121. Best Time to Buy and Sell Stock

You are given an array prices where prices[i] represents the price of a stock on the i-th day.

Your goal is to maximize profit by choosing exactly one day to buy and a different later day to sell the stock.

Return the maximum profit you can achieve from the transaction.
If no profit is possible, return 0.

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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: You cannot buy on day 2 and sell on day 1 because the selling day must come after the buying day.

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Example 2:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: The price always drops, so no transaction yields a profit.

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Constraints:

• 1 <= prices.length <= 10^5
• 0 <= prices[i] <= 10^4

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Instead of checking every possible buy-sell pair, which would be O(n^2) , we track two things in one pass:

1. The lowest price seen so far (minPrice)
2. The maximum profit we can make so far (maxProfit)

On each day:

• If the current price is lower than minPrice, update minPrice (this could be a better buy day).
• Otherwise, calculate price - minPrice (profit if sold today), and if it's larger than maxProfit, update maxProfit.

JavaScript Solution:

var maxProfit = function(prices) {
    let minPrice = Number.MAX_SAFE_INTEGER;
    let maxProfit = 0;
    
    for (let price of prices) {
        if (price < minPrice) {
            minPrice = price;
        } else if (price - minPrice > maxProfit) {
            maxProfit = price - minPrice;
        }
    }
    
    return maxProfit;
};

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TypeScript Solution:

function maxProfit(prices: number[]): number {
    let minPrice = Number.MAX_SAFE_INTEGER;
    let maxProfit = 0;
    
    for (let price of prices) {
        if (price < minPrice) {
            minPrice = price;
        } else if (price - minPrice > maxProfit) {
            maxProfit = price - minPrice;
        }
    }
    
    return maxProfit;
}