<p>You are given two integer arrays <code>energyDrinkA</code> and <code>energyDrinkB</code> of the same length <code>n</code> by a futuristic sports scientist. These arrays represent the energy boosts per hour provided by two different energy drinks, A and B, respectively.</p>

<p>You want to <em>maximize</em> your total energy boost by drinking one energy drink <em>per hour</em>. However, if you want to switch from consuming one energy drink to the other, you need to wait for <em>one hour</em> to cleanse your system (meaning you won&#39;t get any energy boost in that hour).</p>

<p>Return the <strong>maximum</strong> total energy boost you can gain in the next <code>n</code> hours.</p>

<p><strong>Note</strong> that you can start consuming <em>either</em> of the two energy drinks.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> energyDrinkA<span class="example-io"> = [1,3,1], </span>energyDrinkB<span class="example-io"> = [3,1,1]</span></p>

<p><strong>Output:</strong> <span class="example-io">5</span></p>

<p><strong>Explanation:</strong></p>

<p>To gain an energy boost of 5, drink only the energy drink A (or only B).</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> energyDrinkA<span class="example-io"> = [4,1,1], </span>energyDrinkB<span class="example-io"> = [1,1,3]</span></p>

<p><strong>Output:</strong> <span class="example-io">7</span></p>

<p><strong>Explanation:</strong></p>

<p>To gain an energy boost of 7:</p>

<ul>
	<li>Drink the energy drink A for the first hour.</li>
	<li>Switch to the energy drink B and we lose the energy boost of the second hour.</li>
	<li>Gain the energy boost of the drink B in the third hour.</li>
</ul>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == energyDrinkA.length == energyDrinkB.length</code></li>
	<li><code>3 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= energyDrinkA[i], energyDrinkB[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    dpA = []
    dpB = []
    def DP(self, A, B, flag, hour):
        # print(flag,hour)
        if hour == len(A)-1:
            val = A[hour] if flag == 'A' else B[hour]
            # print(val)
            return val
        if hour == len(A):
            val = B[hour-1] if flag == 'A' else A[hour-1]
            # print(val)
            return val
        if flag == 'A':
            if self.dpA[hour] != -1:
                return self.dpA[hour]
            self.dpA[hour] = A[hour]+max(self.DP(A,B,'A',hour+1),self.DP(A,B,'B',hour+2))
            return self.dpA[hour]
        else:
            if self.dpB[hour] != -1:
                return self.dpB[hour]
            self.dpB[hour] = B[hour]+max(self.DP(A,B,'A',hour+2),self.DP(A,B,'B',hour+1))
            return self.dpB[hour]
    def maxEnergyBoost(self, energyDrinkA: List[int], energyDrinkB: List[int]) -> int:
        self.dpA = [-1 for _ in range(len(energyDrinkA)+2)]
        self.dpB = [-1 for _ in range(len(energyDrinkB)+2)]
        return max(self.DP(energyDrinkA, energyDrinkB, 'A', 0,), self.DP(energyDrinkA, energyDrinkB, 'B', 0))
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