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Given an array A of N positive integers and a budget B. Your task is to decide the maximum number of elements to be picked from the array such that the cumulative cost of all picked elements is less than or equal to budget B. Cost of picking the ith element is given by : A[i] + (i * K) where, K is a constant whose value is equal to the number of elements picked. The indexing(i) is 1 based. Print the maximum number and its respective cumulative cost.

Examples:

Input : arr[] = { 2, 3, 5 }, B = 11
Output : 2 11
Explanation : Cost of picking maximum elements = {2 + (1 * 2) } + {3 + (2 * 2)} = 4 + 7 = 11 (which is equal to budget)

Input : arr[] = { 1, 2, 5, 6, 3 }, B = 90
Output : 4 54

Prerequisites : Binary Search



Approach: The idea here is to use binary search on all possible values of K i.e. the optimal number of elements to be picked. Start with zero as lower bound and End with total number of elements i.e. N as upper bound. Check if by setting K as current Mid, obtained cumulative cost is less than or equal to budget. If it satisfies the condition, then try to increase K by setting Start as (Mid + 1), otherwise try to decrease K by setting End as (Mid – 1).
Checking of the condition can be done in a brute force manner by simply modifying the array according to the given formula and adding the K (current number of elements to be picked) smallest modified values to get the cumulative cost.

Below is the implementation of above approach.

// CPP Program to find the optimal number of
// elements such that the cumulative value
// should be less than given number
#include <bits/stdc++.h>
  
using namespace std;
  
// This function returns true if the value cumulative
// according to received integer K is less than budget
// B, otherwise returns false
bool canBeOptimalValue(int K, int arr[], int N, int B,
                       int& value)
{
    // Initialize a temporary array which stores
    // the cumulative value of the original array
    int tmp[N];
  
    for (int i = 0; i < N; i++)
        tmp[i] = (arr[i] + K * (i + 1));
  
    // Sort the array to find the smallest K values
    sort(tmp, tmp + N);
  
    value = 0;
    for (int i = 0; i < K; i++)
        value += tmp[i];
  
    // Check if the value is less than budget
    return value <= B;
}
  
// This function prints the optimal number of elements
// and respective cumulative value which is less than
// the given number
void findNoOfElementsandValue(int arr[], int N, int B)
{
    int start = 0; // Min Value or lower bound
  
    int end = N; // Max Value or upper bound
  
    // Initialize answer as zero as optimal value
    // may not exists
    int ans = 0;
  
    int cumulativeValue = 0;
  
    while (start <= end) {
        int mid = (start + end) / 2;
  
        // If the current Mid Value is an optimal
        // value, then try to maximize it
        if (canBeOptimalValue(mid, arr, N, B,
                              cumulativeValue)) {
            ans = mid;
            start = mid + 1;
        }
        else
            end = mid - 1;
    }
    // Call Again to set the corresponding cumulative
    // value for the optimal ans
    canBeOptimalValue(ans, arr, N, B, cumulativeValue);
  
    cout << ans << " " << cumulativeValue << endl;
}
  
// Driver Code
int main()
{
    int arr[] = { 1, 2, 5, 6, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);
  
    // Budget
    int B = 90;
    findNoOfElementsandValue(arr, N, B);
    return 0;
}

Output:

4 54

Time Complexity: O(N * (log N)2), where N is the number of elements in the given array.



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