Maximum element in an array such that its previous and next element product is maximum

Given an array arr[] of N integers, the task is to print the largest element among the array such that its previous and next element product is maximum.
Examples:

Input: arr[] = {5, 6, 4, 3, 2} 
Output: 6 
The product of the next and the previous elements 
for every element of the given array are: 
5 -> 2 * 6 = 12 
6 -> 5 * 4 = 20 
4 -> 6 * 3 = 18 
3 -> 4 * 2 = 8 
2 -> 3 * 5 = 15 
Out of these 20 is the maximum. 
Hence, 6 is the answer.
Input: arr[] = {9, 2, 3, 1, 5, 17} 
Output: 17 

Approach: For every element of the array, find the product of its previous and next element. The element which has the maximum product is the result. If two elements have an equal product of next and previous elements then choose the greater element among them.
Below is the implementation of the above approach: 

C++

#include<bits/stdc++.h>
using namespace std;

// Function to return the largest element
// such that its previous and next
// element product is maximum
int maxElement(int a[], int n)
{
    if (n < 3)
        return -1;

    int maxElement = a[0];
    int maxProd = a[n - 1] * a[1];

    for (int i = 1; i < n; i++)
    {

        // Calculate the product of the previous
        // and the next element for
        // the current element
        int currProd = a[i - 1] * a[(i + 1) % n];

        // Update the maximum product
        if (currProd > maxProd)
        {
            maxProd = currProd;
            maxElement = a[i];
        }

        // If current product is equal to the
        // current maximum product then
        // choose the maximum element
        else if (currProd == maxProd)
        {
            maxElement = max(maxElement, a[i]);
        }
    }

    return maxElement;
}

// Driver code
int main()
{
    int a[] = { 5, 6, 4, 3, 2};
    int n = sizeof(a)/sizeof(a[0]);
    cout << maxElement(a, n);
    return 0;
}

Java

// Java implementation of the approach
class GFG {

    // Function to return the largest element
    // such that its previous and next
    // element product is maximum
    static int maxElement(int a[], int n)
    {
        if (n < 3)
            return -1;

        int maxElement = a[0];
        int maxProd = a[n - 1] * a[1];

        for (int i = 1; i < n; i++) {

            // Calculate the product of the previous
            // and the next element for
            // the current element
            int currProd = a[i - 1] * a[(i + 1) % n];

            // Update the maximum product
            if (currProd > maxProd) {
                maxProd = currProd;
                maxElement = a[i];
            }

            // If current product is equal to the
            // current maximum product then
            // choose the maximum element
            else if (currProd == maxProd) {
                maxElement = Math.max(maxElement, a[i]);
            }
        }

        return maxElement;
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] a = { 5, 6, 4, 3, 2 };
        int n = a.length;
        System.out.println(maxElement(a, n));
    }
}

Python3

# Function to return the largest element
# such that its previous and next
# element product is maximum
def maxElement(a, n):

    if n < 3:
        return -1
    maxElement = a[0]
    maxProd = a[n - 1] * a[1]

    for i in range(1, n):

        # Calculate the product of the previous
        # and the next element for
        # the current element

        currprod = a[i - 1] * a[(i + 1) % n]

        if currprod > maxProd:
            maxProd = currprod
            maxElement = a[i]

        # If current product is equal to the
        # current maximum product then
        # choose the maximum element
        elif currprod == maxProd:
            maxElement = max(maxElement, a[i])
    return maxElement

# Driver code

a = [5, 6, 4, 3, 2]
n = len(a)#sizeof(a[0])
print(maxElement(a, n))

# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to return the largest element
    // such that its previous and next
    // element product is maximum
    static int maxElement(int []a, int n)
    {
        if (n < 3)
            return -1;

        int maxElement = a[0];
        int maxProd = a[n - 1] * a[1];

        for (int i = 1; i < n; i++)
        {

            // Calculate the product of the previous
            // and the next element for
            // the current element
            int currProd = a[i - 1] * a[(i + 1) % n];

            // Update the maximum product
            if (currProd > maxProd)
            {
                maxProd = currProd;
                maxElement = a[i];
            }

            // If current product is equal to the
            // current maximum product then
            // choose the maximum element
            else if (currProd == maxProd)
            {
                maxElement = Math.Max(maxElement, a[i]);
            }
        }

        return maxElement;
    }

    // Driver code
    public static void Main()
    {
        int[] a = { 5, 6, 4, 3, 2 };
        int n = a.Length;
        Console.WriteLine(maxElement(a, n));
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>
// Java script implementation of the approach

    // Function to return the largest element
    // such that its previous and next
    // element product is maximum
    function maxElement(a,n)
    {
        if (n < 3)
            return -1;

        let maxElement = a[0];
        let maxProd = a[n - 1] * a[1];

        for (let i = 1; i < n; i++) {

            // Calculate the product of the previous
            // and the next element for
            // the current element
            let currProd = a[i - 1] * a[(i + 1) % n];

            // Update the maximum product
            if (currProd > maxProd) {
                maxProd = currProd;
                maxElement = a[i];
            }

            // If current product is equal to the
            // current maximum product then
            // choose the maximum element
            else if (currProd == maxProd) {
                maxElement = Math.max(maxElement, a[i]);
            }
        }

        return maxElement;
    }

    // Driver code
        let a = [ 5, 6, 4, 3, 2 ];
        let n = a.length;
        document.write(maxElement(a, n));

// This code is contributed by sravan kumar G
</script>

Output:

6

Time Complexity : O(n), since there runs a loop for once from 1 to (n – 1).
Auxiliary Space : O(1), since no extra space has been taken.