<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, and an integer <code>k</code>.</p>

<p>Find the maximum profit you can achieve. You may complete at most <code>k</code> transactions: i.e. you may buy at most <code>k</code> times and sell at most <code>k</code> times.</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> k = 2, prices = [2,4,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.
</pre>

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

<pre>
<strong>Input:</strong> k = 2, prices = [3,2,6,5,0,3]
<strong>Output:</strong> 7
<strong>Explanation:</strong> Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
</pre>

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

<ul>
	<li><code>1 &lt;= k &lt;= 100</code></li>
	<li><code>1 &lt;= prices.length &lt;= 1000</code></li>
	<li><code>0 &lt;= prices[i] &lt;= 1000</code></li>
</ul>

class Solution {
public:
    int solve(vector<int> &prices, int n, bool bought, int k, vector<vector<vector<int>>> &dp) {
        if(n == prices.size() || k == 0)
            return 0;
        
        if(dp[n][bought][k] != -1)
            return dp[n][bought][k];

        int buyOrSell;
        if(bought)
            buyOrSell = prices[n] + solve(prices, n + 1, false, k - 1, dp);
        else 
            buyOrSell = -prices[n] + solve(prices, n + 1, true, k, dp);
        int skip = solve(prices, n + 1, bought, k, dp);
        return dp[n][bought][k] = max(buyOrSell, skip);
    }
    int solveTab(vector<int> &prices, int c) {
        int n = prices.size();
        vector<vector<vector<int>>> dp(n + 1, vector<vector<int>>(2, vector<int>(c + 1, 0)));

        for(int i = n - 1; i >= 0; i--) {
            for(int j = 1; j >= 0; j--) {
                for(int k = 1; k <= c; k++) {
                    int buyOrSell;
                    if(j)
                        buyOrSell = prices[i] + dp[i + 1][false][k - 1];
                    else 
                        buyOrSell = -prices[i] + dp[i + 1][true][k];
                    int skip = dp[i + 1][j][k];
                    dp[i][j][k] = max(buyOrSell, skip);
                }
            }
        }
        return dp[0][0][c];
    }
    int maxProfit(int k, vector<int>& prices) {
        int n = prices.size();
        vector<vector<vector<int>>> dp(n + 1, vector<vector<int>>(2, vector<int>(k + 1, -1)));
        // return solve(prices, 0, false, k, dp);

        return solveTab(prices, k);
    }
};
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