<p>There are two types of soup: <strong>type A</strong> and <strong>type B</strong>. Initially, we have <code>n</code> ml of each type of soup. There are four kinds of operations:</p>

<ol>
	<li>Serve <code>100</code> ml of <strong>soup A</strong> and <code>0</code> ml of <strong>soup B</strong>,</li>
	<li>Serve <code>75</code> ml of <strong>soup A</strong> and <code>25</code> ml of <strong>soup B</strong>,</li>
	<li>Serve <code>50</code> ml of <strong>soup A</strong> and <code>50</code> ml of <strong>soup B</strong>, and</li>
	<li>Serve <code>25</code> ml of <strong>soup A</strong> and <code>75</code> ml of <strong>soup B</strong>.</li>
</ol>

<p>When we serve some soup, we give it to someone, and we no longer have it. Each turn, we will choose from the four operations with an equal probability <code>0.25</code>. If the remaining volume of soup is not enough to complete the operation, we will serve as much as possible. We stop once we no longer have some quantity of both types of soup.</p>

<p><strong>Note</strong> that we do not have an operation where all <code>100</code> ml&#39;s of <strong>soup B</strong> are used first.</p>

<p>Return <em>the probability that <strong>soup A</strong> will be empty first, plus half the probability that <strong>A</strong> and <strong>B</strong> become empty at the same time</em>. Answers within <code>10<sup>-5</sup></code> of the actual answer will be accepted.</p>

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

<pre>
<strong>Input:</strong> n = 50
<strong>Output:</strong> 0.62500
<strong>Explanation:</strong> If we choose the first two operations, A will become empty first.
For the third operation, A and B will become empty at the same time.
For the fourth operation, B will become empty first.
So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.25 * (1 + 1 + 0.5 + 0) = 0.625.
</pre>

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

<pre>
<strong>Input:</strong> n = 100
<strong>Output:</strong> 0.71875
</pre>

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

<ul>
	<li><code>0 &lt;= n &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def soupServings(self, n: int) -> float:
        # track the prob of e1 and e2 given current state
        probs = {} # <state, [prob_e1, prob_e2]>

        def serve(amount, dose):
            remain = [0, 0]
            if amount[0] > dose[0]:
                remain[0] = amount[0] - dose[0]
            if amount[1] > dose[1]:
                remain[1] = amount[1] - dose[1]

            return remain[0], remain[1]

        def search(n_a, n_b):

            # print(f'search: {n_a} {n_b}')
            # return two probs: prob of A is empty, prob of both are empty
            # prob: probability of getting into this state
            if n_a == 0 and n_b > 0:
                probs[(n_a, n_b)] = [1.0, 0.0]
                return 1.0, 0.0
            elif n_a == 0 and n_b == 0:
                probs[(n_a, n_b)] = [0.0, 1.0]
                return 0.0, 1.0
            elif n_a > 0 and n_b == 0:
                probs[(n_a, n_b)] = [0.0, 0.0]
                return 0.0, 0.0
            else:  # none is empty
                operations = [[100, 0], [75, 25], [50, 50], [25, 75]]
                prob1, prob2 = 0, 0 # probs of two events given current state

                for op in operations:
                    na1, nb1 = serve([n_a, n_b], op)
                    if (na1, nb1) in probs:
                        p1, p2 = probs[(na1, nb1)]
                    else:
                        p1, p2 = search(na1, nb1)
                    prob1 += 0.25*p1
                    prob2 += 0.25*p2
                probs[(n_a, n_b)] = [prob1, prob2]

                return prob1, prob2
        if n > 5000:
            return 1.0
            
        prob1, prob2 = search(n, n)
        return prob1 + 0.5 * prob2

20900