# 122. Best Time to Buy and Sell Stock II

TAGS: Leetcode easy, C++

You are given an array `prices`

where `prices[i]`

is the price of a given stock on the `ith`

day.

Find the maximum profit you can achieve. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

**Note:** You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

**Example 1:**

**Input:** prices = [7,1,5,3,6,4]

**Output:** 7

**Explanation:** Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.

Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.

**Example 2:**

**Input:** prices = [1,2,3,4,5]

**Output:** 4

**Explanation:** Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.

Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.

**Example 3:**

**Input:** prices = [7,6,4,3,1]

**Output:** 0

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

`class Solution {`

public:

int maxProfit(vector<int>& prices) {

int len = prices.size();

int min_p = 0;

int max_p = 0;

int ans = 0;

for(int i=0; i<len; i++){

while(i+1<len && prices[i+1] < prices[i]){

i++;

}

min_p = prices[i];

while(i+1<len && prices[i+1] > prices[i]){

i++;

}

max_p = prices[i];

ans += max_p - min_p;

}

return ans;

}

};