add solution: palindromic-substrings

Author: dusunax

palindromic-substrings/dusunax.py

@@ -0,0 +1,78 @@
+'''
+# 647. Palindromic Substrings
+
+A. use dynamic programming table to store the palindrome status.
+B. use two pointers to expand around the center.
+
+## Time and Space Complexity
+
+### A. Dynamic Programming Table
+```
+TC: O(n^2)
+SC: O(n^2)
+```
+
+#### TC is O(n^2):
+- filling DP table by iterating through all substrings. 
+- each cell (i, j) checks if a substring is a palindrome & counting the cases = O(n^2)
+
+#### SC is O(n^2):
+- storing the n x n dp table. = O(n^2)
+
+### B. Expand Around Center
+```
+TC: O(n^2)
+SC: O(1)
+```
+
+#### TC is O(n^2):
+- for each char, expand outwards to check for palindromes. = O(n^2)
+
+#### SC is O(1):
+- no additional data structures are used. `count` is a single integer. = O(1)
+'''
+class Solution:
+    def countSubstringsDPTable(self, s: str) -> int:
+        '''
+        A. Dynamic Programming Table
+        '''
+        n = len(s)
+        dp = [[False] * n for _ in range(n)] # List comprehension. = SC: O(n^2)
+        count = 0
+
+        for i in range(n): # TC: O(n)
+            dp[i][i] = True
+            count += 1
+        
+        for i in range(n - 1):
+            if s[i] == s[i + 1]:
+                dp[i][i + 1] = True
+                count += 1
+
+        for s_len in range(3, n + 1): # TC: O(n)
+            for i in range(n - s_len + 1): # TC: O(n) 
+                j = i + s_len - 1
+                
+                if s[i] == s[j] and dp[i + 1][j - 1]:
+                    dp[i][j] = True
+                    count += 1
+
+        return count
+    def countSubstrings(self, s: str) -> int:
+        '''
+        B. Expand Around Center
+        '''
+        count = 0
+
+        def expand(left, right):
+            nonlocal count
+            while left >= 0 and right < len(s) and s[left] == s[right]: # TC: O(n)
+                count += 1
+                left -= 1
+                right += 1
+        
+        for i in range(len(s)): # TC: O(n)
+            expand(i, i)
+            expand(i, i + 1)
+        
+        return count