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: 1
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 = 1-1 = 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
- Make a priority queue(binary max heap) which automatically arrange the element in sorted order.
- Then pick the first element (which is maximum) and 2nd element(2nd max), if both are equal then don’t push anything, if not equal push absolute difference of both in queue.
- Do the above steps till queue size is equal to 1, then return last element. If queue becomes empty before reaching size 1, then return 0.
Below is the implementation of the above approach:
C++
// 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 arrayint 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 Codeint main(){ vector<int> arr = { 2, 7, 4, 1, 8, 1, 1 }; cout << lastElement(arr) << endl; return 0;}// This code is contributed by shinjanpatra |
C
// 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 arrayint 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 Codeint main(){ vector<int> arr = { 2, 7, 4, 1, 8, 1, 1 }; cout << lastElement(arr) << endl; return 0;} |
Java
// Java program to find the array value// by repeatedly replacing max 2 elements// with their absolute differenceimport java.util.*;class GFG{ // Function that return last// value of arraystatic 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 codepublic 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 |
Python3
# 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)): # Multiplying 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 |
C#
// 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 arraystatic 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 Codepublic 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 |
Javascript
<script>// JavaScript program to find the array value// by repeatedly replacing max 2 elements// with their absolute difference// function that return last// value of arrayfunction lastElement(arr) { // Build a binary max_heap. let pq = []; for (let i = 0; i < arr.length; i++) { pq.push(arr[i]); } // For max 2 elements let m1, m2; // Iterate until queue is not empty while (pq.length) { // if only 1 element is left if (pq.length == 1) // return the last // remaining value return pq[pq.length - 1]; pq.sort((a, b) => a - b) m1 = pq[pq.length - 1]; pq.pop(); m2 = pq[pq.length - 1]; pq.pop(); // check that difference // is non zero if (m1 != m2) pq.push(m1 - m2); } // finally return 0 return 0;}// Driver Codelet arr = [2, 7, 4, 1, 8, 1, 1];document.write(lastElement(arr) + "<br>");</script> |
Output:
0
Time Complexity: O(N)
Auxiliary Complexity: O(N)
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