Day 1 Leet code series, today we will be picking the problem best time to buy and sell stocks (https://leetcode.com/problems/best-time-to-buy-and-sell-stock/).

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.

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

Explaination:

1. We will maintain a minimumSoFar to track for the minimum value.
2. ans will be the ans which will be returned in the last.
3. we will run a loop from 0 to prices.size() and calculate the profit.
4. If at anytime we find that prices[i] < minimumSoFar then we will make that as minimumSoFar
5. If curProfit is greater than ans then ans will be curProfit.