<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:
    // space opt
    int maxProfit(int k, vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>> after(2, vector<int>(k + 1, 0));
        vector<vector<int>> curr(2, vector<int>(k + 1, 0));
        int profit = 0;
        for (int i = n - 1; i >= 0; i--) {
            for (int buy = 0; buy <= 1; buy++) {
                for (int cap = 1; cap <= k; cap++) {
                    if (buy) {
                        profit = max(-prices[i] + after[0][cap], after[1][cap]);
                    } else {
                        profit =
                            max(prices[i] + after[1][cap - 1], after[0][cap]);
                    }
                    curr[buy][cap] = profit;
                }
            }
            after = curr;
        }
        return after[1][k];
    }

    // tabulation
    // 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, 0)));
    //     int profit = 0;
    //     for (int i = n - 1; i >= 0; i--) {
    //         for (int buy = 0; buy <= 1; buy++) {
    //             for (int cap = 1; cap <= k; cap++) {
    //                 if (buy) {
    //                     profit = max(-prices[i] + dp[i + 1][0][cap],
    //                                  dp[i + 1][1][cap]);
    //                 } else {
    //                     profit = max(prices[i] + dp[i + 1][1][cap - 1],
    //                                  dp[i + 1][0][cap]);
    //                 }
    //                 dp[i][buy][cap] = profit;
    //             }
    //         }
    //     }
    //     return dp[0][1][k];
    // }

    // memo
    // int rec(vector<int>& prices, int n, int i, int buy,
    //         vector<vector<vector<int>>>& dp, int cap) {
    //     if (i == n || cap == 0) {
    //         return 0;
    //     }
    //     if (dp[i][buy][cap] != -1) {
    //         return dp[i][buy][cap];
    //     }
    //     int profit = 0;
    //     if (buy) {
    //         profit = max(-prices[i] + rec(prices, n, i + 1, 0, dp, cap),
    //                      rec(prices, n, i + 1, 1, dp, cap));
    //     } else {
    //         profit = max(prices[i] + rec(prices, n, i + 1, 1, dp, cap - 1),
    //                      rec(prices, n, i + 1, 0, dp, cap));
    //     }
    //     return dp[i][buy][cap] = profit;
    // }
    // int maxProfit(int k, vector<int>& prices) {
    //     int n = prices.size();
    //     vector<vector<vector<int>>> dp(
    //         n, vector<vector<int>>(2, vector<int>(k + 1, -1)));
    //     return rec(prices, n, 0, 1, dp, k);
    // }

    // recursion
    // int rec(vector<int>& prices, int n, int i, int buy, int cap) {
    //     if (i == n || cap == 0) {
    //         return 0;
    //     }
    //     int profit = 0;
    //     if (buy) {
    //         profit = max(-prices[i] + rec(prices, n, i + 1, 0, cap),
    //                      rec(prices, n, i + 1, 1, cap));
    //     } else {
    //         profit = max(prices[i] + rec(prices, n, i + 1, 1, cap - 1),
    //                      rec(prices, n, i + 1, 0, cap));
    //     }
    //     return profit;
    // }
    // int maxProfit(int k, vector<int>& prices) {
    //     int n = prices.size();
    //     return rec(prices, n, 0, 1, k);
    // }
};
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