Find the Array element left after alternate removal of minimum and maximum

Last Updated : 24 Mar, 2023

Given an array Arr of length N, it is reduced by 1 element at each step. Maximum and Minimum elements will be removed in alternating order from the remaining array until a single element is remaining in the array. The task is to find the remaining element in the given array.

Examples:

Input: arr[] = {1, 5, 4, 2}
Output: 2
Explanation:
Remove Max element i.e., 5 arr[] = {1, 4, 2}
Remove Min element i.e., 1 arr[] = {4, 2}
Remove Max element i.e., 4 arr[] = {2}

Input: arr[] = {5, 10}
Output: 5

Approach:

Follow the below idea to solve the problem:

The idea is to sort the array and return the middle element as all of the right and left elements will be removed in the process.

Follow the below steps to solve this problem:

If N =1, return arr[0]
Sort the array arr[]
Return the middle element of the array i.e., arr[(N-1)/2]

Below is the implementation of the above approach:

C++

// C++ code to implement the approach  #include <bits/stdc++.h> using namespace std;  // Function to find the element left int lastRemaining(int arr[], int N) {     // If single element in array     if (N == 1)         return arr[0];      // Sort the array     sort(arr, arr + N);      // return middle element     return arr[(N - 1) / 2]; }  // Driver Code int main() {     int arr[] = { 1, 5, 4, 2 };     int N = sizeof(arr) / sizeof(arr[0]);      // Function call     cout << lastRemaining(arr, N) << endl;     return 0; }

Java

// Java code to implement the approach import java.util.*;  class GFG {      // Function to find the element left     static int lastRemaining(int arr[], int N)     {         // If single element in array         if (N == 1)             return arr[0];          // Sort the array         Arrays.sort(arr);          // return middle element         return arr[(N - 1) / 2];     }     // Driver Code     public static void main(String[] args)     {         int arr[] = { 1, 5, 4, 2 };         int N = arr.length;          // Function call         System.out.print(lastRemaining(arr, N));     } }  // This code is contributed by sanjoy_62.

Python3

# Python code to implement the approach  # Function to find the element left def lastRemaining(arr, N):          # If single element in array     if (N == 1):         return arr[0]      # Sort the array     arr.sort()      # return middle element     return arr[(N - 1) // 2]  # Driver Code arr = [ 1, 5, 4, 2 ] N = len(arr)  # Function call print(lastRemaining(arr, N))  # This code is contributed by athakur42u.

// C# code to implement the approach using System; class GFG {      // Function to find the element left     static int lastRemaining(int[] arr, int N)     {          // If single element in array         if (N == 1)             return arr[0];          // Sort the array         Array.Sort(arr);          // return middle element         return arr[(N - 1) / 2];     }      // Driver code     public static void Main(string[] args)     {         int[] arr = { 1, 5, 4, 2 };         int N = arr.Length;          // Function call         Console.Write(lastRemaining(arr, N));     } }  // This code is contributed by code_hunt.

JavaScript

 <script>         // JS code to implement the approach          // Function to find the element left         function lastRemaining(arr, N)         {                      // If single element in array             if (N == 1)                 return arr[0];              let x = Math.floor((N - 1) / 2);             // Sort the array             arr.sort(function (a, b) { return a - b })              // return middle element             return arr[x];         }          // Driver Code         let arr = [1, 5, 4, 2];         let N = arr.length;          // Function call         document.write(lastRemaining(arr, N));  // This code is contributed by lokeshpotta20.     </script>

Output

Time Complexity: O(N * log(N)), for sorting the given array of size N.
Auxiliary Space: O(1), as constant extra space is required.

Program to find the minimum (or maximum) element of an array

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Find the Array element left after alternate removal of minimum and maximum

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