给定两个整数N和K以及一个由重复元素组成的数组arr[] ，任务是将 N 个元素拆分为 K 个不同元素的集合。
例子：
Input: N = 5, K = 2, arr[] = {3, 2, 1, 2, 3}
Output:
( 3 2 1 )
( 2 3 )
Input: N = 5, K = 2, arr[] = {2, 1, 1, 2, 1}
Output: -1
Explanation:
It is not possible to split all the elements into K sets of distinct elements as 1 appears more than K times in the array.
方法：为了解决这个问题，我们使用一个映射来存储每个元素的频率。如果任何元素的频率超过K ，则打印-1 。维护另一个地图以存储每个相应频率的集合。如果没有元素的频率大于K ，则将所有对应频率的集合打印为所需的集合。
下面是上述方法的实现：
C++
// C++ Program to split N elements
// into exactly K sets consisting
// of no distinct elements
#include
using namespace std;
// Function to check if possible to
// split N elements into exactly K
// sets consisting of no distinct elements
void splitSets(int a[], int n, int k)
{
// Store the frequency
// of each element
map freq;
// Store the required sets
map > ans;
for (int i = 0; i < n; i++) {
// If frequency of a
// particular element
// exceeds K
if (freq[a[i]] + 1 > k) {
// Not possible
cout << -1 << endl;
return;
}
// Increase the frequency
freq[a[i]] += 1;
// Store the element for the
// respective set
ans[freq[a[i]]].push_back(a[i]);
}
// Display the sets
for (auto it : ans) {
cout << "( ";
for (int i : it.second) {
cout << i << " ";
}
cout << ")\n";
}
}
// Driver code
int main()
{
int arr[] = { 2, 1, 2, 3, 1,
4, 1, 3, 1, 4 };
int n = sizeof(arr) / sizeof(int);
int k = 4;
splitSets(arr, n, k);
return 0;
} Java
// Java program to split N elements
// into exactly K sets consisting
// of no distinct elements
import java.util.*;
class GFG{
// Function to check if possible to
// split N elements into exactly K
// sets consisting of no distinct elements
static void splitSets(int a[], int n, int k)
{
// Store the frequency
// of each element
Map freq = new HashMap<>();
// Store the required sets
Map> ans = new HashMap<>();
for(int i = 0; i < n; i++)
{
// If frequency of a
// particular element
// exceeds K
if(freq.get(a[i]) != null)
{
if (freq.get(a[i]) + 1 > k)
{
// Not possible
System.out.println(-1);
return;
}
}
// Increase the frequency
freq.put(a[i], freq.getOrDefault(a[i], 0) + 1 );
// Store the element for the
// respective set
if( ans.get(freq.get(a[i])) == null)
ans.put(freq.get(a[i]),
new ArrayList());
ans.get(freq.get(a[i])).add(a[i]);
}
// Display the sets
for(ArrayList it : ans.values())
{
System.out.print("( ");
for(int i = 0; i < it.size() - 1; i++)
{
System.out.print(it.get(i) + " ");
}
System.out.print(it.get(it.size() - 1));
System.out.println(" )");
}
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 2, 1, 2, 3, 1,
4, 1, 3, 1, 4 };
int n = arr.length;
int k = 4;
splitSets(arr, n, k);
}
}
// This code is contributed by coder001 Python3
# Python3 Program to split N elements
# into exactly K sets consisting
# of no distinct elements
# Function to check if possible to
# split N elements into exactly K
# sets consisting of no distinct elements
def splitSets(a, n, k) :
# Store the frequency
# of each element
freq = {}
# Store the required sets
ans = {}
for i in range(n) :
# If frequency of a
# particular element
# exceeds K
if a[i] in freq :
if freq[a[i]] + 1 > k :
# Not possible
print(-1)
return
# Increase the frequency
if a[i] in freq :
freq[a[i]] += 1
else :
freq[a[i]] = 1
# Store the element for the
# respective set
if freq[a[i]] in ans :
ans[freq[a[i]]].append(a[i])
else :
ans[freq[a[i]]] = [a[i]]
# Display the sets
for it in ans :
print("( ", end = "")
for i in ans[it] :
print(i , end = " ")
print(")")
arr = [ 2, 1, 2, 3, 1, 4, 1, 3, 1, 4 ]
n = len(arr)
k = 4
splitSets(arr, n, k)
# This code is contributed by divyesh072019C#
// C# Program to split N elements
// into exactly K sets consisting
// of no distinct elements
using System;
using System.Collections.Generic;
class GFG
{
// Function to check if possible to
// split N elements into exactly K
// sets consisting of no distinct elements
static void splitSets(int[] a, int n, int k)
{
// Store the frequency
// of each element
Dictionary freq = new Dictionary();
// Store the required sets
Dictionary> ans = new Dictionary>();
for (int i = 0; i < n; i++)
{
// If frequency of a
// particular element
// exceeds K
if(freq.ContainsKey(a[i]))
{
if(freq[a[i]] + 1 > k)
{
// Not possible
Console.WriteLine(-1);
return;
}
}
else if(1 > k)
{
// Not possible
Console.WriteLine(-1);
return;
}
// Increase the frequency
if(freq.ContainsKey(a[i]))
{
freq[a[i]] += 1;
}
else
{
freq[a[i]] = 1;
}
// Store the element for the
// respective set
if(ans.ContainsKey(freq[a[i]]))
{
ans[freq[a[i]]].Add(a[i]);
}
else
{
ans[freq[a[i]]] = new List();
ans[freq[a[i]]].Add(a[i]);
}
}
// Display the sets
foreach(KeyValuePair> it in ans)
{
Console.Write("( ");
foreach(int i in it.Value)
{
Console.Write(i + " ");
}
Console.WriteLine(")");
}
}
// Driver code
static void Main()
{
int[] arr = { 2, 1, 2, 3, 1,
4, 1, 3, 1, 4 };
int n = arr.Length;
int k = 4;
splitSets(arr, n, k);
}
}
// This code is contributed by divyeshrabadiya07 输出：
( 2 1 3 4 )
( 2 1 3 4 )
( 1 )
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