main.cpp

// Source: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii
// Title: Best Time to Buy and Sell Stock II
// Difficulty: Medium
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// You are given an integer array `prices` where `prices[i]` is the price of a given stock on the `i^th` day.
//
// On each day, you may decide to buy and/or sell the stock. You can only hold **at most one** share of the stock at any time. However, you can sell and buy the stock multiple times on the **same day**, ensuring you never hold more than one share of the stock.
//
// Find and return the **maximum** profit you can achieve.
//
// **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.
// Total profit is 4 + 3 = 7.
// ```
//
// **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.
// Total profit is 4.
// ```
//
// **Example 3:**
//
// ```
// Input: prices = [7,6,4,3,1]
// Output: 0
// Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.
// ```
//
// **Constraints:**
//
// - `1 <= prices.length <= 3 * 10^4`
// - `0 <= prices[i] <= 10^4`
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

#include <algorithm>
#include <climits>
#include <vector>

using namespace std;

// DP
//
// Flat[i] be the maximum profit without holding a stock.
// Hold[i] be the maximum profit with holding a stock.
class Solution {
 public:
  int maxProfit(const vector<int>& prices) {
    int flat = 0;
    int hold = INT_MIN;  // impossible at start

    for (int price : prices) {
      int nextFlat = max(flat, hold + price);
      int nextHold = max(hold, flat - price);
      flat = nextFlat, hold = nextHold;
    }

    return flat;
  }
};