valid-permutations-for-di-sequence.py

# Time:  O(n^2)
# Space: O(n)

# We are given S, a length n string of characters from the set {'D', 'I'}.
# (These letters stand for "decreasing" and "increasing".)
#
# A valid permutation is a permutation P[0], P[1], ..., P[n]
# of integers {0, 1, ..., n}, such that for all i:
#
# If S[i] == 'D', then P[i] > P[i+1], and;
# If S[i] == 'I', then P[i] < P[i+1].
# How many valid permutations are there?
# Since the answer may be large, return your answer modulo 10^9 + 7.
#
# Example 1:
#
# Input: "DID"
# Output: 5
# Explanation: 
# The 5 valid permutations of (0, 1, 2, 3) are:
# (1, 0, 3, 2)
# (2, 0, 3, 1)
# (2, 1, 3, 0)
# (3, 0, 2, 1)
# (3, 1, 2, 0)
#
# Note:
# - 1 <= S.length <= 200
# - S consists only of characters from the set {'D', 'I'}.


class Solution(object):
    def numPermsDISequence(self, S):
        """
        :type S: str
        :rtype: int
        """
        dp = [1]*(len(S)+1)
        for c in S:
            if c == "I":
                dp = dp[:-1]
                for i in xrange(1, len(dp)):
                    dp[i] += dp[i-1]
            else:
                dp = dp[1:]
                for i in reversed(xrange(len(dp)-1)):
                    dp[i] += dp[i+1]
        return dp[0] % (10**9+7)