Encuentre los puntos máximos que se pueden obtener eliminando elementos de la array

Dada una array A que tiene N elementos y dos números enteros L y R donde,  y  . Puede elegir cualquier elemento de la array (digamos una _x ) y eliminarlo , y también eliminar todos los elementos iguales a _x +1 , _x +2 … a _x + R y _x -1 , _x -2 … a _x -L de la array. Este paso costará _x puntos. La tarea es maximizar el costo total después de eliminar todos los elementos de la array. Ejemplos:
 
 

Input : 2 1 2 3 2 2 1
        L = 1, R = 1
Output : 8
We select 2 to delete, then (2-1)=1 and (2+1)=3 will need to be deleted, 
for given L and R range respectively.
Repeat this until 2 is completely removed. So, total cost = 2*4 = 8.

Input : 2 4 2 9 5
        L = 1, R = 2
Output : 18
We select 2 to delete, then 5 and then 9.
So total cost = 2*2 + 5 + 9 = 18.

Planteamiento: Encontraremos el conteo de todos los elementos. Ahora supongamos que se selecciona un elemento X y luego se eliminarán todos los elementos en el rango [XL, X+R]. Ahora seleccionamos el rango mínimo de L y R y encontramos hasta qué elementos se eliminarán cuando se seleccione el elemento X. Nuestros resultados serán el máximo de elementos eliminados previamente y cuando se elimine el elemento X. Usaremos la programación dinámica para almacenar el resultado de los elementos eliminados previamente y usarlos más. 
 

// C++ program to find maximum cost after
// deleting all the elements form the array
#include <bits/stdc++.h>
using namespace std;
 
// function to return maximum cost obtained
int maxCost(int a[], int n, int l, int r)
{
 
    int mx = 0, k;
    // find maximum element of the array.
    for (int i = 0; i < n; ++i)
        mx = max(mx, a[i]);
 
    // initialize count of all elements to zero.
    int count[mx + 1];
    memset(count, 0, sizeof(count));
 
    // calculate frequency of all elements of array.
    for (int i = 0; i < n; i++)
        count[a[i]]++;
 
    // stores cost of deleted elements.
    int res[mx + 1];
    res[0] = 0;
 
    // selecting minimum range from L and R.
    l = min(l, r);
 
    for (int num = 1; num <= mx; num++) {
 
        // finds upto which elements are to be
        // deleted when element num is selected.
        k = max(num - l - 1, 0);
 
        // get maximum when selecting element num or not.
        res[num] = max(res[num - 1], num * count[num] + res[k]);
    }
 
    return res[mx];
}
 
// Driver program
int main()
{
    int a[] = { 2, 1, 2, 3, 2, 2, 1 }, l = 1, r = 1;
 
    // size of array
    int n = sizeof(a) / sizeof(a[0]);
 
    // function call to find maximum cost
    cout << maxCost(a, n, l, r);
 
    return 0;
}

//Java program to find maximum cost after
//deleting all the elements form the array
 
public class GFG {
     
    //function to return maximum cost obtained
    static int maxCost(int a[], int n, int l, int r)
    {
 
     int mx = 0, k;
     // find maximum element of the array.
     for (int i = 0; i < n; ++i)
         mx = Math.max(mx, a[i]);
 
     // initialize count of all elements to zero.
     int[] count = new int[mx + 1];
     for(int i = 0; i < count.length; i++)
         count[i] = 0;
 
     // calculate frequency of all elements of array.
     for (int i = 0; i < n; i++)
         count[a[i]]++;
 
     // stores cost of deleted elements.
     int[] res = new int[mx + 1];
     res[0] = 0;
 
     // selecting minimum range from L and R.
     l = Math.min(l, r);
 
     for (int num = 1; num <= mx; num++) {
 
         // finds upto which elements are to be
         // deleted when element num is selected.
         k = Math.max(num - l - 1, 0);
 
         // get maximum when selecting element num or not.
         res[num] = Math.max(res[num - 1], num * count[num] + res[k]);
     }
 
     return res[mx];
    }
 
    //Driver program
    public static void main(String[] args) {
         
        int a[] = { 2, 1, 2, 3, 2, 2, 1 }, l = 1, r = 1;
 
         // size of array
         int n = a.length;
 
         // function call to find maximum cost
         System.out.println(maxCost(a, n, l, r));
    }
}

# Python 3 Program to find maximum cost after
# deleting all the elements form the array
 
# function to return maximum cost obtained
def maxCost(a, n, l, r) :
 
    mx = 0
 
    # find maximum element of the array.
    for i in range(n) :
        mx = max(mx, a[i])
 
    # create and initialize count of all elements to zero.
    count = [0] * (mx + 1)
 
    # calculate frequency of all elements of array.
    for i in range(n) :
        count[a[i]] += 1
 
    # stores cost of deleted elements.
    res = [0] * (mx + 1)
    res[0] = 0
 
    # selecting minimum range from L and R.
    l = min(l, r)
 
    for num in range(1, mx + 1) :
 
        # finds upto which elements are to be
        # deleted when element num is selected.
        k = max(num - l - 1, 0)
 
        # get maximum when selecting element num or not.
        res[num] = max(res[num - 1], num * count[num] + res[k])
 
    return res[mx]
 
# Driver code
if __name__ == "__main__" :
 
    a = [2, 1, 2, 3, 2, 2, 1 ]
    l, r = 1, 1
 
    # size of array
    n =  len(a)
 
    # function call to find maximum cost
    print(maxCost(a, n, l, r))
 
# This code is contributed by ANKITRAI1

// C# program to find maximum cost
// after deleting all the elements
// form the array
using System;
 
class GFG
{
 
// function to return maximum
// cost obtained
static int maxCost(int []a, int n,
                   int l, int r)
{
    int mx = 0, k;
     
    // find maximum element
    // of the array.
    for (int i = 0; i < n; ++i)
        mx = Math.Max(mx, a[i]);
     
    // initialize count of all
    // elements to zero.
    int[] count = new int[mx + 1];
    for(int i = 0; i < count.Length; i++)
        count[i] = 0;
     
    // calculate frequency of all
    // elements of array.
    for (int i = 0; i < n; i++)
        count[a[i]]++;
     
    // stores cost of deleted elements.
    int[] res = new int[mx + 1];
    res[0] = 0;
     
    // selecting minimum range
    // from L and R.
    l = Math.Min(l, r);
     
    for (int num = 1; num <= mx; num++)
    {
     
        // finds upto which elements
        // are to be deleted when
        // element num is selected.
        k = Math.Max(num - l - 1, 0);
     
        // get maximum when selecting
        // element num or not.
        res[num] = Math.Max(res[num - 1], num *
                          count[num] + res[k]);
    }
 
return res[mx];
}
 
// Driver Code
public static void Main()
{
    int []a = { 2, 1, 2, 3, 2, 2, 1 };
    int l = 1, r = 1;
 
    // size of array
    int n = a.Length;
 
    // function call to find maximum cost
    Console.WriteLine(maxCost(a, n, l, r));
}
}
 
// This code is contributed
// by inder_verma

<script>
 
// Javascript program to find maximum cost after
// deleting all the elements form the array
 
// function to return maximum cost obtained
function maxCost(a, n, l, r)
{
 
    var mx = 0, k;
    // find maximum element of the array.
    for (var i = 0; i < n; ++i)
        mx = Math.max(mx, a[i]);
 
    // initialize count of all elements to zero.
    var count = new Array(mx + 1);
    count.fill(0);
 
    // calculate frequency of all elements of array.
    for (var i = 0; i < n; i++)
        count[a[i]]++;
 
    // stores cost of deleted elements.
    var res = new Array(mx + 1);
    res[0] = 0;
 
    // selecting minimum range from L and R.
    l = Math.min(l, r);
 
    for (var num = 1; num <= mx; num++) {
 
        // finds upto which elements are to be
        // deleted when element num is selected.
        k = Math.max(num - l - 1, 0);
 
        // get maximum when selecting element num or not.
        res[num] = Math.max(res[num - 1],
        num * count[num] + res[k]);
    }
 
    return res[mx];
}
 
var a = [ 2, 1, 2, 3, 2, 2, 1 ];
var l = 1, r = 1;
// size of array
var n = a.length;
 
// function call to find maximum cost
document.write(maxCost(a, n, l, r));
 
// This code is contributed by SoumikMondal
 
</script>

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Complejidad del tiempo: O(max(A))