Skip to content
Related Articles

Related Articles

Last element remaining by deleting two largest elements and replacing by their absolute difference if they are unequal
  • Last Updated : 19 Jan, 2021

Given an array arr[] of N elements, the task is to perform the following operation: 

  • Pick the two largest element from the array and remove these element. If the elements are unequal then insert the absolute difference of the elements into the array.
  • Perform the above operations untill array has 1 or no element in it. If the array has only one element left then print that element, else print “-1”.

Examples: 

Input: arr[] = { 3, 5, 2, 7 } 
Output:
Explanation: 
The two largest elements are 7 and 5. Discard them. Since both are not equal, insert 7 – 5 = 2 into the array. Hence, arr[] = { 3, 2, 2 } 
The two largest elements are 3 and 2. Discard them. Since both are not equal, insert 3 – 2 = 1 into the array. Hence, arr[] = { 1, 2 } 
The two largest elements are 2 and 1. Discard them. Since both are not equal, insert 2 – 1 = 1 into the array. Hence, arr[] = { 1 } 
The only element left is 1. Print the value of the only element left.

Input: arr[] = { 3, 3 } 
Output: -1 
Explanation: 
The two largest elements are 3 and 3. Discard them. Now the array has no elements. So, print -1. 

Approach: To solve the above problem we will use Priority Queue Data Structure. Below are the steps: 



  1. Insert all the array elements in the Priority Queue.
  2. As priority queue is based on the implementation of Max-Heap. Therefore removing element from it gives the maximum element.
  3. Till the size of priority queue is not less than 2, remove two elements(say X & Y) from it and do the following: 
    • If X and Y are not same then insert the absolute value of X and Y into the priority queue.
    • Else return to step 3.
  4. If the heap has only one element then print that element.
  5. Else print “-1”.

Below is the implementation of the above approach:  

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the remaining element
int final_element(int arr[], int n)
{
 
    // Priority queue can be used
    // to construct max-heap
    priority_queue<int> heap;
 
    // Insert all element of arr[]
    // into priority queue
    for (int i = 0; i < n; i++)
        heap.push(arr[i]);
 
    // Perform operation until heap
    // size becomes 0 or 1
    while (heap.size() > 1) {
 
        // Remove largest element
        int X = heap.top();
        heap.pop();
 
        // Remove 2nd largest element
        int Y = heap.top();
        heap.pop();
 
        // If extracted element
        // are not equal
        if (X != Y) {
 
            // Find X - Y and push
            // it to heap
            int diff = abs(X - Y);
            heap.push(diff);
        }
    }
 
    // If heap size is 1, then
    // print the remaining element
    if (heap.size() == 1) {
 
        cout << heap.top();
    }
 
    // Else print "-1"
    else {
        cout << "-1";
    }
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 3, 5, 2, 7 };
 
    // Size of array arr[]
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    final_element(arr, n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for the above approach
import java.io.*;
import java.util.Collections;
import java.util.*;
 
class GFG{
     
// Function to print the remaining element    
static public int final_element(Integer[] arr, int n)
{
    if(arr == null)
    {
        return 0;
    }
     
    // Priority queue can be used
    // to construct max-heap
    PriorityQueue<Integer> heap = new
    PriorityQueue<Integer>(Collections.reverseOrder());
     
    // Insert all element of arr[]
    // into priority queue
    for(int i = 0; i < n; i++)
    {
       heap.offer(arr[i]);
    }
     
    // Perform operation until heap
    // size becomes 0 or 1
    while (heap.size() > 1)
    {
         
        // Remove largest element
        int X = heap.poll();
         
        // Remove 2nd largest element
        int Y = heap.poll();
         
        // If extracted element
        // are not equal
        if (X != Y)
        {
            // Find X - Y and push
            // it to heap
            int diff = Math.abs(X - Y);
            heap.offer(diff);
        }
    }
     
    // If heap size is 1, then
    // print the remaining element
    // Else print "-1"
    return heap.size() == 1 ? heap.poll() : -1;
}
 
// Driver code
public static void main (String[] args)
{
    // Given array arr[]
    Integer arr[] = new Integer[] { 3, 5, 2, 7};
     
    // Size of array arr[]
    int n = arr.length;
     
    // Function Call
    System.out.println(final_element(arr, n)); 
}
}
 
// This code is contributed by deepika_sharma

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program for the above approach
from queue import PriorityQueue
 
# Function to print the remaining element
def final_element(arr, n):
     
    # Priority queue can be used
    # to construct max-heap
    heap = PriorityQueue()
     
    # Insert all element of
    # arr[] into priority queue.
    # Default priority queue in Python
    # is min-heap so use -1*arr[i]
    for i in range(n):
        heap.put(-1 * arr[i])
     
    # Perform operation until heap
    # size becomes 0 or 1
    while (heap.qsize() > 1):
 
        # Remove largest element
        X = -1 * heap.get()
 
        # Remove 2nd largest element
        Y = -1 * heap.get()
 
        # If extracted elements
        # are not equal
        if (X != Y):
 
            # Find X - Y and push
            # it to heap
            diff = abs(X - Y)
            heap.put(-1 * diff)
 
    # If heap size is 1, then
    # print the remaining element
    if (heap.qsize() == 1):
        print(-1 * heap.get())
 
    # Else print "-1"
    else:
        print("-1")
 
# Driver Code
if __name__ == '__main__':
     
    # Given array arr[]
    arr = [ 3, 5, 2, 7 ]
 
    # Size of array arr[]
    n = len(arr)
 
    # Function call
    final_element(arr, n)
 
# This code is contributed by himanshu77

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to print the remaining element
static void final_element(int[] arr, int n)
{
     
    // Priority queue can be used
    // to construct max-heap
    List<int> heap = new List<int>();
   
    // Insert all element of arr[]
    // into priority queue
    for(int i = 0; i < n; i++)
        heap.Add(arr[i]);
   
    // Perform operation until heap
    // size becomes 0 or 1
    while (heap.Count > 1)
    {
         
        // Remove largest element
        heap.Sort();
        heap.Reverse();
        int X = heap[0];
        heap.RemoveAt(0);
   
        // Remove 2nd largest element
        int Y = heap[0];
        heap.RemoveAt(0);
   
        // If extracted element
        // are not equal
        if (X != Y)
        {
             
            // Find X - Y and push
            // it to heap
            int diff = Math.Abs(X - Y);
            heap.Add(diff);
        }
    }
   
    // If heap size is 1, then
    // print the remaining element
    if (heap.Count == 1)
    {
        heap.Sort();
        heap.Reverse();
        Console.Write(heap[0]);
    }
   
    // Else print "-1"
    else
    {
        Console.Write(-1);
    }
}
 
// Driver code
static void Main()
{
     
    // Given array arr[]
    int[] arr = { 3, 5, 2, 7 };
     
    // Size of array arr[]
    int n = arr.Length;
     
    // Function Call
    final_element(arr, n);
}
}
 
// This code is contributed by divyeshrabadiya07

chevron_right


Output: 

1

 

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :