Array value by repeatedly replacing max 2 elements with their absolute difference

Given an array arr size N, the task is to print the final array value remaining in the array when the maximum and second maximum element of the array is replaced by their absolute difference in the array, repeatedly.

Note: If the maximum two elements are same, then both are removed from the array, without replacing any value.

Examples: 

Input: arr = [2, 7, 4, 1, 8, 1] 
Output:
Explanations: 
Merging 7 and 8: absolute difference = 7 – 8 = 1. So the array converted into [2, 4, 1, 1, 1]. 
Merging 2 and 4: absolute difference = 4 – 2 = 2. So the array converted into [2, 1, 1, 1]. 
Merging 2 and 1: absolute difference = 2 – 1 = 1. So the array converted into [1, 1, 1]. 
Merging 1 and 1: absolute difference = 4 – 2 = 0. So nothing will be Merged. 
So final array = [1].

Input: arr = [7, 10, 5, 4, 11, 25] 
Output: 2



Efficient Approach: Using Priority Queue

Below is the implementation of the above approach:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find the array value
// by repeatedly replacing max 2 elements
// with their absolute difference
 
#include <bits/stdc++.h>
using namespace std;
 
// function that return last
// value of array
int lastElement(vector<int>& arr)
{
    // Build a binary max_heap.
    priority_queue<int> pq;
    for (int i = 0; i < arr.size(); i++) {
        pq.push(arr[i]);
    }
 
    // For max 2 elements
    int m1, m2;
 
    // Iterate until queue is not empty
    while (!pq.empty()) {
 
        // if only 1 element is left
        if (pq.size() == 1)
 
// return the last
// remaining value
            return pq.top();
 
        m1 = pq.top();
        pq.pop();
        m2 = pq.top();
        pq.pop();
 
        // check that difference
        // is non zero
        if (m1 != m2)
            pq.push(m1 - m2);
    }
 
    // finally return 0
    return 0;
}
 
// Driver Code
int main()
{
    vector<int> arr = { 2, 7, 4, 1, 8, 1, 1 };
 
    cout << lastElement(arr) << endl;
    return 0;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find the array value
// by repeatedly replacing max 2 elements
// with their absolute difference
import java.util.*;
 
class GFG{
     
// Function that return last
// value of array
static int lastElement(int[] arr)
{
     
    // Build a binary max_heap
    PriorityQueue<Integer> pq = new PriorityQueue<>(
                                (a, b) -> b - a);
     
    for(int i = 0; i < arr.length; i++)
        pq.add(arr[i]);
     
    // For max 2 elements
    int m1, m2;
     
    // Iterate until queue is not empty
    while(!pq.isEmpty())
    {
         
        // If only 1 element is left
        if (pq.size() == 1)
        {
             
            // Return the last
            // remaining value
            return pq.poll();
        }
         
        m1 = pq.poll();
        m2 = pq.poll();
         
        // Check that difference
        // is non zero
        if (m1 != m2)
            pq.add(m1 - m2);
    }
     
    // Finally return 0
    return 0;
}
 
// Driver code
public static void main(String[] args)
{
    int[] arr = new int[]{2, 7, 4, 1, 8, 1, 1 };
     
    System.out.println(lastElement(arr));
}
}
 
// This code is contributed by dadi madhav
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to find the array value
# by repeatedly replacing max 2 elements
# with their absolute difference
from queue import PriorityQueue
 
# Function that return last
# value of array
def lastElement(arr):
     
    # Build a binary max_heap.
    pq = PriorityQueue()
    for i in range(len(arr)):
         
        # Multipying by -1 for
        # max heap
        pq.put(-1 * arr[i])
     
    # For max 2 elements
    m1 = 0
    m2 = 0
     
    # Iterate until queue is not empty
    while not pq.empty():
     
        # If only 1 element is left
        if pq.qsize() == 1:
             
            # Return the last
            # remaining value
            return -1 * pq.get()
        else:
            m1 = -1 * pq.get()
            m2 = -1 * pq.get()
             
        # Check that difference
        # is non zero
        if m1 != m2 :
            pq.put(-1 * abs(m1 - m2))
             
    return 0
     
# Driver Code
arr = [ 2, 7, 4, 1, 8, 1, 1 ]
 
print(lastElement(arr))
 
# This code is contributed by ishayadav181
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find the array value
// by repeatedly replacing max 2 elements
// with their absolute difference
using System;
using System.Collections.Generic;
  
class GFG{
  
// Function that return last
// value of array
static int lastElement(int[] arr)
{
     
    // Build a binary max_heap
    Queue<int> pq = new Queue<int>();
      
    for(int i = 0; i < arr.Length; i++)
        pq.Enqueue(arr[i]);
      
    // For max 2 elements
    int m1, m2;
      
    // Iterate until queue is not empty
    while (pq.Contains(0))
    {
         
        // If only 1 element is left
        if (pq.Count == 1)
        {
             
            // Return the last
            // remaining value
            return pq.Peek();
        }
          
        m1 = pq.Dequeue();
        m2 = pq.Peek();
          
        // Check that difference
        // is non zero
        if (m1 != m2)
            pq.Enqueue(m1 - m2);
    }
     
    // Finally return 0
    return 0;
}
  
// Driver Code
public static void Main(String[] args)
{
    int[] arr = { 2, 7, 4, 1, 8, 1, 1 };
      
    Console.WriteLine(lastElement(arr));
}
}
 
// This code is contributed by sanjoy_62
chevron_right

Output: 
0

Time Complexity: O(N) 
Auxiliary 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.




Article Tags :