Buy Maximum Stocks if i stocks can be bought on i-th day Last Updated : 25 Oct, 2024 Comments Improve Suggest changes Like Article Like Report Try it on GfG Practice In a stock market, there is a product with its infinite stocks. The stock prices are given for N days, where arr[i] denotes the price of the stock on the ith day. There is a rule that a customer can buy at most i stock on the ith day. If the customer has k amount of money initially, find out the maximum number of stocks a customer can buy. For example, for 3 days the price of a stock is given as 7, 10, 4. You can buy 1 stock worth 7 rs on day 1, 2 stocks worth 10 rs each on day 2 and 3 stock worth 4 rs each on day 3.Examples: Input : price[] = { 10, 7, 19 }, k = 45.Output : 4A customer purchases 1 stock on day 1 for 10 rs, 2 stocks on day 2 for 7 rs each and 1 stock on day 3 for 19 rs.Therefore total of10, 7 * 2 = 14 and 19 respectively. Hence, total amount is 10 + 14 + 19 = 43 and number of stocks purchased is 4.Input : price[] = { 7, 10, 4 }, k = 100.Output : 6The idea is to use greedy approach, where we should start buying product from the day when the stock price is least and so on. So, we will sort the pair of two values i.e { stock price, day } according to the stock price, and if stock prices are same, then we sort according to the day. Now, we will traverse along the sorted list of pairs, and start buying as follows: Let say, we have R rs remaining till now, and the cost of product on this day be C, and we can buy atmost L products on this day then, total purchase on this day (P) = min(L, R/C) Now, add this value to the answer total_purchase = total_purchase + P, where P =min(L, R/C) Now, subtract the cost of buying P items from remaining money, R = R - P*C. Total number of products that we can buy is equal to total_purchase.Below is the implementation of this approach: C++ // C++ program to find maximum number of stocks that // can be bought with given constraints. #include <bits/stdc++.h> using namespace std; // Return the maximum stocks int buyMaximumProducts(int n, int k, int price[]) { vector<pair<int, int> > v; // Making pair of product cost and number // of day.. for (int i = 0; i < n; ++i) v.push_back(make_pair(price[i], i + 1)); // Sorting the vector pair. sort(v.begin(), v.end()); // Calculating the maximum number of stock // count. int ans = 0; for (int i = 0; i < n; ++i) { ans += min(v[i].second, k / v[i].first); k -= v[i].first * min(v[i].second, (k / v[i].first)); } return ans; } // Driven Program int main() { int price[] = { 10, 7, 19 }; int n = sizeof(price)/sizeof(price[0]); int k = 45; cout << buyMaximumProducts(n, k, price) << endl; return 0; } Java // Java program to find maximum number of stocks that // can be bought with given constraints. import java.util.*; public class GFG { // Return the maximum stocks static int buyMaximumProducts(int[] price, int K, int n) { Pair[] arr = new Pair[n]; // Making pair of product cost and number of day.. for (int i = 0; i < n; i++) arr[i] = new Pair(price[i], i + 1); // Sorting the pair array. Arrays.sort(arr, new SortPair()); // Calculating the maximum number of stock // count. int ans = 0; for (int i = 0; i < n; i++) { ans += Math.min(arr[i].second, K / arr[i].first); K -= arr[i].first * Math.min(arr[i].second, K / arr[i].first); } return ans; } // Driver code public static void main(String[] args) { int[] price = { 10, 7, 19 }; int K = 45; // int []price = { 7, 10, 4 }; // int K = 100; System.out.println( buyMaximumProducts(price, K, price.length)); } } // Helper class class Pair { int first, second; Pair(int first, int second) { this.first = first; this.second = second; } } // For Sorting using Pair.first value class SortPair implements Comparator<Pair> { public int compare(Pair a, Pair b) { return a.first - b.first; } } // This code is contributed by Aakash Choudhary Python # Python3 program to find maximum number of stocks # that can be bought with given constraints. # Returns the maximum stocks def buyMaximumProducts(n, k, price): # Making pair of stock cost and day number arr = [] for i in range(n): arr.append([i + 1, price[i]]) # Sort based on the price of stock arr.sort(key = lambda x: x[1]) # Calculating the max stocks purchased total_purchase = 0 for i in range(n): P = min(arr[i][0], k//arr[i][1]) total_purchase += P k -= (P * arr[i][1]) return total_purchase # Driver code price = [ 10, 7, 19 ] n = len(price) k = 45 print(buyMaximumProducts(n, k, price)) # This code is contributed by Tharun Reddy C# // C# program to find maximum number of stocks that // can be bought with given constraints. using System; class GFG { // Driver code static void Main(string[] args) { int[] price = { 10, 7, 19 }; int K = 45; // int []price = { 7, 10, 4 }; // int K = 100; Console.WriteLine( buyMaximumProducts(price, K, price.Length)); } // Return the maximum stocks static int buyMaximumProducts(int[] price, int K, int n) { Pair[] arr = new Pair[n]; // Making pair of product cost and number of day.. for (int i = 0; i < n; i++) arr[i] = new Pair(price[i], i + 1); // Sorting the pair array. Array.Sort(arr); // Calculating the maximum number of stock count. int ans = 0; for (int i = 0; i < n; i++) { ans += Math.Min(arr[i].second, K / arr[i].first); K -= arr[i].first * Math.Min(arr[i].second, K / arr[i].first); } return ans; } } // Helper class class Pair : IComparable<Pair> { public int first, second; public Pair(int first, int second) { this.first = first; this.second = second; } // For Sorting using Pair.first value public int CompareTo(Pair other) { return this.first - other.first; } } // This code is contributed by Tapesh (tapeshdua420) JavaScript // javascript program to find maximum number of stocks that // can be bought with given constraints. function buyMaximumProducts(n, k, price) { let v = []; // Making pair of product cost and number // of day.. for (let i = 0; i < n; ++i) { v.push([price[i], i + 1]); } // Sorting the vector pair. v.sort((a, b) => a[0] - b[0]); // Calculating the maximum number of stock // count. let ans = 0; for (let i = 0; i < n; ++i) { ans += Math.min(v[i][1], Math.floor(k / v[i][0])); k -= v[i][0] * Math.min(v[i][1], Math.floor(k / v[i][0])); } return ans; } let price = [10, 7, 19]; let n = price.length; let k = 45; console.log(buyMaximumProducts(n, k, price)); Output4Time Complexity: O(n log n).Auxiliary Space: O(n) Buy Maximum Stocks if i stocks can be bought on i-th day | DSA Problem Comment More infoAdvertise with us A Anuj Chauhan Improve Article Tags : Misc Greedy DSA Practice Tags : GreedyMisc Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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