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++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to print the remaining elementint 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 Codeint 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;} |
Java
// 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 codepublic 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 |
Python3
# Python3 program for the above approach from queue import PriorityQueue# Function to print the remaining elementdef 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 Codeif __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 |
C#
// 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 codestatic 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 |
Javascript
<script>// Javascript program for the above approach// Function to print the remaining elementfunction final_element(arr, n){ // Priority queue can be used // to construct max-heap var heap = []; // Insert all element of arr[] // into priority queue for (var i = 0; i < n; i++) heap.push(arr[i]); heap.sort((a,b)=>a-b); // Perform operation until heap // size becomes 0 or 1 while (heap.length > 1) { // Remove largest element var X = heap[heap.length-1]; heap.pop(); // Remove 2nd largest element var Y = heap[heap.length-1]; heap.pop(); // If extracted element // are not equal if (X != Y) { // Find X - Y and push // it to heap var diff = Math.abs(X - Y); heap.push(diff); } heap.sort((a,b)=>a-b); } // If heap size is 1, then // print the remaining element if (heap.length == 1) { document.write(heap[heap.length-1]); } // Else print "-1" else { document.write("-1"); }}// Driver Code// Given array arr[]var arr = [3, 5, 2, 7 ];// Size of array arr[]var n = arr.length;// Function Callfinal_element(arr, n);// This code is contributed by rutvik_56.</script> |
Output:
1
Time Complexity: O(N*log(N))
Auxiliary Space Complexity: O(N)
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