The problem
You are given an array prices where prices[i] is the price of a given stock on the i-th day.
You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.
Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.
Examples
Input: prices = [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.
Constraints
- 1 <= prices.length <= 10⁵
- 0 <= prices[i] <= 10⁴
Brute force solution
func maxProfit(_ prices: [Int]) -> Int {
let n = prices.count
var res = 0
for i in 0 ..< n {
for j in i + 1 ..< n {
let profit = prices[j] - prices[i]
if profit > 0 {
res = max(res, profit)
}
}
}
return res
}
Explanation
We can start with a brute force solution and find a way to a more optimal solution as we go.
By visualizing the problem using input from example 1 [7,1,5,3,6,4]:
We can see that the maximum profit is possible if you buy on day 2 for price 1 and sell on day 5 at price 6, resulting in a profit of 5.
We can iterate over all prices, compare the current price with the next one, and calculate the profit. This will work with a time complexity of O(n²), but we can do better by using the two-pointer technique.
Time/Space Complexity
- Time complexity: O(n²)
- Space complexity: O(1)
Solution 2 - Two Pointers
func maxProfit(_ prices: [Int]) -> Int {
let n = prices.count
var res = 0
var l = 0
var r = 1
while r < n {
let profit = prices[r] - prices[l]
if profit > 0 {
res = max(res, profit)
} else {
l = r
}
r += 1
}
return res
}
Explanation
From the brute force approach, we learned that it is possible to solve this problem in a more optimal way by using the two-pointer technique. The calculation of profit remains the same, but the solution is optimized to O(n) time complexity.
Time/Space Complexity
- Time complexity: O(n)
- Space complexity: O(1)