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



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:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
  
// Function to return the maximum
// integer after performing the operations
int reduceOne(int N)
{
    priority_queue<ll> 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;
}

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Output:

170


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