# Replace every array element by multiplication of previous and next

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.

```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 C++ implementation of this idea.

## C

```// 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 interation
int curr = arr[i];

// Update current value using previos 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
int 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

```// Java program to update every array element with
// multiplication of previous and next numbers in array
import 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[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 interation
int curr = arr[i];

// Update current value using previos 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 */
```

Output:
`6 8 15 24 30`

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