<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(int ind, int trans, int n, vector<int>& prices,
              vector<vector<int>>& dp) {
        if (ind == n || trans == 0)
            return 0;
        if (dp[ind][trans] != -1)
            return dp[ind][trans];
        if (trans % 2 == 0) {
            int take = -prices[ind] + Solve(ind + 1, trans - 1, n, prices, dp);
            int nottake = Solve(ind + 1, trans, n, prices, dp);
            return dp[ind][trans] = max(take, nottake);
        } else {
            int take = prices[ind] + Solve(ind + 1, trans - 1, n, prices, dp);
            int nottake = Solve(ind + 1, trans, n, prices, dp);
            return dp[ind][trans] = max(take, nottake);
        }
    }

    int maxProfit(int k, vector<int>& prices) {
        int n = prices.size();
        // vector<vector<int>> dp(n+1, vector<int>(201, -1));
        // return Solve(0, 2 * k, n, prices, dp);

        //tabulation method
        vector<vector<int>> dp(n+1, vector<int>(201, 0));
        vector<int> curr(2*k+1,0),after(2*k+1,0);
        for (int ind = n - 1; ind >= 0; ind--) {
            for (int trans = 1; trans <= 2 * k; trans++) {
                if (trans % 2 == 0) {
                    int take =
                        -prices[ind] + after[trans-1];
                    int nottake = after[trans];
                     curr[trans] = max(take, nottake);
                } else {
                    int take =
                        prices[ind] + after[trans-1];
                    int nottake = after[trans];
                     curr[trans] = max(take, nottake);
                }
            }
            after = curr;
        }

        return after[2*k];
    }
};
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