Remaining array element after repeated removal of last element and subtraction of each element from next adjacent element

Given an array arr[] consisting of N integers, the task is to find the remaining array element after subtracting each element from its next adjacent element and removing the last array element repeatedly.

Examples:

Input: arr[] = {3, 4, 2, 1}
Output: 4
Explanation:
Operation 1: The array arr[] modifies to {4 – 3, 2 – 4, 1 – 2} = {1, -2, -1}.
Operation 2: The array arr[] modifies to {-2 – 1, -1 + 2} = {-3, 1}.
Operation 3: The array arr[] modifies to {1 + 3} = {4}.
Therefore, the last remaining array element is 4.

Input: arr[] = {1, 8, 4}
Output: -11
Explanation:
Operation 1: The array arr[] modifies to {1 – 8, 4 – 8} = {7, -4}.
Operation 2: The array arr[] modifies to {-4 – 7 } = {-11}.
Therefore, the last remaining array element is -11.

Naive Approach: The simplest approach is to traverse the array until its size reduces to 1 and perform the given operations on the array. After completing the traversal, print the remaining elements. Below is the implementation of the above approach.

C++

#include <iostream>
#include <vector>

using namespace std;

// Function to find the remaining array element
int findRemainingElement(vector<int>& arr) {
    int n = arr.size();

    while (n > 1) {
        for (int i = 0; i < n - 1; i++) {
            arr[i] = arr[i+1] - arr[i];
        }
        n--;
    }

    return arr[0];
}

// Driver code
int main() {
    // Given input
    vector<int> arr = {3, 4, 2, 1};

    // Function call
    int remainingElement = findRemainingElement(arr);

    // Print the remaining element
    cout << "Remaining element: " << remainingElement << endl;

    return 0;
}

C#

using System;
using System.Collections.Generic;

class MainClass {
    // Function to find the remaining array element
    static int FindRemainingElement(List<int> arr) {
        int n = arr.Count;

        while (n > 1) {
            for (int i = 0; i < n - 1; i++) {
                arr[i] = arr[i+1] - arr[i];
            }
            n--;
        }

        return arr[0];
    }

    // Driver code
    static void Main() {
        // Given input
        List<int> arr = new List<int> {3, 4, 2, 1};

        // Function call
        int remainingElement = FindRemainingElement(arr);

        // Print the remaining element
        Console.WriteLine("Remaining element: " + remainingElement);
    }
}
// This code is contributed by sarojmcy2e

Javascript

function findRemainingElement(arr) {
    let n = arr.length;

    while (n > 1) {
        for (let i = 0; i < n - 1; i++) {
            arr[i] = arr[i+1] - arr[i];
        }
        n--;
    }

    return arr[0];
}

// Given input
let arr = [3, 4, 2, 1];

// Function call
let remainingElement = findRemainingElement(arr);

// Print the remaining element
console.log("Remaining element: " + remainingElement);

Output

Remaining element: 4

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

Efficient Approach: The above approach can be optimized based on the following observations:

• Suppose the given array is arr[] = {a, b, c, d}. Then, performing the operations:

• Now, suppose the array arr[] = {a, b, c, d, e}. Then, performing the operations:

• From the above two observations, it can be concluded that the answer is the sum of multiplication of coefficients of terms in the expansion of (x – y)^{(N – 1)} and each array element arr[i].
• Therefore, the idea is to find the sum of the array arr[] after updating each array element as (arr[i]* ^{(N – 1)}C_(i-1)* (-1)ⁱ).

Follow the steps below to solve the problem:

• Traverse the array arr[] and update arr[i] as arr[i] = arr[i]* ^{(N – 1)}C_{(i – 1)}* (-1)ⁱ after calculating the ^NC_r using Pascal’s triangle.
• Print the sum of array arr[].

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 performing
// the given operations repeatedly
int lastElement(const int arr[], int n)
{
    // Stores the resultant sum
    int sum = 0;

    int multiplier = n % 2 == 0 ? -1 : 1;

    // Traverse the array
    for (int i = 0; i < n; i++) {

        // Increment sum by arr[i]
        // * coefficient of i-th term
        // in (x - y) ^ (N - 1)
        sum += arr[i] * multiplier;

        // Update multiplier
        multiplier
            = multiplier * (n - 1 - i)
              / (i + 1) * (-1);
    }

    // Return the resultant sum
    return sum;
}

// Driver Code
int main()
{
    int arr[] = { 3, 4, 2, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << lastElement(arr, N);

    return 0;
}

Java

/*package whatever //do not write package name here */

import java.io.*;

class GFG {

    // Function to find the last remaining
    // array element after performing
    // the given operations repeatedly
    public static int lastElement(int arr[], int n)
    {
        // Stores the resultant sum
        int sum = 0;

        int multiplier = n % 2 == 0 ? -1 : 1;

        // Traverse the array
        for (int i = 0; i < n; i++) {

            // Increment sum by arr[i]
            // * coefficient of i-th term
            // in (x - y) ^ (N - 1)
            sum += arr[i] * multiplier;

            // Update multiplier
            multiplier
                = multiplier * (n - 1 - i) / (i + 1) * (-1);
        }

        // Return the resultant sum
        return sum;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 3, 4, 2, 1 };
        int N = 4;
        System.out.println(lastElement(arr, N));
    }
}

// This code is contributed by aditya7409.

Python3

# Python 3 program for the above approach

# Function to find the last remaining
# array element after performing
# the given operations repeatedly
def lastElement(arr, n):

    # Stores the resultant sum
    sum = 0
    if n % 2 == 0:
        multiplier = -1
    else:
        multiplier = 1

    # Traverse the array
    for i in range(n):

        # Increment sum by arr[i]
        # * coefficient of i-th term
        # in (x - y) ^ (N - 1)
        sum += arr[i] * multiplier

        # Update multiplier
        multiplier = multiplier * (n - 1 - i) / (i + 1) * (-1)

    # Return the resultant sum
    return sum

# Driver Code
if __name__ == '__main__':
    arr = [3, 4, 2, 1]
    N = len(arr)
    print(int(lastElement(arr, N)))

    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# program for the above approach
using System;
class GFG
{

  // Function to find the last remaining
  // array element after performing
  // the given operations repeatedly
  public static int lastElement(int[] arr, int n)
  {

    // Stores the resultant sum
    int sum = 0;

    int multiplier = n % 2 == 0 ? -1 : 1;

    // Traverse the array
    for (int i = 0; i < n; i++) {

      // Increment sum by arr[i]
      // * coefficient of i-th term
      // in (x - y) ^ (N - 1)
      sum += arr[i] * multiplier;

      // Update multiplier
      multiplier
        = multiplier * (n - 1 - i) / (i + 1) * (-1);
    }

    // Return the resultant sum
    return sum;
  }

  // Driver code
  static void Main()
  {
    int[] arr = { 3, 4, 2, 1 };
    int N = 4;
    Console.WriteLine(lastElement(arr, N));
  }
}

// This code is contributed by susmitakundugoaldanga.

Javascript

<script>
// JavaScript program for the above approach

// Function to find the last remaining
// array element after performing
// the given operations repeatedly
function lastElement(arr, n)
{

    // Stores the resultant sum
    let sum = 0;
    let multiplier = n % 2 == 0 ? -1 : 1;

    // Traverse the array
    for (let i = 0; i < n; i++)
    {

        // Increment sum by arr[i]
        // * coefficient of i-th term
        // in (x - y) ^ (N - 1)
        sum += arr[i] * multiplier;

        // Update multiplier
        multiplier
            = multiplier * (n - 1 - i)
            / (i + 1) * (-1);
    }

    // Return the resultant sum
    return sum;
}

// Driver Code
    let arr = [ 3, 4, 2, 1 ];
    let N = arr.length;
    document.write(lastElement(arr, N));

// This code is contributed by Surbhi Tyagi.
</script>

Output

4

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