Best Time To Buy and Sell Stock LeetCode Problem

 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.

Optimal Approach;


class Solution {
    public int maxProfit(int[] prices) {

 int maxPro = 0;
    int minPrice = Integer.MAX_VALUE;
    for (int i = 0; i < prices.length; i++) {
        minPrice = Math.min(minPrice, prices[i]);
        maxPro = Math.max(maxPro, prices[i] - minPrice);
    }
    return maxPro;
    }
}



BruteForce Approach 

But In this Approach You are geeting Time Limit Exceed Problem .

class Solution {
    public int maxProfit(int[] prices) {

int n=prices.length;

int maxi=0;
        for(int i=0;i<n;i++)
        {

            for(int j=i+1;j<n;j++)
            {
 if (prices[j] > prices[i]) {
                    maxi = Math.max(prices[j] - prices[i], maxi);
                }
            }
         
        }
       
        return maxi;
    }
}


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