122. Best Time to Buy and Sell Stock II

Problem statement:

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

Solution:

Compare with 121. Best Time to Buy and Sell Stock, this problem allows multiple transactions. Instead of only finding the biggest difference, we should find all local lowest value and the highest value and add their difference in the final answer.

Keep a min price and a max price, when a price is lower than the max price, update the max profit along with min and max prices.

Time complexity is also 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] >= max_val){
max_val = prices[i];
} else {
max_profit += max_val - min_val;
max_val = prices[i];
min_val = prices[i];
}
}
max_profit += max_val - min_val;
return max_profit;
}
};

A more concise version from Leetcode discussion part is as follows:

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