# ---
# title: 122. Best Time to Buy and Sell Stock II
# id: problem122
# author: AquaIndigo
# date: 2020-11-08
# difficulty: Easy
# categories: Array, Greedy
# link: <https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/>
# hidden: true
# ---
# 
# Say you have an array `prices` for which the _i_ th 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 (i.e., buy one and sell one share of the stock
# multiple times).
# 
# **Note:** You may not engage in multiple transactions at the same time (i.e.,
# you must sell the stock before you buy again).
# 
# **Example 1:**
# 
#     
#     
#     Input: [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: [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: [7,6,4,3,1]
#     Output: 0
#     Explanation: In this case, no transaction is done, i.e. max profit = 0.
# 
# 
# 
# **Constraints:**
# 
#   * `1 <= prices.length <= 3 * 10 ^ 4`
#   * `0 <= prices[i] <= 10 ^ 4`
# 
# 
## @lc code=start
using LeetCode

function max_profit(prices::Vector{Int})::Int
    res = 0
    for i in 2:length(prices)
        res += (prices[i] > prices[i - 1]) ? (prices[i] - prices[i - 1]) : 0
    end
    return res
end
## @lc code=end
