Create valid-permutations-for-di-sequence.py

Python/valid-permutations-for-di-sequence.py

@@ -0,0 +1,48 @@
+# 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)