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# C++ program to find maximum how many chocolates we can buy with at most k rupees

Suppose we have an array A with n elements, and other values l, r and k are there. Amal wants to buy chocolates and he will not buy too expensive chocolates, and not also too cheap chocolates. In the store, there are n different chocolate bars and the prices are represented in A. A chocolate bar is too expensive if its price is larger than r and too cheap if its price is less than l. He wants to spend at most k rupees. We have to find the maximum amount of chocolates he can buy.

So, if the input is like A = [1, 2, 3, 4, 5, 6]; l = 3; r = 5; k = 10, then the output will be 2, because he can buy chocolates worth rupees 3 and 4 by 7 rupees.

## Steps

To solve this, we will follow these steps −

n := size of A ans := 0 sort the array A for initialize i := 0, when i < n, update (increase i by 1), do: if A[i] > k, then: Come out from the loop if A[i] >= l and A[i] <= r, then: k := k - A[i] (increase ans by 1) return ans

## Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h> using namespace std; int solve(vector<int> A, int l, int r, int k) { int n = A.size(); int ans = 0; sort(A.begin(), A.end()); for (int i = 0; i < n; ++i) { if (A[i] > k) break; if (A[i] >= l && A[i] <= r) { k -= A[i]; ++ans; } } return ans; } int main() { vector<int> A = { 1, 2, 3, 4, 5, 6 }; int l = 3; int r = 5; int k = 10; cout << solve(A, l, r, k) << endl; }

## Input

{ 1, 2, 3, 4, 5, 6 }, 3, 5, 10

## Output

2

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