Best Time to Buy and Sell Stock – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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 <= 105
• 0 <= prices[i] <= 104

C++ Best Time to Buy and Sell StockLeetCode Solution

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int lsf = INT_MAX;
        int op = 0;
        int pist = 0;
        
        for(int i = 0; i < prices.size(); i++){
            if(prices[i] < lsf){
                lsf = prices[i];
            }
            pist = prices[i] - lsf;
            if(op < pist){
                op = pist;
            }
        }
        return op;
    }
};

Java Best Time to Buy and Sell StockLeetCode Solution

class Solution {
    public int maxProfit(int[] prices) {
        int lsf = Integer.MAX_VALUE;
        int op = 0;
        int pist = 0;
        
        for(int i = 0; i < prices.length; i++){
            if(prices[i] < lsf){
                lsf = prices[i];
            }
            pist = prices[i] - lsf;
            if(op < pist){
                op = pist;
            }
        }
        return op;
    }
}
 

Python 3 Best Time to Buy and Sell Stock LeetCode Solution

class Solution:
    def maxProfit(self,prices):
        left = 0 #Buy
        right = 1 #Sell
        max_profit = 0
        while right < len(prices):
            currentProfit = prices[right] - prices[left] #our current Profit
            if prices[left] < prices[right]:
                max_profit =max(currentProfit,max_profit)
            else:
                left = right
            right += 1
        return max_profit

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