<p>You are given an integer array <code>prices</code> where <code>prices[i]</code> is the price of a given stock on the <code>i<sup>th</sup></code> day.</p>

<p>On each day, you may decide to buy and/or sell the stock. You can only hold <strong>at most one</strong> share of the stock at any time. However, you can buy it then immediately sell it on the <strong>same day</strong>.</p>

<p>Find and return <em>the <strong>maximum</strong> profit you can achieve</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> prices = [7,1,5,3,6,4]
<strong>Output:</strong> 7
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> prices = [1,2,3,4,5]
<strong>Output:</strong> 4
<strong>Explanation:</strong> Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Total profit is 4.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> prices = [7,6,4,3,1]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= prices.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= prices[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int helper(vector<int>& prices, int idx, bool have_it, vector<pair<int,int>>& dp){
        if(idx == prices.size()){
            return 0;
        }
        if(have_it){
            if(dp[idx].second != -1){
                return dp[idx].second;
            }
            // 3 options:
            // 1st option: do nothing:
            int res1 = helper(prices, idx+1, true, dp);
            // 2nd option: sell it:
            int res2 = prices[idx] + helper(prices, idx+1, false, dp);
            // 3rd option: sell it and buy it right away:
            int res3 = helper(prices, idx+1, true, dp);
            // best option:
            int best_deal = max(max(res1,res2), max(res2,res3));
            dp[idx].second = best_deal;
            return best_deal;
        }
        else{
            if(dp[idx].first != -1){
                return dp[idx].first;
            }
            // 2 options:
            // 1st option: Do nothing:
            int res1 = helper(prices, idx+1, false, dp);
            // 2nd option: Buy it:
            int res2 = helper(prices, idx+1, true, dp) - prices[idx];
            // best option:
            int best_deal = max(res1, res2);
            dp[idx].first = best_deal;
            return best_deal;
        }
    }
    int maxProfit(vector<int>& prices) {
        vector<pair<int,int>> dp(prices.size());
        for(int d=0; d<prices.size(); ++d){
            dp[d].first = -1;
            dp[d].second = -1;
        }
        return helper(prices, 0, 0, dp);
    }
};
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