Replace every array element by multiplication of previous and next

• Difficulty Level : Easy
• Last Updated : 07 Jul, 2021

Given an array of integers, update every element with multiplication of previous and next elements with following exceptions.
a) First element is replaced by multiplication of first and second.
b) Last element is replaced by multiplication of last and second last.

Example:

Input: arr[] = {2, 3, 4, 5, 6}
Output: arr[] = {6, 8, 15, 24, 30}

// We get the above output using following
// arr[] = {2*3, 2*4, 3*5, 4*6, 5*6}

Source: Top 25 Interview Questions
A Simple Solution is to create an auxiliary array, copy contents of given array to auxiliary array. Finally traverse the auxiliary array and update given array using copied values. Time complexity of this solution is O(n), but it requires O(n) extra space.
An efficient solution can solve the problem in O(n) time and O(1) space. The idea is to keep track of previous element in loop.

Below is the implementation of this idea.

C++

 // C++ program to update every array element with// multiplication of previous and next numbers in array#includeusing namespace std; void modify(int arr[], int n){    // Nothing to do when array size is 1    if (n <= 1)      return;     // store current value of arr and update it    int prev = arr;    arr = arr * arr;     // Update rest of the array elements    for (int i=1; i

Java

 // Java program to update every array element with// multiplication of previous and next numbers in arrayimport java.io.*;import java.util.*;import java.lang.Math; class Multipy{   static void modify(int arr[], int n)    {        // Nothing to do when array size is 1        if (n <= 1)            return;         // store current value of arr and update it        int prev = arr;        arr = arr * arr;         // Update rest of the array elements        for (int i=1; i

Python3

 # Python program to update every array element with# multiplication of previous and next numbers in array def modify(arr, n):  # Nothing to do when array size is 1    if n <= 1:        return     # store current value of arr and update it    prev = arr    arr = arr * arr     # Update rest of the array elements    for i in range(1, n-1):             # Store current value of next interaction        curr = arr[i];         # Update current value using previous value        arr[i] = prev * arr[i+1]            # Update previous value        prev = curr          # Update last array element    arr[n-1] = prev * arr[n-1]  # Driver programarr = [2, 3, 4, 5, 6]n = len(arr)modify(arr, n)for i in range (0, n):    print(arr[i],end=" ") # This code is contributed by# Smitha Dinesh Semwal

C#

 // C# program to update every array// element with multiplication of// previous and next numbers in arrayusing System; class GFG{    static void modify(int []arr, int n)    {        // Nothing to do when array size is 1        if (n <= 1)            return;         // store current value of arr and update it        int prev = arr;        arr = arr * arr;         // Update rest of the array elements        for (int i=1; i





Output:

6 8 15 24 30