Find maximum points which can be obtained by deleting elements from array in C++

C++Server Side ProgrammingProgramming

Concept

With respect of a given array A having N elements and two integers l and r where, 1≤ ax ≤ 105 and 1≤ l≤ r≤ N. We can select any element of the array (let’s say ax) and delete it, and also delete all elements equal to ax+1, ax+2 … ax+R and ax-1, ax-2 … ax-L from the array. This step will cost ax points. Our task is to maximize the total cost after deleting all the elements from the array.

Input

2 1 2 3 2 2 1
l = 1, r = 1

Output

8

Here, we choose 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 10 5
l = 1, r = 2

Output

18

Here, we choose 2 to delete, then 5 and then 10.

So total cost = 2*2 + 5 + 10 = 19.

Method

Now, we will determine the count of all the elements. Assume an element X is chosen then, all elements in the range [X-l, X+r] will be deleted. At present, we choose the minimum range from l and r and determines up to which elements are to be deleted when element X is chosen. The results will be maximum of previously deleted elements and when element X is deleted. Now, we will implement dynamic programming to store the result of previously deleted elements and implement it further.

Example

 Live Demo

// C++ program to find maximum cost after
// deleting all the elements form the array
#include <bits/stdc++.h>
using namespace std;
// Shows function to return maximum cost obtained
int maxCost(int a[], int m, int L, int R){
   int mx1 = 0, k1;
   // Determine maximum element of the array.
   for (int p = 0; p < m; ++p)
      mx1 = max(mx1, a[p]);
   // Used to initialize count of all elements to zero.
   int count1[mx1 + 1];
   memset(count1, 0, sizeof(count1));
   // Compute frequency of all elements of array.
   for (int p = 0; p < m; p++)
      count1[a[p]]++;
   // Used to store cost of deleted elements.
   int res1[mx1 + 1];
   res1[0] = 0;
   // Choosing minimum range from L and R.
   L = min(L, R);
   for (int num1 = 1; num1 <= mx1; num1++) {
      // Determines upto which elements are to be
      // deleted when element num is selected.
      k1 = max(num1 - L - 1, 0);
      // Obtain maximum when selecting element num or not.
      res1[num1] = max(res1[num1 - 1], num1 * count1[num1] +
      res1[k1]);
   }
   return res1[mx1];
}
// Driver program
int main(){
   int a1[] = { 1, 1, 3, 3, 3, 2, 4 }, l1 = 1, r1 = 1;
   int a2[] = { 2, 4, 2, 10, 5 }, l2 = 1, r2 = 2;
   // size of array
   int n1 = sizeof(a1) / sizeof(a1[0]);
   int n2 = sizeof(a2) / sizeof(a2[0]);
   // function call to find maximum cost
   cout<<"Maximum Cost for First Example:" << maxCost(a1, n1, l1,r1)<<endl;
   cout<<"Maximum Cost for Second Example:" << maxCost(a2, n2, l2,r2);
   return 0;
}

Output

Maximum Cost for First Example:11
Maximum Cost for Second Example:19
raja
Published on 25-Jul-2020 12:21:28
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