# Remaining array element after repeated removal of the smallest element from pairs with absolute difference of 2 or 0

• Last Updated : 01 Apr, 2021

Given an array arr[] consisting of N positive integers, the task is to check if it is possible to reduce the size of the array to 1 by repeatedly removing the smallest element from a pair having absolute difference of 2 or 0 between them. If it is not possible to reduce, then print “-1”. Otherwise, print the last remaining element in the array.

Examples:

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Input: arr[] = {2, 4, 6, 8, 0, 8}
Output: 8
Explanation:
arr[] = {2, 4, 6, 8, 0, 8}, Remove 0 from the pair (2, 0).
arr[] = {2, 4, 6, 8, 8}. Remove 2 from the pair (2, 4).
arr[] = {4, 6, 8, 8}, Remove 4 from the pair (4, 6).
arr[] = {6, 8, 8}. Remove 6 from the pair (6, 8).
arr[] = {8, 8}. Remove 8.
arr[] = {8}

Input: arr[] = {1, 7, 3, 3}
Output: -1
Explanation:
arr[] = {1, 7, 3, 3}. Remove 1 from the pair (1, 3).
arr[] = {7, 3, 3}. Remove 3 from the pair (3, 3).
arr[] = {7, 3}. No more removals possible.

Approach: Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find the last remaining``// array element after repeatedly removing``// the smallest from pairs having absolute``// difference 2 or 0``void` `findLastElement(``int` `arr[], ``int` `N)``{``    ``// Sort the given array in``    ``// ascending order``    ``sort(arr, arr + N);``    ``int` `i = 0;` `    ``// Traverse the array``    ``for` `(i = 1; i < N; i++) {` `        ``// If difference between``        ``// adjacent elements is``        ``// not equal to 0 or 2``        ``if` `(arr[i] - arr[i - 1] != 0``            ``&& arr[i] - arr[i - 1] != 2) {` `            ``cout << ``"-1"` `<< endl;``            ``return``;``        ``}``    ``}` `    ``// If operations can be performed``    ``cout << arr[N - 1] << endl;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 4, 6, 8, 0, 8 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``findLastElement(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG``{` `  ``// Function to find the last remaining``  ``// array element after repeatedly removing``  ``// the smallest from pairs having absolute``  ``// difference 2 or 0``  ``static` `void` `findLastElement(``int` `arr[], ``int` `N)``  ``{``    ``// Sort the given array in``    ``// ascending order``    ``Arrays.sort(arr);``    ``int` `i = ``0``;` `    ``// Traverse the array``    ``for` `(i = ``1``; i < N; i++) {` `      ``// If difference between``      ``// adjacent elements is``      ``// not equal to 0 or 2``      ``if` `(arr[i] - arr[i - ``1``] != ``0``          ``&& arr[i] - arr[i - ``1``] != ``2``)``      ``{` `        ``System.out.println(``"-1"``);``        ``return``;``      ``}``    ``}` `    ``// If operations can be performed``    ``System.out.println( arr[N - ``1``]);``  ``}`  `  ``// Driver Code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``int` `arr[] = { ``2``, ``4``, ``6``, ``8``, ``0``, ``8` `};``    ``int` `N = arr.length;``    ``findLastElement(arr, N);``  ``}``}` `// This code is contributed by code_hunt.`

## Python3

 `# Python program for the above approach` `# Function to find the last remaining``# array element after repeatedly removing``# the smallest from pairs having absolute``# difference 2 or 0``def` `findLastElement(arr, N):``  ` `    ``# Sort the given array in``    ``# ascending order``    ``arr.sort();``    ``i ``=` `0``;` `    ``# Traverse the array``    ``for` `i ``in` `range``(``1``, N):` `        ``# If difference between``        ``# adjacent elements is``        ``# not equal to 0 or 2``        ``if` `(arr[i] ``-` `arr[i ``-` `1``] !``=` `0``\``                ``and` `arr[i] ``-` `arr[i ``-` `1``] !``=` `2``):``            ``print``(``"-1"``);``            ``return``;` `    ``# If operations can be performed``    ``print``(arr[N ``-` `1``]);` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``2``, ``4``, ``6``, ``8``, ``0``, ``8``];``    ``N ``=` `len``(arr);``    ``findLastElement(arr, N);` `# This code is contributed by 29AjayKumar.`

## C#

 `// C# program for the above approach``using` `System;``public` `class` `GFG``{` `  ``// Function to find the last remaining``  ``// array element after repeatedly removing``  ``// the smallest from pairs having absolute``  ``// difference 2 or 0``  ``static` `void` `findLastElement(``int` `[]arr, ``int` `N)``  ``{``    ` `    ``// Sort the given array in``    ``// ascending order``    ``Array.Sort(arr);``    ``int` `i = 0;` `    ``// Traverse the array``    ``for` `(i = 1; i < N; i++)``    ``{` `      ``// If difference between``      ``// adjacent elements is``      ``// not equal to 0 or 2``      ``if` `(arr[i] - arr[i - 1] != 0``          ``&& arr[i] - arr[i - 1] != 2)``      ``{``        ``Console.WriteLine(``"-1"``);``        ``return``;``      ``}``    ``}` `    ``// If operations can be performed``    ``Console.WriteLine(arr[N - 1]);``  ``}`  `  ``// Driver Code``  ``public` `static` `void` `Main(String[] args)``  ``{``    ``int` `[]arr = { 2, 4, 6, 8, 0, 8 };``    ``int` `N = arr.Length;``    ``findLastElement(arr, N);``  ``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

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Output:
`8`

Time Complexity: O(N*log N)
Auxiliary Space: O(1)

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