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

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.
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:

Below is the implementation of the above approach:

C++

#include "bits/stdc++.h" using namespace std;    int lastElement(const int arr[], int n) {          int sum = 0;        int multiplier = n % 2 == 0 ? -1 : 1;             for (int i = 0; i < n; i++) {                                       sum += arr[i] * multiplier;                     multiplier             = multiplier * (n - 1 - i)               / (i + 1) * (-1);     }             return sum; }    int main() {     int arr[] = { 3, 4, 2, 1 };     int N = sizeof(arr) / sizeof(arr[0]);     cout << lastElement(arr, N);        return 0; }

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

Let's block ads! (Why?)