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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 <bits/stdc++.h>
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[0]);
    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




<script>
// JavaScript 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
function findLastElement(arr, N)
{
 
    // Sort the given array in
    // ascending order
    arr.sort();
    let 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)
        {
 
            document.write("-1" + "<br>");
            return;
        }
    }
 
    // If operations can be performed
    document.write(arr[N - 1] + "<br>");
}
 
// Driver Code
    let arr = [ 2, 4, 6, 8, 0, 8 ];
    let N = arr.length;
    findLastElement(arr, N);
 
// This code is contributed by Surbhi Tyagi.
</script>

 
 

Output: 
8

 

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




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