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Reduce the array to atmost one element by the given operations
  • Difficulty Level : Easy
  • Last Updated : 15 Jun, 2021

Given an array of integers arr[], the task is to find the remaining element in the array after performing the following operations:
 

  • In each turn, choose the two maximum integers X and Y from the array.
  • If X == Y, remove both elements from the array.
  • If X != Y, insert an element into the array equal to abs(X – Y).

Note: If no element remains, print 0. 
Examples: 
 

Input: arr[] = [3, 4, 6, 2, 7, 1] 
Output:
Explanation: 
Elements 7 and 6 are reduced to 1 so the array converts to [3, 4, 2, 1, 1] 
Elements 3 and 4 are reduced to 2 so the array converts to [2, 1, 1, 1] 
Elements 2 and 1 are reduced to 1 so the array converts to [1, 1, 1] 
Element 1 is completely destroyed by another 1 so array is [1] at the end.
Input: arr[] = [1, 2, 3, 4, 5] 
Output:
Explanation: 
Elements 4 and 5 are reduced to 1 so the array converts to [1, 2, 3, 1] 
Elements 2 and 3 are reduced to 1 so the array converts to [1, 1, 1] 
Element 1 is completely destroyed by another 1 so array is [1] at the end. 
 

Approach: 
To solve the problem mentioned above, we need to use Heap Data Structure. Heaps are used because we only require the current maximum element for every instant and are not concerned about the rest of the elements. 
 

  • Create a Max Heap from the elements of the given array.
  • Keep extracting the top element twice for every iteration. If their absolute difference is non-zero, push their difference back to the queue.
  • Continue until only one or no elements are remaining.
  • If no elements remain, print 0. Otherwise, print the remaining element.

Below is the implementation of above approach: 
 



C++




// C++ Program to reduce the
// array to almost one element
// by the given operations
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the remaining
// element of array
int reduceArray(vector<int>& arr)
{
    priority_queue<int> p;
 
    // Build a priority queue
    // using the array
    for (int i = 0; i < arr.size(); ++i)
        p.push(arr[i]);
 
    // Continue until the priority
    // queue has one or no elements
    // remaining
    while (p.size() > 1) {
 
        // Top-most integer from heap
        int y = p.top();
        p.pop();
 
        // Second integer from heap
        int x = p.top();
        p.pop();
 
        // Resultant value
        int val = y - x;
        if (val != 0)
            p.push(val);
    }
 
    // Return 0 if no elements are left
    if (p.size() == 0)
        return 0;
 
    // Return top value of the heap
    return p.top();
}
 
// Driver code
int main()
{
 
    vector<int> arr
        = { 3, 4, 6, 2, 7, 1 };
    cout << reduceArray(arr)
         << "\n";
    return 0;
}

Java




// Java program to reduce the
// array to almost one element
// by the given operations
import java.util.*;
 
class GFG{
 
// Function to find the remaining
// element of array
static int reduceArray(int[] arr)
{
    PriorityQueue<Integer> p = new PriorityQueue<>(
        11, Collections.reverseOrder());
 
    // Build a priority queue
    // using the array
    for(int i = 0; i < arr.length; ++i)
        p.add(arr[i]);
         
    // Continue until the priority
    // queue has one or no elements
    // remaining
    while (p.size() > 1)
    {
 
        // Top-most integer from heap
        int y = p.poll();
 
        // Second integer from heap
        int x = p.poll();
         
        // Resultant value
        int val = y - x;
        if (val != 0)
            p.add(val);
    }
     
    // Return 0 if no elements are left
    if (p.size() == 0)
        return 0;
 
    // Return top value of the heap
    return p.poll();
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 3, 4, 6, 2, 7, 1 };
     
    System.out.println(reduceArray(arr));
}
}
 
// This code is contributed by jrishabh99

Python3




# Python3 program to reduce the
# array to almost one element
# by the given operations
 
# Function to find the remaining
# element of array
def reduceArray(arr):
     
    p = []
 
    # Build a priority queue
    # using the array
    for i in range(len(arr)):
        p.append(arr[i])
    p.sort(reverse = True)
 
    # Continue until the priority
    # queue has one or no elements
    # remaining
    while (len(p) > 1):
         
        # Top-most integer from heap
        y = p[0]
        p = p[1:]
 
        # Second integer from heap
        x = p[0]
        p = p[1:]
 
        # Resultant value
        val = y - x
         
        if (val != 0):
            p.append(val)
        p.sort(reverse = True)
 
    # Return 0 if no elements are left
    if (len(p) == 0):
        return 0
 
    # Return top value of the heap
    return p[0]
 
# Driver code
if __name__ == '__main__':
     
    arr = [ 3, 4, 6, 2, 7, 1 ]
     
    print(reduceArray(arr))
 
# This code is contributed by Surendra_Gangwar
Output: 
1

 

Time Complexity:O(NlogN)
 

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