Merge pull request #55 from krishnak3101/leet_code_que_5

Leet code que no 5

Author: pawangeek

0005/README.MD

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+## Longest Palindromic substring
+
+
+Find the longest palindromic substring in a given string S
+* Ex1 :
+For S = cadab, answer is ada 
+
+* Ex2 : For S = cadac, answer is cadac
+
+Link to the problem :- https://leetcode.com/problems/longest-palindromic-substring/

0005/longest_palindromic_substring.py

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+class Solution:
+    def longestPalindrome(self, s: str) -> str:
+        # Two pointer approach , time complexity O(n), space complexity O(1)
+        max_len = -float("inf")
+        max_ind = [-1, -1]
+
+        # odd lenth palindromes
+        for k in range(len(s)):
+            i = k
+            j = k
+            while(i >= 0 and j <= len(s) - 1):
+                if s[i] == s[j]:
+                    if (j - i + 1) > max_len:
+                        max_len = j - i + 1
+                        max_ind = [i, j]
+
+                    i -= 1
+                    j += 1
+
+                else:
+                    break
+
+        # even lenth palindromes
+        for k in range(len(s) - 1):
+            i = k
+            j = k + 1
+            while(i >= 0 and j <= len(s) - 1):
+                if s[i] == s[j]:
+                    if (j - i + 1) > max_len:
+                        max_len = j - i + 1
+                        max_ind = [i, j]
+
+                    i -= 1
+                    j += 1
+
+                else:
+                    break
+
+        return s[max_ind[0]:max_ind[1] + 1]
+
+        # Dynamic Programming approach ,
+        # Time complexity O(n^2), Space complexity O(n^2)
+
+        dp = [[0 for _ in range(len(s))] for _ in range(len(s))]
+
+        # sub string of length 1
+        for i in range(len(s)):
+            dp[i][i] = 1
+
+        # substring of length 1 id always a palindrome
+        max_len = 1
+        max_ind = [0, 0]
+
+        # sub string of length 2
+        for i in range(len(s) - 1):
+            if s[i] == s[i + 1]:
+                dp[i][i + 1] = 1
+                if max_len == 1:
+                    max_len = 2
+                    max_ind[0] = i
+                    max_ind[1] = i + 1
+
+        # sub string of length greater than 2
+        for j in range(2, len(s)):
+            for i in range(0, len(s) - j):
+                k = j + i
+                if (s[k] == s[i]) and (dp[i + 1][k - 1] == 1):
+                    dp[i][k] = 1
+                    if (k - i + 1) > max_len:
+                        max_len = k - i + 1
+                        max_ind = [i, k]
+
+        return s[max_ind[0]:max_ind[1] + 1]