121. Best Time to Buy and Sell Stock

Problem statement:

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 (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:

Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 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

In this case, no transaction is done, i.e. max profit = 0.

Solution one:

This is array problem. The most simple solution is brute force with two level loop.

For a price, loop till the end to find any price which is higher than it and update the max profit. Otherwise, return 0. Time complexity is O(n * n)

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.size() < 2){
            return 0;
        }
        int min = prices[0];
        int max_price = 0;
        int max_profit = 0;
        for(size_t i=0; i<prices.size()-1; i++){
            max_price = prices[i];
            for(size_t j=i+1; j<prices.size(); j++){
                if( prices[j] > max_price){
                    max_price = prices[j];    
                }
            }
            if((max_price - prices[i]) > max_profit){
                max_profit = max_price - prices[i];
            }
        }
        return max_profit;
    }
};

Solution two: 

In order to maximize the profit, we need a deep thinking. Actually, this problem asks the biggest difference between two numbers with the limitation that the lower number should in the front of the higher number in the array.

We need to buy it at a low price and sell at high price. Keep two value: min price and max price. 

For an element which is less than min value, it means we already find a local down and peak value. update the max profit, min price and max price.

The time complexity is O(n)

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.empty()){
            return 0;
        }
        int max_val = prices[0];
        int min_val = prices[0];
        int max_profit = 0;
        for(int i = 1; i < prices.size(); i++){
            if(prices[i] < min_val){
                max_profit = max(max_profit, max_val - min_val);
                min_val = prices[i];
                max_val = prices[i];
            } else {
                max_val = max(max_val, prices[i]);
            }
        }
        max_profit = max(max_profit, max_val - min_val);
        return max_profit;
    }
};

The following code is a more concise version, but it is the same idea.

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int min_val = INT_MAX;
        int max_profit = 0;
        for(auto price : prices){
            min_val = min(min_val, price);
            max_profit = max(max_profit, price - min_val);
        }
        return max_profit;
    }
};