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: 
 

  • Store the frequency of all the numbers in a Map.
  • Make a max priority queue ans to store pairs in it where the first element in pair is the frequency of some element and the second element in pair is the element itself.
  • Now repeatedly extract the top 3 elements from the priority queue, make triplets using those 3 elements, decrease their frequency by 1 and again insert elements in priority queue using frequency is greater than 0.

Below is the implementation of the above approach:
 

C++

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// 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;
}

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Java

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// 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

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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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