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:1Explanations:

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**

- 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 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;` `}` |

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## Java

`// 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` |

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## 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)):` ` ` ` ` `# 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` |

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## 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 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` |

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**Output:**

0

**Time Complexity:** O(N)**Auxiliary Complexity:** O(N)

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