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 until 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: 1
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:
- Insert all the array elements in the Priority Queue.
- As priority queue is based on the implementation of Max-Heap. Therefore removing element from it gives the maximum element.
- 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.
- If the heap has only one element then print that element.
- Else print “-1”.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int final_element(int arr[], int n)
{
priority_queue<int> heap;
for (int i = 0; i < n; i++)
heap.push(arr[i]);
while (heap.size() > 1) {
int X = heap.top();
heap.pop();
int Y = heap.top();
heap.pop();
if (X != Y) {
int diff = abs(X - Y);
heap.push(diff);
}
}
if (heap.size() == 1) {
cout << heap.top();
}
else {
cout << "-1";
}
}
int main()
{
int arr[] = { 3, 5, 2, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
final_element(arr, n);
return 0;
}
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Java
import java.io.*;
import java.util.Collections;
import java.util.*;
class GFG{
static public int final_element(Integer[] arr, int n)
{
if(arr == null)
{
return 0;
}
PriorityQueue<Integer> heap = new
PriorityQueue<Integer>(Collections.reverseOrder());
for(int i = 0; i < n; i++)
{
heap.offer(arr[i]);
}
while (heap.size() > 1)
{
int X = heap.poll();
int Y = heap.poll();
if (X != Y)
{
int diff = Math.abs(X - Y);
heap.offer(diff);
}
}
return heap.size() == 1 ? heap.poll() : -1;
}
public static void main (String[] args)
{
Integer arr[] = new Integer[] { 3, 5, 2, 7};
int n = arr.length;
System.out.println(final_element(arr, n));
}
}
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Python3
from queue import PriorityQueue
def final_element(arr, n):
heap = PriorityQueue()
for i in range(n):
heap.put(-1 * arr[i])
while (heap.qsize() > 1):
X = -1 * heap.get()
Y = -1 * heap.get()
if (X != Y):
diff = abs(X - Y)
heap.put(-1 * diff)
if (heap.qsize() == 1):
print(-1 * heap.get())
else:
print("-1")
if __name__ == '__main__':
arr = [ 3, 5, 2, 7 ]
n = len(arr)
final_element(arr, n)
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C#
using System;
using System.Collections.Generic;
class GFG{
static void final_element(int[] arr, int n)
{
List<int> heap = new List<int>();
for(int i = 0; i < n; i++)
heap.Add(arr[i]);
while (heap.Count > 1)
{
heap.Sort();
heap.Reverse();
int X = heap[0];
heap.RemoveAt(0);
int Y = heap[0];
heap.RemoveAt(0);
if (X != Y)
{
int diff = Math.Abs(X - Y);
heap.Add(diff);
}
}
if (heap.Count == 1)
{
heap.Sort();
heap.Reverse();
Console.Write(heap[0]);
}
else
{
Console.Write(-1);
}
}
static void Main()
{
int[] arr = { 3, 5, 2, 7 };
int n = arr.Length;
final_element(arr, n);
}
}
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Javascript
<script>
function final_element(arr, n)
{
var heap = [];
for (var i = 0; i < n; i++)
heap.push(arr[i]);
heap.sort((a,b)=>a-b);
while (heap.length > 1) {
var X = heap[heap.length-1];
heap.pop();
var Y = heap[heap.length-1];
heap.pop();
if (X != Y) {
var diff = Math.abs(X - Y);
heap.push(diff);
}
heap.sort((a,b)=>a-b);
}
if (heap.length == 1) {
document.write(heap[heap.length-1]);
}
else {
document.write("-1");
}
}
var arr = [3, 5, 2, 7 ];
var n = arr.length;
final_element(arr, n);
</script>
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Time Complexity: O(N*log(N))
Auxiliary Space Complexity: O(N)