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Merge first two minimum elements of the array until all the elements are greater than K
  • Last Updated : 13 Jan, 2021
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Given an array arr[] and an integer K, the task is to find the number of merge operation required such that all the elements of the array is greater than or equal to K.

Merge Process of the Element – 

New Element = 
   1 * (First Minimum element) + 
   2 * (Second Minimum element)

Examples:  

Input: arr[] = {1, 2, 3, 9, 10, 12}, K = 7 
Output:
Explanation: 
After the first merge operation elements 1 and 2 is removed, 
and the element (1*1 + 2*2 = 5) is inserted into the array 
{3, 5, 9, 10, 12} 
After the second merge operation elements 3 and 5 is removed, 
and the element (3*1 + 5*2 = 13) is inserted into the array 
{9, 10, 12, 13} 
Thus, 2 operations are required such that all elements are greater than K.

Input: arr[] = {52, 96, 13, 37}, K = 10 
Output:
Explanation: 
All the elements of the array are greater than K already. 
Therefore, no merge operation is required. 



Approach: The idea is to use Min-Heap to store the elements and then merge the two minimum elements of the array until the minimum element of the array is greater than or equal to K. 

Below is the implementation of the above approach: 

C++




// C++ implementation to merge the
// elements of the array untill all
// the array element of the array
// greater than or equal to K
 
#include <bits/stdc++.h>
 
using namespace std;
 
// Function to find the minimum
// operation required to merge
// elements of the array
int minOperations(int arr[], int K,
                          int size)
{
    int least, second_least,
       min_operations = 0,
       new_ele = 0, flag = 0;
 
    // Heap to store the elements
    // of the array and to extract
    // minimum elements of O(logN)
    priority_queue<int, vector<int>,
                 greater<int> > heap;
                  
    // Loop to push all the elements
    // of the array into heap
    for (int i = 0; i < size; i++) {
        heap.push(arr[i]);
    }
     
    // Loop to merge the minimum
    // elements untill there is only
    // all the elements greater than K
    while (heap.size() != 1) {
         
        // Condition to check minimum
        // element of the array is
        // greater than the K
        if (heap.top() >= K) {
            flag = 1;
            break;
        }
         
        // Merge the two minimum
        // elements of the heap
        least = heap.top();
        heap.pop();
        second_least = heap.top();
        heap.pop();
        new_ele = (1 * least) +
            (2 * second_least);
        min_operations++;
        heap.push(new_ele);
    }
    if (heap.top() >= K) {
        flag = 1;
    }
    if (flag == 1) {
        return min_operations;
    }
    return -1;
}
 
// Driver Code
int main()
{
    int N = 6, K = 7;
    int arr[] = { 1, 2, 3, 9, 10, 12 };
    int size = sizeof(arr) / sizeof(arr[0]);
    cout << minOperations(arr, K, size);
    return 0;
}

Java




// Java implementation to merge the
// elements of the array untill all
// the array element of the array
// greater than or equal to K
import java.util.Collections;
import java.util.PriorityQueue;
 
class GFG{
 
// Function to find the minimum
// operation required to merge
// elements of the array
static int minOperations(int arr[], int K,
                         int size)
{
    int least, second_least,
        min_operations = 0,
        new_ele = 0, flag = 0;
 
    // Heap to store the elements
    // of the array and to extract
    // minimum elements of O(logN)
    PriorityQueue<Integer> heap = new PriorityQueue<>();
     
    // priority_queue<int, vector<int>,
    // greater<int> > heap;
 
    // Loop to push all the elements
    // of the array into heap
    for(int i = 0; i < size; i++)
    {
        heap.add(arr[i]);
    }
 
    // Loop to merge the minimum
    // elements untill there is only
    // all the elements greater than K
    while (heap.size() != 1)
    {
         
        // Condition to check minimum
        // element of the array is
        // greater than the K
        if (heap.peek() >= K)
        {
            flag = 1;
            break;
        }
 
        // Merge the two minimum
        // elements of the heap
        least = heap.poll();
        second_least = heap.poll();
        new_ele = (1 * least) +
                  (2 * second_least);
        min_operations++;
        heap.add(new_ele);
    }
    if (heap.peek() >= K)
    {
        flag = 1;
    }
    if (flag == 1)
    {
        return min_operations;
    }
    return -1;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 6, K = 7;
    int arr[] = { 1, 2, 3, 9, 10, 12 };
    int size = arr.length;
     
    System.out.println(minOperations(arr, K, size));
}
}
 
// This code is contributed by sanjeev2552

Python3




# Python3 implementation to merge the
# elements of the array untill all
# the array element of the array
# greater than or equal to K
 
# Function to find the minimum
# operation required to merge
# elements of the array
def minOperations(arr, K, size):
    least, second_least = 0, 0
    min_operations = 0
    new_ele, flag = 0, 0
 
    # Heap to store the elements
    # of the array and to extract
    # minimum elements of O(logN)
    heap = []
 
    # Loop to append all the elements
    # of the array into heap
    for i in range(size):
        heap.append(arr[i])
 
    heap  = sorted(heap)[::-1]
 
    # Loop to merge the minimum
    # elements untill there is only
    # all the elements greater than K
    while (len(heap) > 0):
 
        # Condition to check minimum
        # element of the array is
        # greater than the K
        if (heap[-1] >= K):
            flag = 1
            break
 
        # Merge the two minimum
        # elements of the heap
        least = heap[-1]
        del heap[-1]
        second_least = heap[-1]
        del heap[-1]
        new_ele = (1 * least) + (2 * second_least)
        min_operations += 1
        heap.append(new_ele)
 
        heap = sorted(heap)[::-1]
 
    if (heap[-1] >= K):
        flag = 1
 
    if (flag == 1):
        return min_operations
    return -1
 
# Driver Code
if __name__ == '__main__':
    N, K = 6, 7
    arr = [1, 2, 3, 9, 10, 12]
    size = len(arr)
    print(minOperations(arr, K, size))
 
# This code is contributed by mohit kumar 29

C#




// C# implementation to merge the
// elements of the array untill all
// the array element of the array
// greater than or equal to K
using System;
using System.Collections.Generic;
class GFG
{
 
  // Function to find the minimum
  // operation required to merge
  // elements of the array
  static int minOperations(int[] arr, int K, int size)
  {
    int least, second_least, min_operations = 0,
    new_ele = 0, flag = 0;
 
    // Heap to store the elements
    // of the array and to extract
    // minimum elements of O(logN)
    List<int> heap = new List<int>();
 
    // priority_queue<int, vector<int>,
    // greater<int> > heap;
 
    // Loop to push all the elements
    // of the array into heap
    for(int i = 0; i < size; i++)
    {
      heap.Add(arr[i]);
    }
    heap.Sort();
    heap.Reverse();
 
    // Loop to merge the minimum
    // elements untill there is only
    // all the elements greater than K
    while(heap.Count != 1)
    {
 
      // Condition to check minimum
      // element of the array is
      // greater than the K
      if(heap[heap.Count - 1] >= K)
      {
        flag = 1;
        break;
      }
 
      // Merge the two minimum
      // elements of the heap
      least = heap[heap.Count - 1];
      heap.RemoveAt(heap.Count - 1);
      second_least = heap[heap.Count - 1];
      heap.RemoveAt(heap.Count - 1);
      new_ele = (1 * least) +(2 * second_least);
      min_operations++;
      heap.Add(new_ele);
      heap.Sort();
      heap.Reverse();          
    }
    if(heap[heap.Count - 1] >= K)
    {
      flag = 1;
    }
    if(flag == 1)
    {
      return min_operations;
    }
    return -1;       
  }
 
  // Driver Code
  static public void Main ()
  {
    int K = 7;
    int[] arr = { 1, 2, 3, 9, 10, 12 };
    int size = arr.Length;
    Console.WriteLine(minOperations(arr, K, size));
  }
}
 
// This code is contributed by avanitrachhadiya2155
Output: 
2

 

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