# Reduce the array to a single integer with the given operation

Given an array arr[] of N integers from 1 to N. The task is to perform the following operations N – 1 times.

1. Select two elements X and Y from the array.
2. Delete the chosen elements from the array.
3. Add X2 + Y2in the array.

After performing above operations N – 1 times only one integer will be left in the array. The task is to print the maximum possible value of that integer.

Examples:

Input: N = 3
Output: 170
Initial array: arr[] = {1, 2, 3}
Choose 2 and 3 and the array becomes arr[] = {1, 13}
Performing the operation again by choosing the only two elements left,
the array becomes arr[] = {170} which is the maximum possible value.

Input: N = 4
Output: 395642

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

Approach: To maximize the value of final integer we have to maximize the value of (X2 + Y2). So each time we have to choose the maximum two values from the array. Store all integers in a priority queue. Each time pop top 2 elements and push the result of (X2 + Y2) in the priority queue. The last remaining element will be the maximum possible value of the required integer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define ll long long int ` ` `  `// Function to return the maximum ` `// integer after performing the operations ` `int` `reduceOne(``int` `N) ` `{ ` `    ``priority_queue pq; ` ` `  `    ``// Initialize priority queue with ` `    ``// 1 to N ` `    ``for` `(``int` `i = 1; i <= N; i++) ` `        ``pq.push(i); ` ` `  `    ``// Perform the operations while ` `    ``// there are at least 2 elements ` `    ``while` `(pq.size() > 1) { ` ` `  `        ``// Get the maximum and ` `        ``// the second maximum ` `        ``ll x = pq.top(); ` `        ``pq.pop(); ` `        ``ll y = pq.top(); ` `        ``pq.pop(); ` ` `  `        ``// Push (x^2 + y^2) ` `        ``pq.push(x * x + y * y); ` `    ``} ` ` `  `    ``// Return the only element left ` `    ``return` `pq.top(); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 3; ` ` `  `    ``cout << reduceOne(N); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function to return the maximum ` `    ``// integer after performing the operations ` `    ``static` `long` `reduceOne(``int` `N)  ` `    ``{ ` `        ``// To create Max-Heap ` `        ``PriorityQueue pq = ``new`  `        ``PriorityQueue(Collections.reverseOrder()); ` ` `  `        ``// Initialize priority queue with ` `        ``// 1 to N ` `        ``for` `(``long` `i = ``1``; i <= N; i++) ` `            ``pq.add(i); ` ` `  `        ``// Perform the operations while ` `        ``// there are at least 2 elements ` `        ``while` `(pq.size() > ``1``) ` `        ``{ ` ` `  `            ``// Get the maximum and ` `            ``// the second maximum ` `            ``long` `x = pq.poll(); ` `            ``long` `y = pq.poll(); ` ` `  `            ``// Push (x^2 + y^2) ` `            ``pq.add(x * x + y * y); ` `        ``} ` ` `  `        ``// Return the only element left ` `        ``return` `pq.peek(); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `N = ``3``; ` `        ``System.out.println(reduceOne(N)); ` `    ``} ` `} ` ` `  `// This code is contributed by ` `// sanjeev2552 `

Output:

```170
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : sanjeev2552

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.