<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:
    // Recursive + Memoization Solution: T.C = O(n), S.C = O(n)
    int solveMemo(int index, int isBuy, int transactions, vector<int>& prices, vector<vector<vector<int>>>& dp) {
        // Base Condition
        if ( index == prices.size() || transactions == 0 )
            return 0;

        // Return alredy calculated result
        if ( dp[index][isBuy][transactions] != -1 )
            return dp[index][isBuy][transactions];

        int profit = 0;
        // When we can buy or ignore the stock
        if ( isBuy )
            profit = max( -prices[index] + solveMemo(index+1, 0, transactions, prices, dp),
                            0 + solveMemo(index+1, 1, transactions, prices, dp) );
        // When we can sell or ignore the stock
        else 
            profit = max( +prices[index] + solveMemo(index+1, 1, transactions-1, prices, dp),
                            0 + solveMemo(index+1, 0, transactions, prices, dp) );
        
        return dp[index][isBuy][transactions] = profit;
    }

    // Tabulation Solution: T.C = O(n), S.C = O(n)
    int solveTabu(vector<int>& prices, int k, int n) {
        vector<vector<vector<int>>> dp(n+1,vector<vector<int>>(3,vector<int>(k+1,0)));

        for ( int index=n-1; index>=0; index-- )
        for ( int isBuy = 0; isBuy<=1; isBuy++ )
        for ( int limit = 1; limit<=k; limit++ ) {
            int profit = 0;
            // When we can buy or ignore the stock
            if ( isBuy )
                profit = max( -prices[index] + dp[index+1][0][limit], 0 + dp[index+1][1][limit] );
            // When we can sell or ignore the stock
            else 
                profit = max( +prices[index] + dp[index+1][1][limit-1], 0 + dp[index+1][0][limit]);
            
            dp[index][isBuy][limit] = profit;
        }

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

    int maxProfit(int k, vector<int>& prices) {
        int n = prices.size();
        // vector<vector<vector<int>>> dp(n+1,vector<vector<int>>(3,vector<int>(k+1,-1)));
        // return solveMemo(0, 1, k, prices, dp);
        return solveTabu(prices, k, n);
    }
};
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