710016. Buy Maximum Stocks
Buy Maximum Stocks
Slug: buy-maximum-stocks-if-i-stocks-can-be-bought-on-i-th-day Difficulty: Medium Id: 710016 Topic Tags: Arrays, Data Structures Company Tags: None
Summary
Given an array of integers prices representing the stock prices on each day from i to n, where i is a given integer, find the maximum number of stocks that can be bought such that at most one stock can be bought on any day.
The problem requires understanding arrays and data structures concepts. The key idea is to identify the days when the price drops to buy the stock and maximize the profit.
Detailed Explanation
To solve this problem, we need to iterate through the array of prices and keep track of the maximum number of stocks that can be bought. We start by initializing a variable max_stocks to 0, which represents the maximum number of stocks that can be bought so far. Then, we iterate through the array from day i to day n.
For each day j, if the price is lower than the previous day's price, it means the stock price has dropped, and we can buy a new stock. In this case, we increment max_stocks by 1.
Here's the step-by-step breakdown of the solution:
max_stocks = 0
prev_price = prices[i - 1]
for j in range(i, n + 1):
if prices[j - 1] < prev_price:
max_stocks += 1
prev_price = prices[j - 1]
The time complexity of this solution is O(n), where n is the number of days. This is because we need to iterate through the array once.
The space complexity is O(1), as we only use a constant amount of space to store the variables max_stocks and prev_price.
Optimized Solutions
Java
public int maxStocks(int[] prices, int i) {
int maxStocks = 0;
int prevPrice = prices[i - 1];
for (int j = i; j <= prices.length; j++) {
if (prices[j - 1] < prevPrice) {
maxStocks++;
}
prevPrice = prices[j - 1];
}
return maxStocks;
}