Remaining array element after repeated removal of last element and subtraction of each element from next adjacent element
Last Updated :
25 Apr, 2023
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;
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];
}
int main() {
vector< int > arr = {3, 4, 2, 1};
int remainingElement = findRemainingElement(arr);
cout << "Remaining element: " << remainingElement << endl;
return 0;
}
|
Java
import java.util.*;
public class Main {
public static int
findRemainingElement(List<Integer> arr)
{
int n = arr.size();
while (n > 1 ) {
for ( int i = 0 ; i < n - 1 ; i++) {
arr.set(i, arr.get(i + 1 ) - arr.get(i));
}
n--;
}
return arr.get( 0 );
}
public static void main(String[] args)
{
List<Integer> arr = Arrays.asList( 3 , 4 , 2 , 1 );
int remainingElement = findRemainingElement(arr);
System.out.println( "Remaining element: "
+ remainingElement);
}
}
|
Python3
def find_remaining_element(arr):
n = len (arr)
while n > 1 :
for i in range (n - 1 ):
arr[i] = arr[i + 1 ] - arr[i]
n - = 1
return arr[ 0 ]
arr = [ 3 , 4 , 2 , 1 ]
remaining_element = find_remaining_element(arr)
print ( "Remaining element:" , remaining_element)
|
C#
using System;
using System.Collections.Generic;
class MainClass {
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];
}
static void Main() {
List< int > arr = new List< int > {3, 4, 2, 1};
int remainingElement = FindRemainingElement(arr);
Console.WriteLine( "Remaining element: " + remainingElement);
}
}
|
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];
}
let arr = [3, 4, 2, 1];
let remainingElement = findRemainingElement(arr);
console.log( "Remaining element: " + remainingElement);
|
OutputRemaining 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:
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;
}
|
Java
import java.io.*;
class GFG {
public static int lastElement( 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;
}
public static void main(String[] args)
{
int arr[] = { 3 , 4 , 2 , 1 };
int N = 4 ;
System.out.println(lastElement(arr, N));
}
}
|
Python3
def lastElement(arr, n):
sum = 0
if n % 2 = = 0 :
multiplier = - 1
else :
multiplier = 1
for i in range (n):
sum + = arr[i] * multiplier
multiplier = multiplier * (n - 1 - i) / (i + 1 ) * ( - 1 )
return sum
if __name__ = = '__main__' :
arr = [ 3 , 4 , 2 , 1 ]
N = len (arr)
print ( int (lastElement(arr, N)))
|
Javascript
<script>
function lastElement(arr, n)
{
let sum = 0;
let multiplier = n % 2 == 0 ? -1 : 1;
for (let i = 0; i < n; i++)
{
sum += arr[i] * multiplier;
multiplier
= multiplier * (n - 1 - i)
/ (i + 1) * (-1);
}
return sum;
}
let arr = [ 3, 4, 2, 1 ];
let N = arr.length;
document.write(lastElement(arr, N));
</script>
|
C#
using System;
class GFG
{
public static int lastElement( 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;
}
static void Main()
{
int [] arr = { 3, 4, 2, 1 };
int N = 4;
Console.WriteLine(lastElement(arr, N));
}
}
|
Time Complexity: O(N)
Auxiliary Space: O(1)
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