From dacf1d0375dcfd4f8b294220c03df833a6d8ecac Mon Sep 17 00:00:00 2001
From: faizan2700 <46817346+faizan2700@users.noreply.github.com>
Date: Wed, 5 Feb 2020 16:57:43 +0530
Subject: [PATCH] Implement Manacher's algorithm (#1721)

* manacher's algorithm updated
---
 strings/manacher.py | 94 +++++++++++++++++++++++++++++++++------------
 1 file changed, 70 insertions(+), 24 deletions(-)

diff --git a/strings/manacher.py b/strings/manacher.py
index 4193def2f..95aba1fbe 100644
--- a/strings/manacher.py
+++ b/strings/manacher.py
@@ -1,25 +1,18 @@
-# calculate palindromic length from center with incrementing difference
-def palindromic_length(center, diff, string):
-    if (
-        center - diff == -1
-        or center + diff == len(string)
-        or string[center - diff] != string[center + diff]
-    ):
-        return 0
-    return 1 + palindromic_length(center, diff + 1, string)
-
-
 def palindromic_string(input_string):
     """
-    Manacher’s algorithm which finds Longest Palindromic Substring in linear time.
+    >>> palindromic_string('abbbaba')
+    'abbba'
+    >>> palindromic_string('ababa')
+    'ababa'
+
+    Manacher’s algorithm which finds Longest palindromic Substring in linear time.
 
     1. first this convert input_string("xyx") into new_string("x|y|x") where odd
         positions are actual input characters.
-    2. for each character in new_string it find corresponding length and store,
-        a. max_length
-        b. max_length's center
-    3. return output_string from center - max_length to center + max_length and remove
-        all "|"
+    2. for each character in new_string it find corresponding length and store the length
+        and l,r to store previously calculated info.(please look the explanation for details)
+        
+    3. return corresponding output_string by removing all "|"
     """
     max_length = 0
 
@@ -33,19 +26,38 @@ def palindromic_string(input_string):
     # append last character
     new_input_string += input_string[-1]
 
+    # we will store the starting and ending of previous furthest ending palindromic substring
+    l, r = 0, 0
+
+    # length[i] shows the length of palindromic substring with center i
+    length = [1 for i in range(len(new_input_string))]
+
     # for each character in new_string find corresponding palindromic string
     for i in range(len(new_input_string)):
+        k = 1 if i > r else min(length[l + r - i] // 2, r - i + 1)
+        while (
+            i - k >= 0
+            and i + k < len(new_input_string)
+            and new_input_string[k + i] == new_input_string[i - k]
+        ):
+            k += 1
 
-        # get palindromic length from i-th position
-        length = palindromic_length(i, 1, new_input_string)
+        length[i] = 2 * k - 1
+
+        # does this string is ending after the previously explored end (that is r) ?
+        # if yes the update the new r to the last index of this
+        if i + k - 1 > r:
+            l = i - k + 1
+            r = i + k - 1
 
         # update max_length and start position
-        if max_length < length:
-            max_length = length
+        if max_length < length[i]:
+            max_length = length[i]
             start = i
 
     # create that string
-    for i in new_input_string[start - max_length : start + max_length + 1]:
+    s = new_input_string[start - max_length // 2 : start + max_length // 2 + 1]
+    for i in s:
         if i != "|":
             output_string += i
 
@@ -53,5 +65,39 @@ def palindromic_string(input_string):
 
 
 if __name__ == "__main__":
-    n = input()
-    print(palindromic_string(n))
+    import doctest
+
+    doctest.testmod()
+
+"""
+...a0...a1...a2.....a3......a4...a5...a6....
+
+consider the string for which we are calculating the longest palindromic substring is shown above where ...
+are some characters in between and right now we are calculating the length of palindromic substring with
+center at a5 with following conditions :
+i) we have stored the length of palindromic substring which has center at a3 (starts at l ends at r) and it
+    is the furthest ending till now, and it has ending after a6
+ii) a2 and a4 are equally distant from a3 so char(a2) == char(a4)
+iii) a0 and a6 are equally distant from a3 so char(a0) == char(a6)
+iv) a1 is corresponding equal character of a5 in palindrome with center a3 (remember that in below derivation of a4==a6)
+
+now for a5 we will calculate the length of palindromic substring with center as a5 but can we use previously
+calculated information in some way?
+Yes, look the above string we know that a5 is inside the palindrome with center a3 and previously we have
+have calculated that
+a0==a2 (palindrome of center a1)
+a2==a4 (palindrome of center a3)
+a0==a6 (palindrome of center a3)
+so a4==a6
+
+so we can say that palindrome at center a5 is at least as long as palindrome at center a1
+but this only holds if a0 and a6 are inside the limits of palindrome centered at a3 so finally ..
+
+len_of_palindrome__at(a5) = min(len_of_palindrome_at(a1), r-a5)
+where a3 lies from l to r and we have to keep updating that
+
+and if the a5 lies outside of l,r boundary we calculate length of palindrome with bruteforce and update
+l,r.
+
+it gives the linear time complexity just like z-function
+"""