# 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.

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++ program to update every array element with ` `// multiplication of previous and next numbers in array ` `#include ` `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 and update it ` `    ``int` `prev = arr; ` `    ``arr = arr * arr; ` ` `  `    ``// Update rest of the array elements ` `    ``for` `(``int` `i=1; i

 `// 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 and update it ` `        ``int` `prev = arr[``0``]; ` `        ``arr[``0``] = arr[``0``] * arr[``1``]; ` ` `  `        ``// Update rest of the array elements ` `        ``for` `(``int` `i=``1``; i

 `# 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[``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 interation ` `        ``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 ` `arr ``=` `[``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 array ` `using` `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`