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

# Consider the string s to be the infinite wraparound string of
# "abcdefghijklmnopqrstuvwxyz", so s will look like this:
# "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....".
#
# Now we have another string p. Your job is to find out
# how many unique non-empty substrings of p are present in s.
# In particular, your input is the string p and you need to output
# the number of different non-empty substrings of p in the string s.
#
# Note: p consists of only lowercase English letters and the size of p might be over 10000.
#
# Example 1:
# Input: "a"
# Output: 1
#
# Explanation: Only the substring "a" of string "a" is in the string s.
# Example 2:
# Input: "cac"
# Output: 2
# Explanation: There are two substrings "a", "c" of string "cac" in the string s.
# Example 3:
# Input: "zab"
# Output: 6
# Explanation: There are six substrings "z", "a", "b", "za", "ab", "zab" of string "zab" in the string s.

class Solution(object):
    def findSubstringInWraproundString(self, p):
        """
        :type p: str
        :rtype: int
        """
        letters = [0] * 26
        result, length = 0, 0
        for i in xrange(len(p)):
            curr = ord(p[i]) - ord('a')
            if i > 0 and ord(p[i-1]) != (curr-1)%26 + ord('a'):
                length = 0
            length += 1
            if length > letters[curr]:
                result += length - letters[curr]
                letters[curr] = length
        return result

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

# Consider the string s to be the infinite wraparound string of
# "abcdefghijklmnopqrstuvwxyz", so s will look like this:
# "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....".
#
# Now we have another string p. Your job is to find out
# how many unique non-empty substrings of p are present in s.
# In particular, your input is the string p and you need to output
# the number of different non-empty substrings of p in the string s.
#
# Note: p consists of only lowercase English letters and the size of p might be over 10000.
#
# Example 1:
# Input: "a"
# Output: 1
#
# Explanation: Only the substring "a" of string "a" is in the string s.
# Example 2:
# Input: "cac"
# Output: 2
# Explanation: There are two substrings "a", "c" of string "cac" in the string s.
# Example 3:
# Input: "zab"
# Output: 6
# Explanation: There are six substrings "z", "a", "b", "za", "ab", "zab" of string "zab" in the string s.

class Solution(object):
    def findSubstringInWraproundString(self, p):
        """
        :type p: str
        :rtype: int
        """
        letters = [0] * 26
        result, length = 0, 0
        for i in xrange(len(p)):
            curr = ord(p[i]) - ord('a')
            if i > 0 and ord(p[i-1]) != (curr-1)%26 + ord('a'):
                length = 0
            length += 1
            if length > letters[curr]:
                result += length - letters[curr]
                letters[curr] = length
        return result

# Time:  O(n) # Space: O(1) # Consider the string s to be the infinite wraparound string of # "abcdefghijklmnopqrstuvwxyz", so s will look like this: # "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....". # # Now we have another string p. Your job is to find out # how many unique non-empty substrings of p are present in s. # In particular, your input is the string p and you need to output # the number of different non-empty substrings of p in the string s. # # Note: p consists of only lowercase English letters and the size of p might be over 10000. # # Example 1: # Input: "a" # Output: 1 # # Explanation: Only the substring "a" of string "a" is in the string s. # Example 2: # Input: "cac" # Output: 2 # Explanation: There are two substrings "a", "c" of string "cac" in the string s. # Example 3: # Input: "zab" # Output: 6 # Explanation: There are six substrings "z", "a", "b", "za", "ab", "zab" of string "zab" in the string s. :type p: str         :rtype: int