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Maximum number of unique Triplets such that each element is selected only once

Given an array arr[] of size, N. Find the maximum number of triplets that can be made using array elements such that all elements in each triplet are different. Print the maximum number of possible triplets along with a list of the triplets.
Note: Each element of the array can belong to only 1 triplet.
Examples: 

Input: arr[] = {2, 2, 3, 3, 4, 4, 4, 4, 5} 
Output: 
Maximum number of possible triples : 2 
2 3 4 
3 4 5 
Explanation: 
We can form at most 2 triples using the given array such that each triple contains different elements. 
Input: arr[] = {1, 2, 3, 4, 5, 6, 7 } 
Output: 
Maximum number of possible triples : 2 
5 6 7 
2 3 4 
Explanation: 
We can form at most 2 triples using the given array such that each triple contains different elements. 

Naive Approach: The idea is to run three nested loops to generate all triplets and for every triplet, check if they are pairwise distinct and also check if each element of the array belongs to exactly 1 triplet.
Time Complexity: O(N3
Auxiliary Space: O(1) 
Efficient Approach: The problem can be solved Greedy Approach and keep taking triplets having a maximum frequency. Below are the steps: 
 

Below is the implementation of the above approach:
 




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that finds maximum number
// of triplets with different elements
void findTriplets(int ar[], int n)
{
 
    // Map M will store the frequency
    // of each element in the array
    unordered_map<int, int> mp;
 
    for (int x = 0; x < n; x++)
        mp[ar[x]]++;
 
    // Priority queue of pairs
    // {frequency, value}
    priority_queue<pair<int, int> > pq;
 
    for (auto& pa : mp)
        pq.push({ pa.second, pa.first });
 
    // ans will store possible triplets
    vector<array<int, 3> > ans;
 
    while (pq.size() >= 3) {
 
        // Extract top 3 elements
        pair<int, int> ar[3];
        for (int x = 0; x < 3; x++) {
            ar[x] = pq.top();
            pq.pop();
        }
 
        // Make a triplet
        ans.push_back({ ar[0].second,
                        ar[1].second,
                        ar[2].second });
 
        // Decrease frequency and push
        // back into priority queue if
        // non-zero frequency
        for (int x = 0; x < 3; x++) {
            ar[x].first--;
            if (ar[x].first)
                pq.push(ar[x]);
        }
    }
 
    // Print the triplets
    cout << "Maximum number of "
         << "possible triples: ";
    cout << ans.size() << endl;
 
    for (auto& pa : ans) {
 
        // Print the triplets
        for (int v : pa)
            cout << v << " ";
        cout << endl;
    }
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 2, 2, 3, 3, 4, 4, 4, 4, 5 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    findTriplets(arr, n);
    return 0;
}




// Java program for the
// above approach
import java.util.*;
import java.lang.*;
class GFG{
 
static class pair
{
  int first, second;
  pair(int first, int second)
  {
    this.first = first;
    this.second = second;
  }
}
 
// Function that finds maximum
// number of triplets with
// different elements
static void findTriplets(int arr[],
                         int n)
{
  // Map M will store the frequency
  // of each element in the array
  Map<Integer,
      Integer> mp = new HashMap<>();
 
  for (int x = 0; x < n; x++)
    mp.put(arr[x],
    mp.getOrDefault(arr[x], 0) + 1);
 
  // Priority queue of pairs
  // {frequency, value}
  PriorityQueue<pair> pq =
          new PriorityQueue<>((a, b) ->
                               a.first -
                               b.first);
 
  for (Map.Entry<Integer,
                 Integer> k : mp.entrySet())
    pq.add(new pair(k.getValue(),
                    k.getKey()));
 
  // ans will store possible
  // triplets
  ArrayList<List<Integer> > ans =
                 new ArrayList<>();
 
  while (pq.size() >= 3)
  {
    // Extract top 3 elements
    pair[] ar = new pair[3];
    for (int x = 0; x < 3; x++)
    {
      ar[x] = pq.peek();
      pq.poll();
    }
 
    // Make a triplet
    ans.add(Arrays.asList(ar[0].second,
                          ar[1].second,
                          ar[2].second));
 
    // Decrease frequency and push
    // back into priority queue if
    // non-zero frequency
    for (int x = 0; x < 3; x++)
    {
      ar[x].first--;
      if (ar[x].first != 0)
        pq.add(ar[x]);
    }
  }
 
  // Print the triplets
  System.out.println("Maximum number of " +
                     "possible triples: " +
                      ans.size());
 
  for (List<Integer> pa : ans)
  {
    // Print the triplets
    for (Integer v : pa)
      System.out.print(v + " ");
 
    System.out.println();
  }
}
 
// Driver function
public static void main(String[] args)
{
  // Given array arr[]
  int arr[] = {2, 2, 3, 3, 4,
               4, 4, 4, 5};
 
  int n = arr.length;
 
  // Function Call
  findTriplets(arr, n);
}
}
 
// This code is contributed by offbeat




# Python3 program for the above approach
from functools import cmp_to_key
def mycmp(a, b):
           
    if(a[0] == b[0]):
        return a[1] - b[1]
    return a[0] - b[0]
 
# Function that finds maximum number
# of triplets with different elements
def findTriplets(ar, n):
 
    # Map M will store the frequency
    # of each element in the array
    mp = {}
 
    for x in range(n):
 
        if(ar[x] in mp):
          mp[ar[x]] += 1
        else:
            mp[ar[x]] = 1
 
    # Priority queue of pairs
    # {frequency, value}
    pq = []
 
    for pa,pa2 in mp.items():
        pq.append([pa2, pa])
 
    # ans will store possible triplets
    ans = []
    pq.sort(key = cmp_to_key(mycmp))
 
    while (len(pq) >= 3):
 
        # Extract top 3 elements
        ar = [[0 for i in range(2)]for j in range(3)]
        for x in range(3):
            ar[x] = pq[len(pq)-1]
            pq.pop()
 
        # Make a triplet
        ans.append([ar[0][1],ar[1][1],ar[2][1]])
 
        # Decrease frequency and append
        # back into priority queue if
        # non-zero frequency
        for x in range(3):
            ar[x][0] -= 1
            if (ar[x][0]):
                pq.append(ar[x])
         
        pq.sort(key = cmp_to_key(mycmp))
 
    # Print the triplets
    print("Maximum number of " + "possible triples:",end=" ")
    print(len(ans))
 
    for pa in ans:
 
        # Print the triplets
        for v in pa:
            print(v ,end = " ")
        print()
     
 
# Driver Code
# Given array arr[]
arr = [2, 2, 3, 3, 4, 4, 4, 4, 5]
n = len(arr)
 
# Function Call
findTriplets(arr, n)
 
# This code is contributed by Shinjan Patra




// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
 
  // Function that finds maximum number
  // of triplets with different elements
  static void findTriplets(int[] arr, int n)
  {
 
    // Map M will store the frequency
    // of each element in the array
    Dictionary<int, int> mp = new Dictionary<int, int>();
    for (int x = 0; x < n; x++)
    {
      if(mp.ContainsKey(arr[x]))
      {
        mp[arr[x]]++;
      }
      else
      {
        mp[arr[x]] = 1;
      }
    }
 
    // Priority queue of pairs
    // {frequency, value}
    List<Tuple<int, int> > pq = new List<Tuple<int,int>>();
    int cnt = 0;
    foreach(KeyValuePair<int, int> pa in mp)
      pq.Add(new Tuple<int,int>(pa.Value, pa.Key));
 
    // ans will store possible triplets
    List<List<int>> ans = new List<List<int>>();
    pq.Sort();
    pq.Reverse();
    while (pq.Count >= 3)
    {
 
      // Extract top 3 elements
      Tuple<int, int>[] ar = new Tuple<int,int>[3];
      for (int x = 0; x < 3; x++)
      {
        ar[x] = pq[0];
        pq.RemoveAt(0);
      }
      ans.Add(new List<int>());
      ans[cnt].Add(ar[0].Item2);
      ans[cnt].Add(ar[1].Item2);
      ans[cnt].Add(ar[2].Item2);
 
      // Decrease frequency and push
      // back into priority queue if
      // non-zero frequency
      for (int x = 0; x < 3; x++)
      {
        ar[x] = new Tuple<int,int>(ar[x].Item1 - 1, ar[x].Item2);
        if (ar[x].Item1 != 0)
        {
          pq.Add(ar[x]);
          pq.Sort();
          pq.Reverse();
        }
      }
      cnt++;
    }
 
    // Print the triplets
    Console.Write("Maximum number of possible triples: ");
    Console.WriteLine(ans.Count);
    foreach(List<int> pa in ans)
    {
 
      // Print the triplets
      foreach(int v in pa)
        Console.Write(v + " ");
      Console.WriteLine();
    }
  }
 
  // Driver code
  static void Main()
  {
 
    // Given array arr[]
    int[] arr = { 2, 2, 3, 3, 4, 4, 4, 4, 5 };
 
    int n = arr.Length;
 
    // Function Call
    findTriplets(arr, n);
  }
}
 
// This code is contributed by divyeshrabadiya07.




<script>
 
// Javascript program for the above approach
 
// Function that finds maximum number
// of triplets with different elements
function findTriplets(ar, n)
{
 
    // Map M will store the frequency
    // of each element in the array
    var mp = new Map();
 
    for (var x = 0; x < n; x++)
    {
        if(mp.has(ar[x]))
          mp.set(ar[x], mp.get(ar[x])+1)
        else
            mp.set(ar[x],1)
    }
 
    // Priority queue of pairs
    // {frequency, value}
    var pq = [];
 
    for(var pa of mp)
        pq.push([pa[1], pa[0]]);
 
    // ans will store possible triplets
    var ans = [];
    pq.sort((a,b)=>{
          if(a[0]==b[0])
            return a[1]-b[1];
         return a[0]-b[0];
      });
 
    while (pq.length >= 3) {
 
        // Extract top 3 elements
        var ar = Array.from(Array(3), ()=>Array(2).fill(0));
        for (var x = 0; x < 3; x++) {
            ar[x] = pq[pq.length-1];
            pq.pop();
        }
 
        // Make a triplet
        ans.push([ar[0][1],
                        ar[1][1],
                        ar[2][1] ]);
 
        // Decrease frequency and push
        // back into priority queue if
        // non-zero frequency
        for (var x = 0; x < 3; x++) {
            ar[x][0]--;
            if (ar[x][0])
                pq.push(ar[x]);
        }
      pq.sort((a,b)=>{
          if(a[0]==b[0])
            return a[1]-b[1];
         return a[0]-b[0];
      });
    }
 
    // Print the triplets
    document.write("Maximum number of "
          + "possible triples: ");
    document.write(ans.length + "<br>");
 
    for (var pa of ans) {
 
        // Print the triplets
        for (var v of pa)
            document.write( v + " ");
        document.write("<br>");
    }
}
 
// Driver Code
// Given array arr[]
var arr = [2, 2, 3, 3, 4, 4, 4, 4, 5];
var n = arr.length;
 
// Function Call
findTriplets(arr, n);
 
// This code is contributed by noob20000
</script>

Output: 
Maximum number of possible triples: 2
4 3 2 
4 5 3

 

Time Complexity: O(N*log N) 
Auxiliary Space: O(N) 


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