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 elementint 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 codeint 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;} |
Java
import java.util.*;public class Main { // Function to find the remaining array element 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); } // Driver code public static void main(String[] args) { // Given input List<Integer> arr = Arrays.asList(3, 4, 2, 1); // Function call int remainingElement = findRemainingElement(arr); // Print the remaining element System.out.println("Remaining element: " + remainingElement); }}// This code is contributed by user_dtewbxkn77n |
Python3
# Python3 impelementationdef 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]# Driver codearr = [3, 4, 2, 1]# Function callremaining_element = find_remaining_element(arr)# Print the remaining elementprint("Remaining element:", remaining_element)# written by kk |
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 inputlet arr = [3, 4, 2, 1];// Function calllet remainingElement = findRemainingElement(arr);// Print the remaining elementconsole.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 NCr 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 repeatedlyint 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 Codeint 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 repeatedlydef 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 Codeif __name__ == '__main__': arr = [3, 4, 2, 1] N = len(arr) print(int(lastElement(arr, N))) # This code is contributed by SURENDRA_GANGWAR. |
Javascript
<script>// JavaScript program for the above approach// Function to find the last remaining// array element after performing// the given operations repeatedlyfunction 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> |
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. |
Output
4
Time Complexity: O(N)
Auxiliary Space: O(1)
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