Given an array of integers, Task is to print the Maximum Product among the array such that its previous and next element product is maximum.**Note:**Array can be considered in the cyclic order. The previous element of the first element would be equal to the last element and the next element for the last element would be the first element.

**Examples:**

Input :a[ ] = { 5, 6, 4, 3, 2}Output :20

For 5 :previous element is 2 and next element is 6 so, product will be 12.

For 6 :previous element is 5 and next element is 4 so, product will be 20.

For 4 :previous element is 6 and next element is 3 so, product will be 18.

For 3 :previous element is 4 and next element is 2 so, product will be 8.

For 2 :previous element is 3 and next element is 5 so, product will be 15.

maximum possible product is 20

and maximum element in an array is 6.

Input :a[ ] = {9, 2, 3, 1, 5, 17}Output :45

**Approach:**

- Idea is to firstly find previous element and next element .
- After finding both element, take product and find the maximum product among them .

Below is the implementation of the above approach:

## C++

` ` `// C++ program to print maximum product` `// such that its previous and next` `// element product is maximum.` `#include<bits/stdc++.h>` `using` `namespace` `std;` ` ` `// function to return largest element` ` ` `// such that its previous and next` ` ` `// element product is maximum.` ` ` `int` `maxProduct(` `int` `a[], ` `int` `n)` ` ` `{` ` ` ` ` `int` `product[n];` ` ` ` ` `int` `maxA[n]; ` ` ` `int` `maxProd = 0;` ` ` `int` `maxArr = 0;` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` ` ` `// product of previous and next` ` ` `// element and stored into an ` ` ` `// array product[i]` ` ` `product[i] = a[(i + 1) % n] *` ` ` `a[(i + (n - 1)) % n];` ` ` ` ` `// find maximum product ` ` ` `// in product[i] array` ` ` `if` `(maxProd < product[i]) {` ` ` `maxProd = product[i];` ` ` `}` ` ` `}` ` ` `return` `maxProd;` ` ` `}` ` ` ` ` `// Driver program` ` ` `int` `main()` ` ` `{` ` ` `int` `a[] = { 5, 6, 4, 3, 2 };` ` ` ` ` `int` `n = ` `sizeof` `(a)/` `sizeof` `(a[0]);` ` ` ` ` `cout<<(maxProduct(a, n));` ` ` `}` ` ` `// This code contributed by Rajput-Ji` |

## Java

`// Java program to print maximum product` `// such that its previous and next` `// element product is maximum.` `import` `java.io.*;` ` ` `class` `GFG` `{` ` ` `// function to return largest element` ` ` `// such that its previous and next` ` ` `// element product is maximum.` ` ` `static` `int` `maxProduct(` `int` `a[], ` `int` `n)` ` ` `{` ` ` ` ` `int` `[] product = ` `new` `int` `[n];` ` ` ` ` `int` `maxA[] = ` `new` `int` `[n];` ` ` ` ` `int` `maxProd = ` `0` `;` ` ` `int` `maxArr = ` `0` `;` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) ` ` ` `{` ` ` ` ` `// product of previous and next` ` ` `// element and stored into an ` ` ` `// array product[i]` ` ` `product[i] = a[(i + ` `1` `) % n] *` ` ` `a[(i + (n - ` `1` `)) % n];` ` ` ` ` `// find maximum product ` ` ` `// in product[i] array` ` ` `if` `(maxProd < product[i]) ` ` ` `{` ` ` `maxProd = product[i];` ` ` `}` ` ` `}` ` ` `return` `maxProd;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` ` ` `{` ` ` `int` `[] a = { ` `5` `, ` `6` `, ` `4` `, ` `3` `, ` `2` `};` ` ` ` ` `int` `n = a.length;` ` ` ` ` `System.out.println(maxProduct(a, n));` ` ` `}` `}` ` ` `// This code is contributed by Rajput-Ji` |

## Python3

`# Python3 program to print maximum product ` `# such that its previous and next ` `# element product is maximum. ` ` ` `# function to return largest element ` `# such that its previous and next ` `# element product is maximum. ` `def` `maxProduct(a, n) : ` ` ` ` ` `product ` `=` `[` `0` `]` `*` `n; ` ` ` `maxA ` `=` `[` `0` `]` `*` `n; ` ` ` `maxProd ` `=` `0` `; ` ` ` `maxArr ` `=` `0` `; ` ` ` ` ` `for` `i ` `in` `range` `(n) :` ` ` ` ` `# product of previous and next ` ` ` `# element and stored into an ` ` ` `# array product[i] ` ` ` `product[i] ` `=` `a[(i ` `+` `1` `) ` `%` `n] ` `*` `a[(i ` `+` `(n ` `-` `1` `)) ` `%` `n]; ` ` ` ` ` `# find maximum product ` ` ` `# in product[i] array ` ` ` `if` `(maxProd < product[i]) :` ` ` `maxProd ` `=` `product[i]; ` ` ` ` ` `return` `maxProd; ` ` ` ` ` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `a ` `=` `[ ` `5` `, ` `6` `, ` `4` `, ` `3` `, ` `2` `]; ` ` ` `n ` `=` `len` `(a);` ` ` ` ` `print` `(maxProduct(a, n)); ` ` ` `# This code is contributed by AnkitRai01` |

## C#

`// C# program to print maximum product` `// such that its previous and next` `// element product is maximum.` `using` `System;` ` ` `class` `GFG` `{` ` ` `// function to return largest element` ` ` `// such that its previous and next` ` ` `// element product is maximum.` ` ` `static` `int` `maxProduct(` `int` `[]a, ` `int` `n)` ` ` `{` ` ` ` ` `int` `[] product = ` `new` `int` `[n];` ` ` ` ` `//int []maxA = new int[n];` ` ` ` ` `int` `maxProd = 0;` ` ` `//int maxArr = 0;` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `{` ` ` ` ` `// product of previous and next` ` ` `// element and stored into an ` ` ` `// array product[i]` ` ` `product[i] = a[(i + 1) % n] *` ` ` `a[(i + (n - 1)) % n];` ` ` ` ` `// find maximum product ` ` ` `// in product[i] array` ` ` `if` `(maxProd < product[i]) ` ` ` `{` ` ` `maxProd = product[i];` ` ` `}` ` ` `}` ` ` `return` `maxProd;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` ` ` `{` ` ` `int` `[] a = { 5, 6, 4, 3, 2 };` ` ` ` ` `int` `n = a.Length;` ` ` ` ` `Console.WriteLine(maxProduct(a, n));` ` ` `}` `}` ` ` `// This code is contributed by anuj_67..` |

**Output:**

20

**Time Complexity :** O(N)

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