In this approach, we traverse the prices array while keeping track of the minimum price seen so far and calculate the maximum profit we could achieve if we sold on that day. The maximum profit is updated accordingly through each iteration.
This approach makes a single pass through the array (O(n) time complexity) and uses constant space (O(1) space complexity).
Time Complexity: O(n), where n is the number of days.
Space Complexity: O(1).
1function maxProfit(prices) {
2 let minPrice = Infinity;
3 let maxProfit = 0;
4 for (let price of prices) {
5 if (price < minPrice) {
6 minPrice = price;
7 } else if (price - minPrice > maxProfit) {
8 maxProfit = price - minPrice;
9 }
10 }
11 return maxProfit;
12}
13
14const prices = [7, 1, 5, 3, 6, 4];
15console.log("Max Profit:", maxProfit(prices));The JavaScript approach here uses similar logic, iterating over the prices array and updating the minimum price and maximum potential profit for each day's price examined.
This approach considers all possible pairs of buy and sell days, calculating the profit for each combination. It is straightforward but inefficient due to its O(n^2) time complexity, which is impractical for large inputs.
Time Complexity: O(n^2), where n is the number of days.
Space Complexity: O(1).
1#include <iostream>
2#include <vector>
3
4int maxProfit(std::vector<int>& prices) {
5 int max_profit = 0;
6 for (int i = 0; i < prices.size(); i++) {
7 for (int j = i + 1; j < prices.size(); j++) {
8 int profit = prices[j] - prices[i];
9 if (profit > max_profit) {
10 max_profit = profit;
11 }
12 }
13 }
14 return max_profit;
15}
16
17int main() {
18 std::vector<int> prices = {7, 1, 5, 3, 6, 4};
19 std::cout << "Max Profit (Brute Force): " << maxProfit(prices) << std::endl;
20 return 0;
21}This C++ brute force solution implements two nested loops to evaluate each possible selling point for every entry price. It tracks and returns the maximum profit found among these.