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

Last Updated : 25 Feb, 2021

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.
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:

$a, \ b, \ c, \ d\\ b-a, \ c-b, \ d-c\\ (c-b)-(b-a), \ (d-c)-(c-b) = c-2b+a, \ d-2c+b\\ -a+3b-3c+d$

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

$a, \ b, \ c, \ d, \ e\\ \vdots\\ a - 4b + 6c - 4d + e$

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)ⁱafter 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;
}

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.

Output:

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

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