<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 {
    int maxProfitRec(vector<int> &prices, int index, int buy, int maxTransactions){
        if(index == prices.size() || maxTransactions == 0) return 0;

        if(buy){
            return max(maxProfitRec(prices, index + 1, !buy, maxTransactions) - prices[index],
                       maxProfitRec(prices, index +1, buy, maxTransactions));
        }

        return max(maxProfitRec(prices, index + 1, !buy, maxTransactions - 1) + prices[index],
                   maxProfitRec(prices, index +1, buy, maxTransactions));
    }

    int maxProfitMemo(vector<int> &prices, int index, int buy, int maxTransactions, vector<vector<int>> &memo){
        if(index == prices.size() || maxTransactions == 0) return 0;

        if(memo[index][buy] != -1) return memo[index][buy];

        if(buy){
            return memo[index][buy] = max(maxProfitRec(prices, index + 1, !buy, maxTransactions) - prices[index],
                       maxProfitRec(prices, index +1, buy, maxTransactions));
        }

        return memo[index][buy] = max(maxProfitRec(prices, index + 1, !buy, maxTransactions - 1) + prices[index],
                   maxProfitRec(prices, index +1, buy, maxTransactions));
    }

    int maxProfitDP(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)));
        // At last dp[0][0][2] ( maximum profit generated on ith day, when we can buy the stock on 0th day 
        // and can have a total 2 transactions) gives us the final answer.

        // why 0? because of base case where if index == n or cap == 0 return 0;

        for(int i=n-1; i>=0; i--){
            for(int buy=0; buy<=1; buy++){
                for(int maxTransactions = 1; maxTransactions <=k; maxTransactions++){
                    if(buy){
                        dp[i][buy][maxTransactions] = max(dp[i+1][!buy][maxTransactions] - prices[i], 
                                                            dp[i+1][buy][maxTransactions]);
                    } else {
                        dp[i][buy][maxTransactions] = max(dp[i+1][!buy][maxTransactions-1] + prices[i], 
                                                            dp[i+1][buy][maxTransactions]);
                    }
                } 
            }
        }

        return dp[0][1][k];
    }

    int maxProfitDPSpaceOptimised(int k, vector<int> &prices){
        int n = prices.size();

        // vector<vector<vector<int>>> dp(n+1, vector<vector<int>> (2, vector<int> (3,0)));
        vector<vector<int>> nextRow(2, vector<int> (k+1,0));
        vector<vector<int>> currentRow(2, vector<int> (k+1,0));

        for(int i=n-1; i>=0; i--){
            for(int buy=0; buy<=1; buy++){
                for(int maxTransactions = 1; maxTransactions <=k; maxTransactions++){
                    if(buy){
                        currentRow[buy][maxTransactions] = max(nextRow[!buy][maxTransactions] - prices[i], 
                                                            nextRow[buy][maxTransactions]);
                    } else {
                        currentRow[buy][maxTransactions] = max(nextRow[!buy][maxTransactions-1] + prices[i], 
                                                            nextRow[buy][maxTransactions]);
                    }
                } 
            }
            nextRow = currentRow;
        }

        return nextRow[1][k];
    }
public:
    int maxProfit(int k, vector<int>& prices) {
        int n = prices.size();
        // return maxProfitRec(prices, 0, 1, k);

        // vector<vector<int>> memo(n, vector<int> (2,-1));
        // return maxProfitMemo(prices, 0, 1, k, memo);

        // return maxProfitDP(k, prices);
        return maxProfitDPSpaceOptimised(k, prices);
    }
};
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