Reduce the array by replacing 1st and middle element with sum and difference alternatively

Given an array arr[] of size N, the task is to find the last remaining element of the array after consecutively removing the 1^st and the middle element of the array and alternatively appending their sum and difference at the end of the array.

Examples:

Input: A = {2, 4, 1, 5, 7}
Output: 5
Explanation: During the 1st iteration, remove arr[0] and arr[2] from the array 
and append their sum (2 + 1 = 3) to the end of the array. 
Hence, arr[] = {4, 5, 7, 3}. 
Again repeat the same process and remove arr[0] and arr[2] from the array 
and now append their difference (7 – 4) = 3. 
Hence, arr[] = {5, 3, 3}.
 After the next iteration, array will become arr[] = {3, 8}. 
And finally after the last iteration, arr[] = {5} which is the last required value.

Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Output: 15

Approach: The given problem is an implementation-based problem and can be solved by repetitively following the given steps:

• Maintain a variable op, which stores the count of operation.
• Remove the first and middle element of the array till its size is more than 1 and append the array with their addition or subtraction according to the op (i., e if op is even, perform addition, otherwise perform subtraction).

Below is the implementation of the above approach:

C++

// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the last element
// in array after the given operations
int lastElement(vector<int>& arr)
{

    // Maintain operation count
    int op = 0, x = 0;

    // Loop until array has more
    // than 1 element in it
    while (arr.size() != 1) {

        // Stores middle index
        int mid = arr.size() / 2;

        // For even iterations perform
        // addition otherwise subtraction
        if (op % 2 == 0) {
            x = arr[mid] + arr[0];
            arr.erase(arr.begin() + mid);
            arr.erase(arr.begin());
        }
        else {
            x = arr[mid] - arr[0];
            arr.erase(arr.begin() + mid);
            arr.erase(arr.begin());
        }

        // Append in the end
        arr.push_back(x);

        // Increment operation count
        op += 1;
    }

    // Return Answer
    return arr[0];
}

// Driver Code
int main()
{
    vector<int> arr = { 2, 4, 1, 5, 7 };
    cout << lastElement(arr);

    return 0;
}

    // This code is contributed by rakeshsahni

Java

// Java program of the above approach
import java.util.*;

class GFG{

// Function to find the last element
// in array after the given operations
static int lastElement(Vector<Integer> arr)
{

    // Maintain operation count
    int op = 0, x = 0;

    // Loop until array has more
    // than 1 element in it
    while (arr.size() != 1) {

        // Stores middle index
        int mid = arr.size() / 2;

        // For even iterations perform
        // addition otherwise subtraction
        if (op % 2 == 0) {
            x = arr.get(mid) + arr.get(0);
            arr.remove(mid);
            arr.remove(0);
        }
        else {
            x = arr.get(mid) - arr.get(0);
            arr.remove(mid);
            arr.remove(0);
        }

        // Append in the end
        arr.add(x);

        // Increment operation count
        op += 1;
    }

    // Return Answer
    return arr.get(0);
}

// Driver Code
public static void main(String[] args)
{
    Integer []arr = { 2, 4, 1, 5, 7 };
    Vector<Integer> v = new Vector<Integer>();
    Collections.addAll(v, arr);
    System.out.print(lastElement(v));

}
}

// This code is contributed by shikhasingrajput

Python3

# Python program of the above approach

# Function to find the last element
# in array after the given operations
def lastElement(arr):

    # Maintain operation count
    op = 0

    # Loop until array has more
    # than 1 element in it
    while len(arr) != 1:

        # Stores middle index
        mid = len(arr)//2

        # For even iterations perform
        # addition otherwise subtraction
        if op % 2 == 0:
            x = arr.pop(mid) + arr.pop(0)
        else:
            x = arr.pop(mid) - arr.pop(0)

        # Append in the end
        arr.append(x)

        # Increment operation count
        op += 1

    # Return Answer
    return arr[0]


# Driver Code
if __name__ == "__main__":
    arr = [2, 4, 1, 5, 7]
    print(lastElement(arr))

C#

// C# program of the above approach
using System;
using System.Collections.Generic;

public class GFG{

  // Function to find the last element
  // in array after the given operations
  static int lastElement(List<int> arr)
  {

    // Maintain operation count
    int op = 0, x = 0;

    // Loop until array has more
    // than 1 element in it
    while (arr.Count != 1) {

      // Stores middle index
      int mid = arr.Count / 2;

      // For even iterations perform
      // addition otherwise subtraction
      if (op % 2 == 0) {
        x = arr[mid] + arr[0];
        arr.RemoveAt(mid);
        arr.RemoveAt(0);
      }
      else {
        x = arr[mid] - arr[0];
        arr.RemoveAt(mid);
        arr.RemoveAt(0);
      }

      // Append in the end
      arr.Add(x);

      // Increment operation count
      op += 1;
    }

    // Return Answer
    return arr[0];
  }

  // Driver Code
  public static void Main(String[] args)
  {
    int []arr = { 2, 4, 1, 5, 7 };
    List<int> v = new List<int>(arr);
    Console.Write(lastElement(v));

  }
}

// This code is contributed by shikhasingrajput

Javascript

<script>
       // JavaScript code for the above approach

       // Function to find the last element
       // in array after the given operations
       function lastElement(arr) {

           // Maintain operation count
           let op = 0, x = 0;

           // Loop until array has more
           // than 1 element in it
           while (arr.length != 1) {

               // Stores middle index
               let mid = Math.floor(arr.length / 2);

               // For even iterations perform
               // addition otherwise subtraction
               if (op % 2 == 0) {
                   x = arr[mid] + arr[0];
                   let p1 = arr.slice(1, mid)
                   let p2 = arr.slice(mid + 1)
                   arr = [...p1.concat(p2)];
               }
               else {
                   x = arr[mid] - arr[0];
                   let p1 = arr.slice(1, mid)
                   let p2 = arr.slice(mid + 1)
                   arr = [...p1.concat(p2)];


               }

               // Append in the end
               arr.push(x);

               // Increment operation count
               op = op + 1;
           }

           // Return Answer
           return arr[0];
       }

       // Driver Code

       let arr = [2, 4, 1, 5, 7];
       document.write(lastElement(arr));

    // This code is contributed by Potta Lokesh
   </script>

Output

5

Time Complexity: O(N²)
Auxiliary Space: O(1)