121. Best Time to Buy and Sell Stock

Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Note that you cannot sell a stock before you buy one.
Example 1:
Input: [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.
             Not 7-1 = 6, as selling price needs to be larger than buying price.
Example 2:
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.





https://www.youtube.com/watch?v=8pVhUpF1INw
https://www.youtube.com/watch?v=76-CYD0jn7s

// time n^2 , space 1 
class Solution {
    public int maxProfit(int[] prices) {
      int max = 0;
      for(int i = 0; i < prices.length - 1; i++){
        for(int j = i + 1; j< prices.length; j++){
          int profit = prices[j] - prices[i];
          max = Math.max(profit, max);
        }
      }
      return max;
    }
}



// time n, space 1 
// keep a min price met so far, every time we see a new price., update min price if new price is smaller  , otherwise  check the diff, update the max profit if the new diff is bigger 

class Solution {
    public int maxProfit(int[] prices) {
      int max = 0; // max profit 
      int min = Integer.MAX_VALUE; // min_price so far 
      
      for(int i = 0; i < prices.length; i++){
        min = Math.min(min, prices[i]);
        if(prices[i] > min){
          int diff = prices[i] - min;
          max = Math.max(max, diff);
        }
      }
      return max;
    }
}