<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>

// https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/

class Solution {

public:

    int f(const vector<int>& prices, int N, int i, int k) {

        // base case

        if (i == N || k == 0) {
            return 0;
        }

        // recursive case

        // f(i, k) = find the maximum profit you can make by completing <=k transactions starting from the ith day

        // make a decision for the ith day

        // 1. do not buy the stock on the ith day

        int donotBuyOption = f(prices, N, i + 1, k);

        // 2. buy the stock on the ith day

        // decide when to sell the stock bought on the ith day

        int maxBuyOptionSoFar = 0;

        for (int j = i + 1; j <= N - 1; j++ ) {

            maxBuyOptionSoFar = max(maxBuyOptionSoFar,
                                    (prices[j] - prices[i]) + f(prices, N, j + 1, k - 1));

        }

        return max(donotBuyOption, maxBuyOptionSoFar);

    }


    int fTopDown(const vector<int>& prices, int N, int i, int k, vector<vector<int>>& dp) {

        // lookup

        if (dp[i][k] != -1) {
            return dp[i][k];
        }

        // base case

        if (i == N || k == 0) {
            return dp[i][k] = 0;
        }

        // recursive case

        // f(i, k) = find the maximum profit you can make by completing <=k transactions starting from the ith day

        // make a decision for the ith day

        // 1. do not buy the stock on the ith day

        int donotBuyOption = fTopDown(prices, N, i + 1, k, dp);

        // 2. buy the stock on the ith day

        // decide when to sell the stock bought on the ith day

        int maxBuyOptionSoFar = 0;

        for (int j = i + 1; j <= N - 1; j++ ) {

            maxBuyOptionSoFar = max(maxBuyOptionSoFar,
                                    (prices[j] - prices[i]) + fTopDown(prices, N, j + 1, k - 1, dp));

        }

        return dp[i][k] = max(donotBuyOption, maxBuyOptionSoFar);

    }


    int maxProfit(int K, vector<int>& prices) {

        int N = prices.size();

        // return f(prices, N, 0, K);

        vector<vector<int>> dp(N + 1, vector<int>(K + 1, -1));

        return fTopDown(prices, N, 0, K, dp);

    }
};
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