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121. Best Time to Buy and Sell Stock 
Problem Statement
You are given an array prices where prices[i] is the price of a given stock on the ith 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 <= 1050 <= prices[i] <= 104
Solution:
rs
impl Solution {
pub fn max_profit(prices: Vec<i32>) -> i32 {
let (mut profit, mut min) = (0, prices[0]);
// loop through the vector, starting at 1
for i in 1..prices.len() {
// if the current value is less than the minimum value, set the minimum value to the current value
if prices[i] < min {
min = prices[i];
}
// else if the current value minus the minimum value is greater than the current profit, set the profit to the current value minus the minimum value
else if prices[i] - min > profit {
profit = prices[i] - min;
}
}
profit
}
}
go
package main
func maxProfit(prices []int) int {
profit, min := 0, prices[0]
for _, price := range prices {
// if the current price is lower than the min price, update the min price to the current price
// else if the current price minus the min price is greater than the profit, update the profit to the current price minus the min price
if price < min {
min = price
} else if price-min > profit {
profit = price - min
}
}
return profit
}
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