Replace every array element by multiplication of previous and next - GeeksforGeeks

08 Mar, 2022
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 QuestionsA 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++
Java
Python3
C#
PHP
Javascript
// C++ program to update every array element with// multiplication of previous and next numbers in array#include<iostream>using 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[0] and update it    int prev = arr[0];    arr[0] = arr[0] * arr[1];     // Update rest of the array elements    for (int i=1; i<n-1; i++)    {        // Store current value of next interaction        int 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 programint main(){    int arr[] = {2, 3, 4, 5, 6};    int n = sizeof(arr)/sizeof(arr[0]);    modify(arr, n);    for (int i=0; i<n; i++)      cout << arr[i] << " ";    return 0;}
// 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 Multiply{   static void modify(int arr[], int n)    {        // Nothing to do when array size is 1        if (n <= 1)            return;         // store current value of arr[0] and update it        int prev = arr[0];        arr[0] = arr[0] * arr[1];         // Update rest of the array elements        for (int i=1; i<n-1; i++)        {            // Store current value of next interaction            int 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 program to test above function    public static void main(String[] args)    {        int arr[] = {2, 3, 4, 5, 6};        int n = arr.length;        modify(arr, n);        for (int i=0; i<n; i++)         System.out.print(arr[i]+" ");    }}/* This code is contributed by Devesh Agrawal */
# 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[0] and update it    prev = arr[0]    arr[0] = arr[0] * arr[1]     # 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# 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[0] and update it        int prev = arr[0];        arr[0] = arr[0] * arr[1];         // Update rest of the array elements        for (int i=1; i<n-1; i++)        {            // Store current value of next interaction            int 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 program to test above function    public static void Main()    {        int []arr = {2, 3, 4, 5, 6};        int n = arr.Length;        modify(arr, n);        for (int i=0; i<n; i++)        Console.Write(arr[i]+" ");    }} // This code is contributed by Sam007
<?php// PHP program to update every array// element with multiplication of previous// and next numbers in array function modify(&$arr, $n){    // Nothing to do when array size is 1    if ($n <= 1)    return;     // store current value of arr[0]    // and update it    $prev = $arr[0];    $arr[0] = $arr[0] * $arr[1];     // Update rest of the array elements    for ($i = 1; $i < $n - 1; $i++)    {        // 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 Code$arr = array (2, 3, 4, 5, 6);$n = sizeof($arr);modify($arr, $n);for ($i = 0; $i < $n; $i++)echo $arr[$i] ." "; // This code is contributed// by ChitraNayal?>
<script>// Javascript program to update every array element with// multiplication of previous and next numbers in array function modify(arr, n){    // Nothing to do when array size is 1    if (n <= 1)      return;     // store current value of arr[0] and update it    let prev = arr[0];    arr[0] = arr[0] * arr[1];     // Update rest of the array elements    for (let i = 1; i < n - 1; i++)    {        // Store current value of next interaction        let 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 programlet arr = [2, 3, 4, 5, 6];let n = arr.length;modify(arr, n);for (let i = 0; i < n; i++)    document.write(arr[i] + " ");     // This code is contributed by subham348.</script>
Output: 
6 8 15 24 30
This article is contributed by Ravi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above 
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