121. Best Time to Buy and Sell Stock

Problem Description

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.

Examples

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.

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if not prices:
            return 0
            
        min_price = prices[0]
        max_profit = 0
        
        for price in prices[1:]:
            if price < min_price:
                min_price = price
            else:
                max_profit = max(max_profit, price - min_price)
                
        return max_profit

class Solution {
    public int maxProfit(int[] prices) {
        if (prices == null || prices.length < 2)
            return 0;
            
        int minPrice = prices[0];
        int maxProfit = 0;
        
        for (int i = 1; i < prices.length; i++) {
            if (prices[i] < minPrice)
                minPrice = prices[i];
            else
                maxProfit = Math.max(maxProfit, prices[i] - minPrice);
        }
        
        return maxProfit;
    }
}

class Solution {
public:
    int maxProfit(vector& prices) {
        if (prices.empty())
            return 0;
            
        int minPrice = prices[0];
        int maxProfit = 0;
        
        for (int i = 1; i < prices.size(); i++) {
            if (prices[i] < minPrice)
                minPrice = prices[i];
            else
                maxProfit = max(maxProfit, prices[i] - minPrice);
        }
        
        return maxProfit;
    }
};

/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function(prices) {
    if (!prices.length)
        return 0;
        
    let minPrice = prices[0];
    let maxProfit = 0;
    
    for (let i = 1; i < prices.length; i++) {
        if (prices[i] < minPrice)
            minPrice = prices[i];
        else
            maxProfit = Math.max(maxProfit, prices[i] - minPrice);
    }
    
    return maxProfit;
};

public class Solution {
    public int MaxProfit(int[] prices) {
        if (prices == null || prices.Length < 2)
            return 0;
            
        int minPrice = prices[0];
        int maxProfit = 0;
        
        for (int i = 1; i < prices.Length; i++) {
            if (prices[i] < minPrice)
                minPrice = prices[i];
            else
                maxProfit = Math.Max(maxProfit, prices[i] - minPrice);
        }
        
        return maxProfit;
    }
}