# Time:  O(1)
# Space: O(1)

# There are two types of soup: type A and type B.
# Initially we have N ml of each type of soup. There are four kinds of operations:
#
# Serve 100 ml of soup A and 0 ml of soup B
# Serve 75 ml of soup A and 25 ml of soup B
# Serve 50 ml of soup A and 50 ml of soup B
# Serve 25 ml of soup A and 75 ml of soup B
# 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 equal probability 0.25.
# If the remaining volume of soup is not enough to complete the operation,
# we will serve as much as we can.
# We stop once we no longer have some quantity of both types of soup.
#
# Note that we do not have the operation where all 100 ml's of soup B are used first.
#
# Return the probability that soup A will be empty first,
# plus half the probability that A and B become empty at the same time.
#
# Example:
# Input: N = 50
# Output: 0.625
# Explanation:
# 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.
#
# Notes:
# 0 <= N <= 10^9.
# Answers within 10^-6 of the true value will be accepted as correct.


class Solution(object):
    def soupServings(self, N):
        """
        :type N: int
        :rtype: float
        """
        def dp(a, b, lookup):
            if (a, b) in lookup:
                return lookup[a, b]
            if a <= 0 and b <= 0:
                return 0.5
            if a <= 0:
                return 1.0
            if b <= 0:
                return 0.0
            lookup[a, b] = 0.25 * (dp(a-4, b, lookup) +
                                   dp(a-3, b-1, lookup) +
                                   dp(a-2, b-2, lookup) +
                                   dp(a-1, b-3, lookup))
            return lookup[a, b]

        if N >= 5000:
            return 1.0
        lookup = {}
        N = (N-1)//25 + 1
        return dp(N, N, lookup)

# Time:  O(1)
# Space: O(1)

# There are two types of soup: type A and type B.
# Initially we have N ml of each type of soup. There are four kinds of operations:
# 1. Serve 100 ml of soup A and 0 ml of soup B
# 2. Serve 75 ml of soup A and 25 ml of soup B
# 3. Serve 50 ml of soup A and 50 ml of soup B
# 4. Serve 25 ml of soup A and 75 ml of soup B
#
# 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 equal probability 0.25.
# If the remaining volume of soup is not enough to complete the operation,
# we will serve as much as we can.
# We stop once we no longer have some quantity of both types of soup.
#
# Note that we do not have the operation where all 100 ml's of soup B are used first.
#
# Return the probability that soup A will be empty first,
# plus half the probability that A and B become empty at the same time.
#
# Example:
# Input: N = 50
# Output: 0.625
# Explanation:
# 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.
#
# Notes:
# 0 <= N <= 10^9.
# Answers within 10^-6 of the true value will be accepted as correct.


class Solution(object):
    def soupServings(self, N):
        """
        :type N: int
        :rtype: float
        """
        def dp(a, b, lookup):
            if (a, b) in lookup:
                return lookup[a, b]
            if a <= 0 and b <= 0:
                return 0.5
            if a <= 0:
                return 1.0
            if b <= 0:
                return 0.0
            lookup[a, b] = 0.25 * (dp(a-4, b, lookup) +
                                   dp(a-3, b-1, lookup) +
                                   dp(a-2, b-2, lookup) +
                                   dp(a-1, b-3, lookup))
            return lookup[a, b]

        if N >= 5000:
            return 1.0
        lookup = {}
        N = (N-1)//25 + 1
        return dp(N, N, lookup)

Update soup-servings.py