Merge pull request #102 from the-ethan-hunt/master

Added Manachar's algorithm in Python

Author: prateekiiest

Diff for: Competitive Coding/Strings/String Search/Manachar_algorithm/manachar_algorithm.py

@@ -0,0 +1,37 @@
+''' Given a string, find the longest palindromic substring '''
+''' O(n) '''
+def get_palindrome_length(string, index):
+    length = 1
+    while index + length < len(string) and index - length >= 0:
+        if string[index + length] == string[index - length]:
+            length += 1
+        else:
+            break
+        return length - 1
+
+def interleave(string):
+    ret = []
+    for s in string:
+        ret.extend(['#', s])
+    ret.append('#')
+    return ''.join(ret)
+''' Find longest palindrome number '''
+def manacher(string):
+    right = 0
+    center = 0
+    string = interleave(string)
+    P = map(lambda e: 0, xrange(len(string)))
+    for i in xrange(1, len(string)):
+        mirror = 2*center - i
+        if i + P[mirror] <= right and mirror >= len(string) - i:
+            P[i] = P[mirror]
+        else:
+            plength = get_palindrome_length(string, i)
+            P[i] = plength
+            if plength > 1:
+                center = int(i)
+                right = center + plength
+    return [e/2 for e in P]
+''' Return the palindrome sub-string '''
+def get_palindrome_number(string):
+    return sum(manacher(string))

Diff for: Competitive Coding/Strings/String Search/Manachar_algorithm/manchar_algorithm.md

@@ -0,0 +1,34 @@
+# Manachar's Algorithm
+
+Manacher's algorithm is a very handy algorithm with a short implementation that
+can make many programming tasks, such as finding the number of palindromic substrings
+or finding the longest palindromic substring, very easy and efficient. The running
+time of Manacher's algorithm is *O(N)* where *N*  is the length of the input string.
+
+## Implementation
+
+Let:
+
+1. s be a string of N characters
+
+2. s2 be a derived string of s, comprising N * 2 + 1 elements, with each element
+corresponding to one of the following: the N characters in s, the N-1 boundaries
+among characters, and the boundaries before and after the first and last character
+respectively
+
+3. A boundary in s2 is equal to any other boundary in s2 with respect to element
+matching in palindromic length determination
+
+4. p be an array of palindromic span for each element in s2, from center to either
+outermost element, where each boundary is counted towards the length of a palindrome
+(e.g. a palindrome that is three elements long has a palindromic span of 1)
+
+5. c be the position of the center of the palindrome currently known to include a
+boundary closest to the right end of s2 (i.e., the length of the palindrome = p[c]*2+1)
+
+6. r be the position of the right-most boundary of this palindrome (i.e., r = c + p[c])
+
+7. i be the position of an element (i.e., a character or boundary) in s2 whose palindromic
+span is being determined, with i always to the right of c
+
+8. i2 be the mirrored position of i around c (e.g., {i, i2} = {6, 4}, {7, 3}, {8, 2},… when c = 5 (i.e., i2 = c * 2 - i)