Description (leetcode)

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.

Example 1:

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.

Example 2:

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 <= 10^5
  • 0 <= prices[i] <= 10^4

submission

// iterate through prices, keep track of the lowest price so far
// keep track of the maxium difference between lowest price and
// current price, we can do this by fold
impl Solution {
    pub fn max_profit(prices: Vec<i32>) -> i32 {
        prices.into_iter()
            .fold((0, i32::MAX), |(ans, min), price| {
                (ans.max(price - min), min.min(price))
            })
            .0
    }
}