Given an array A having N elements and two integers L and R where, 
and 
. You can choose 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. The task is to maximize the total cost after deleting all the elements from the array.
Examples:
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.
Approach: We will find the count of all the elements. Now let’s say an element X is selected then, all elements in the range [X-L, X+R] will be deleted. Now we select the minimum range from L and R and finds up to which elements are to be deleted when element X is selected. Our results will be the maximum of previously deleted elements and when element X is deleted. We will use dynamic programming to store the result of previously deleted elements and use it further.
Implementation:
C++
// 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 obtainedint 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 programint 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
//Java program to find maximum cost after//deleting all the elements form the arraypublic 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
# 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 codeif __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#
// C# program to find maximum cost // after deleting all the elements // form the arrayusing System;class GFG {// function to return maximum // cost obtainedstatic 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 Codepublic 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 |
Javascript
<script>// Javascript program to find maximum cost after// deleting all the elements form the array// function to return maximum cost obtainedfunction 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 arrayvar n = a.length;// function call to find maximum costdocument.write(maxCost(a, n, l, r));// This code is contributed by SoumikMondal</script> |
Output
8
Time Complexity: O(max(A))
Auxiliary Space: O(max(A))
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