# 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.

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

 #include 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 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));     } }

 # 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# 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

Output:
6

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