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 <= 105
0 <= prices[i] <= 104

Intuition

"taking maximum profit available

Each step we update minimum
Remembering the min along the way"

Links

https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/

Video Links

https://www.youtube.com/watch?v=excAOvwF_Wk&ab_channel=takeUforward

Approach 1:

DP Remembering the minimum along the way

C++

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int mini = prices[0];
        int profit = 0;

        for(int i=1; i<prices.size(); i++){
            /*
            taking maximum profit available
            */
            int cost = prices[i] - mini;
            profit = max(cost, profit);

            /*
            Each step we update minimum

            Remembering the min along the way
            */
            mini = min(mini, prices[i]);
        }

        return profit;
    }
};

Approach 2:

C++

Approach 3:

C++