<p>You are given an 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>Find the maximum profit you can achieve. You may complete <strong>at most two transactions</strong>.</p>

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

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

<pre>
<strong>Input:</strong> prices = [3,3,5,0,0,3,1,4]
<strong>Output:</strong> 6
<strong>Explanation:</strong> Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.</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.
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.
</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> In this case, no transaction is done, i.e. max profit = 0.
</pre>

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

<ul>
	<li><code>1 &lt;= prices.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= prices[i] &lt;= 10<sup>5</sup></code></li>
</ul>

static int __star = []{
    ios_base::sync_with_stdio(0);
    cin.tie(NULL),cout.tie(NULL);
    return 0;
}();
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        int k =2;

        // Solve special cases
        if (n <= 0 || k <= 0) {
            return 0;
        }

        if (k * 2 >= n) {
            int res = 0;
            for (int i = 1; i < n; i++) {
                res += max(0, prices[i] - prices[i - 1]);
            }
            return res;
        }

        int res = 0;
        vector<vector<vector<int>>> dp(
            n, vector<vector<int>>(k+1, vector<int>(2,0))
        );

        // initialize the array with -inf
        // we use INT_MIN/2 here to prevent overflow
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= k; j++) {
                dp[i][j][0] = INT_MIN / 2;
                dp[i][j][1] = INT_MIN / 2;
            }
        }

        // set starting value
        dp[0][0][0] = 0;
        dp[0][1][1] = -prices[0];
        for(int i=1;i<n;i++){
            for(int j=0;j<=k;j++){
                dp[i][j][0] = max(dp[i-1][j][0], dp[i-1][j][1]+prices[i]);
                if(j>0){
                    dp[i][j][1] = max(dp[i-1][j][1], dp[i-1][j-1][0]-prices[i]);
                }
            }
        }

        for (int j = 0; j <= k; j++) {
            res = max(res, dp[n - 1][j][0]);
        }

        return res;
    }
};
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