Maximice la diferencia entre la suma de las dos mitades de la array después de eliminar N elementos

Dado un número entero N y una array arr[] que consta de 3 * N enteros, la tarea es encontrar la diferencia máxima entre la primera mitad y la segunda mitad de la array después de eliminar exactamente N elementos de la array.

Ejemplos:

Entrada: N = 2, arr[] = {3, 1, 4, 1, 5, 9}
Salida: 1
Explicación:
La eliminación de arr[1] y arr[5] del arreglo maximiza la diferencia = (3 + 4 ) – (1 + 5) = 7 – 6 = 1.

Entrada: N = 1, arr[] = {1, 2, 3}
Salida: -1

Enfoque: 
siga los pasos que se indican a continuación para resolver el problema

• Recorra la array desde el principio y siga actualizando la suma de los N elementos más grandes desde el principio de la array.
• Del mismo modo, siga actualizando la suma de los N elementos más pequeños desde el final de la array.
• Recorra estas sumas y calcule las diferencias en cada punto y actualice la máxima diferencia obtenida.
• Imprime la diferencia máxima obtenida.

A continuación se muestra la implementación del enfoque anterior: 

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the maximum difference
// possible between the two halves of the array
long long FindMaxDif(vector<long long> a, int m)
{
    int n = m / 3;
 
    vector<long long> l(m + 5), r(m + 5);
 
    // Stores n maximum values from the start
    multiset<long long> s;
 
    for (int i = 1; i <= m; i++) {
 
        // Insert first n elements
        if (i <= n) {
 
            // Update sum of largest n
            // elements from left
            l[i] = a[i - 1] + l[i - 1];
            s.insert(a[i - 1]);
        }
 
        // For the remaining elements
        else {
            l[i] = l[i - 1];
 
            // Obtain minimum value
            // in the set
            long long d = *s.begin();
 
            // Insert only if it is greater
            // than minimum value
            if (a[i - 1] > d) {
 
                // Update sum from left
                l[i] -= d;
                l[i] += a[i - 1];
 
                // Remove the minimum
                s.erase(s.find(d));
 
                // Insert the current element
                s.insert(a[i - 1]);
            }
        }
    }
 
    // Clear the set
    s.clear();
 
    // Store n minimum elements from the end
    for (int i = m; i >= 1; i--) {
 
        // Insert the last n elements
        if (i >= m - n + 1) {
 
            // Update sum of smallest n
            // elements from right
            r[i] = a[i - 1] + r[i + 1];
            s.insert(a[i - 1]);
        }
 
        // For the remaining elements
        else {
 
            r[i] = r[i + 1];
 
            // Obtain the minimum
            long long d = *s.rbegin();
 
            // Insert only if it is smaller
            // than maximum value
            if (a[i - 1] < d) {
 
                // Update sum from right
                r[i] -= d;
                r[i] += a[i - 1];
 
                // Remove the minimum
                s.erase(s.find(d));
 
                // Insert the new element
                s.insert(a[i - 1]);
            }
        }
    }
 
    long long ans = -9e18L;
 
    for (int i = n; i <= m - n; i++) {
 
        // Compare the difference and
        // store the maximum
        ans = max(ans, l[i] - r[i + 1]);
    }
 
    // Return the maximum
    // possible difference
    return ans;
}
 
// Driver Code
int main()
{
 
    vector<long long> vtr = { 3, 1, 4, 1, 5, 9 };
    int n = vtr.size();
 
    cout << FindMaxDif(vtr, n);
 
    return 0;
}

# Python3 Program to implement
# the above approach
 
# Function to print the maximum difference
# possible between the two halves of the array
def FindMaxDif(a, m) :
 
    n = m // 3
 
    l = [0] * (m + 5)
    r = [0] * (m + 5)
 
    # Stores n maximum values from the start
    s = []
 
    for i in range(1, m + 1) :
 
        # Insert first n elements
        if (i <= n) :
 
            # Update sum of largest n
            # elements from left
            l[i] = a[i - 1] + l[i - 1]
            s.append(a[i - 1])
 
        # For the remaining elements
        else :
            l[i] = l[i - 1]
 
            # Obtain minimum value
            # in the set
            s.sort()
            d = s[0]
 
            # Insert only if it is greater
            # than minimum value
            if (a[i - 1] > d) :
 
                # Update sum from left
                l[i] -= d
                l[i] += a[i - 1]
 
                # Remove the minimum
                s.remove(d)
 
                # Insert the current element
                s.append(a[i - 1])
 
    # Clear the set
    s.clear()
 
    # Store n minimum elements from the end
    for i in range(m, 0, -1) :
 
        # Insert the last n elements
        if (i >= m - n + 1) :
 
            # Update sum of smallest n
            # elements from right
            r[i] = a[i - 1] + r[i + 1]
            s.append(a[i - 1])
 
        # For the remaining elements
        else :
 
            r[i] = r[i + 1]
            s.sort()
             
            # Obtain the minimum
            d = s[-1]
 
            # Insert only if it is smaller
            # than maximum value
            if (a[i - 1] < d) :
 
                # Update sum from right
                r[i] -= d
                r[i] += a[i - 1]
 
                # Remove the minimum
                s.remove(d)
 
                # Insert the new element
                s.append(a[i - 1])
 
    ans = -9e18
 
    for i in range(n, m - n + 1) :
 
        # Compare the difference and
        # store the maximum
        ans = max(ans, l[i] - r[i + 1])
 
    # Return the maximum
    # possible difference
    return ans
 
# Driver code 
vtr = [ 3, 1, 4, 1, 5, 9 ]
n = len(vtr)
 
print(FindMaxDif(vtr, n))
 
# This code is contributed by divyesh072019

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to print the maximum difference
// possible between the two halves of the array
static long FindMaxDif(List<long> a, int m)
{
    int n = m / 3;
     
    long[] l = new long[m + 5];
    long[] r = new long[m + 5];
   
    // Stores n maximum values from the start
    List<long> s = new List<long>();
   
    for(int i = 1; i <= m; i++)
    {
         
        // Insert first n elements
        if (i <= n)
        {
             
            // Update sum of largest n
            // elements from left
            l[i] = a[i - 1] + l[i - 1];
            s.Add(a[i - 1]);
        }
         
        // For the remaining elements
        else
        {
            l[i] = l[i - 1];
             
            s.Sort();
             
            // Obtain minimum value
            // in the set
            long d = s[0];
   
            // Insert only if it is greater
            // than minimum value
            if (a[i - 1] > d)
            {
                 
                // Update sum from left
                l[i] -= d;
                l[i] += a[i - 1];
   
                // Remove the minimum
                s.Remove(d);
   
                // Insert the current element
                s.Add(a[i - 1]);
            }
        }
    }
   
    // Clear the set
    s.Clear();
   
    // Store n minimum elements from the end
    for(int i = m; i >= 1; i--)
    {
         
        // Insert the last n elements
        if (i >= m - n + 1)
        {
             
            // Update sum of smallest n
            // elements from right
            r[i] = a[i - 1] + r[i + 1];
            s.Add(a[i - 1]);
        }
   
        // For the remaining elements
        else
        {
            r[i] = r[i + 1];
             
            s.Sort();
             
            // Obtain the minimum
            long d = s[s.Count - 1];
   
            // Insert only if it is smaller
            // than maximum value
            if (a[i - 1] < d)
            {
                 
                // Update sum from right
                r[i] -= d;
                r[i] += a[i - 1];
   
                // Remove the minimum
                s.Remove(d);
   
                // Insert the new element
                s.Add(a[i - 1]);
            }
        }
    }
   
    long ans = (long)(-9e18);
   
    for(int i = n; i <= m - n; i++)
    {
         
        // Compare the difference and
        // store the maximum
        ans = Math.Max(ans, l[i] - r[i + 1]);
    }
   
    // Return the maximum
    // possible difference
    return ans;
}
 
// Driver Code
static void Main()
{
    List<long> vtr = new List<long>(
        new long[]{ 3, 1, 4, 1, 5, 9 });
    int n = vtr.Count;
     
    Console.Write(FindMaxDif(vtr, n));
}
}
 
// This code is contributed by divyeshrabadiya07

1

Complejidad temporal: O(NlogN)
Espacio auxiliar: O(N)