Daily Dose of DSA - Day 12

//Problem link:https://leetcode.com/problems/best-time-to-buy-and-sell-stock/
//Everyone knows
class Solution 
{
public:
    int maxProfit(vector<int>& prices) //naive solution 
    {

        int diff;
        int profit=0;
        for(int i=0;i<prices.size();i++)
        {
            for(int j=i+1;j<prices.size();j++)
            {
                diff = prices[j]-prices[i];
                profit=max(profit,diff);
            }
        }
        return profit;

    }
};
/*In this the intuition is that for a particular price in array the maximum profit will be that (element - min element at its left). 
so while traversing we will maintain the minimum element at the left.*/
class Solution {
public:
    int maxProfit(vector<int>& prices) {
    int res = 0;
    int minleft=INT_MAX;
    for(int i=0;i<prices.size();i++)
    {
        minleft=min(minleft,prices[i]);
        res=max(res,prices[i]-minleft);
    }
    return res;
}
};