (len(v[j]) > 0):
            v[j].append(arr[i])
 
            # Recursively do the same
            # for remaining elements
            PartitionSub(arr, i + 1, N, K, nos, v)
             
            # Backtrack
            v[j].remove(v[j][len(v[j]) - 1])
 
        # Otherwise, push it in an empty
        # subset and increase the
        # subset count by 1
        else:
            v[j].append(arr[i])
            PartitionSub(arr, i + 1, N, K, nos + 1, v)
            v[j].remove(v[j][len(v[j]) - 1])
 
            # Break to avoid the case of going in
            # other empty subsets, if available,
            # and forming the same combination
            break
 
# Function to to find all possibe ways to
# split array into K subsets
def partKSubsets(arr, N, K):
   
    # Stores K subset by splitting array
    # into K subsets
    v = [[] for i in range(K)]
 
    # Size of each subset must
    # be less than the number of elements
    if (K == 0 or K > N):
        print("Not Possible", end = "")
    else:
        print("The Subset Combinations are: ")
        PartitionSub(arr, 0, N, K, 0, v)
 
# Driver Code
if __name__ == '__main__':
   
    # Given array
    arr =  [1, 2, 3, 4]
 
    # Given K
    K = 2
 
    # Size of the array
    N = len(arr)
 
    # Prints all possible
    # splits into subsets
    partKSubsets(arr, N, K)
     
    # This code is contributed by bgangwar59.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
 
  // arr: Store input array
  // i: Stores current index of arr
  // N: Stores length of arr
  // K: Stores count of subsets
  // nos: Stores count of feasible subsets formed
  // v: Store K subsets of the given array
 
  // Utility function to find all possibe
  // ways to split array into K subsets
  static void PartitionSub(int []arr, int i,
                           int N, int K, int nos,
                           List>v)
{
 
  // If count of elements in K subsets
  // are greater than or equal to N
  if (i >= N) {
 
    // If count of subsets
    // formed is equal to K
    if (nos == K) {
 
      // Print K subsets by splitting
      // array into K subsets
      for (int x = 0; x < v.Count; x++) {
 
        Console.Write("{ ");
 
        // Print current subset
        for (int y = 0; y < v[x].Count; y++) {
          Console.Write(v[x][y]);
 
          // If current element is the last
          // element of the subset
          if (y == v[x].Count - 1) {
 
            Console.Write(" ");
          }
 
          // Otherwise
          else {
 
            Console.Write(", ");
          }
        }
 
        if (x == v.Count - 1) {
 
          Console.Write("}");
        }
        else {
 
          Console.Write("}, ");
        }
      }
      Console.Write("\n");
    }
 
    return;
  }
 
  for (int j = 0; j < K; j++) {
 
    // If any subset is occupied,
    // then push the element
    // in that first
    if (v[j].Count > 0) {
      v[j].Add(arr[i]);
 
      // Recursively do the same
      // for remaining elements
      PartitionSub(arr, i + 1, N, K, nos, v);
 
      // Backtrack
      v[j].RemoveAt(v[j].Count - 1);
    }
 
    // Otherwise, push it in an empty
    // subset and increase the
    // subset count by 1
    else {
 
      v[j].Add(arr[i]);
      PartitionSub(arr, i + 1, N, K, nos + 1, v);
      v[j].RemoveAt(v[j].Count - 1);
 
      // Break to avoid the case of going in
      // other empty subsets, if available,
      // and forming the same combination
      break;
    }
  }
}
 
 // Function to to find all possibe ways to
 // split array into K subsets
 static void partKSubsets(int []arr, int N, int K)
 {
 
   // Stores K subset by splitting array
   // into K subsets
   List > v = new List>();
   for(int i=0;i());
 
   // Size of each subset must
   // be less than the number of elements
   if (K == 0 || K > N) {
 
     Console.WriteLine("Not Possible");
   }
   else {
 
     Console.WriteLine("The Subset Combinations are: ");
     PartitionSub(arr, 0, N, K, 0, v);
   }
 }
 
 // Driver Code
 public static void Main()
 {
 
   // Given array
   int []arr = {1, 2, 3, 4};
 
   // Given K
   int K = 2;
 
   // Size of the array
   int N = arr.Length;
 
   // Prints all possible
   // splits into subsets
   partKSubsets(arr, N, K);
 }
}
 
// This code is contributed by SURENDRA_GANGWAR.

输出

The Subset Combinations are: 
{ 1, 2, 3 }, { 4 }
{ 1, 2, 4 }, { 3 }
{ 1, 2 }, { 3, 4 }
{ 1, 3, 4 }, { 2 }
{ 1, 3 }, { 2, 4 }
{ 1, 4 }, { 2, 3 }
{ 1 }, { 2, 3, 4 }