feat: add palindromic-substrings, lcs solutions

Author: shinsj4653

longest-common-subsequence/shinsj4653.py

@@ -13,3 +13,18 @@
 """
 
 
+class Solution:
+    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
+        n, m = len(text1), len(text2)
+        dp = [[0 for _ in range(m + 1)] for _ in range(n + 1)]
+
+        for i in range(1, n + 1):
+            for j in range(1, m + 1):
+                if text1[i - 1] == text2[j - 1]:
+                    dp[i][j] = dp[i - 1][j - 1] + 1
+
+                else:
+                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
+
+        return dp[n][m]
+

palindromic-substrings/shinsj4653.py

@@ -1,15 +1,74 @@
 """
 [문제풀이]
 # Inputs
+string  s
 
 # Outputs
+the number of palindromic substrings
 
 # Constraints
+1 <= s.length <= 1000
+s consists of lowercase English letters.
 
 # Ideas
+부분 문자열 중 팰린드롬 인거
+순열?
+10^3 => 시초 예상
+
+코드를 짜보니 O(n^3) 나오긴하는데 우선 정답
 
 [회고]
 
 """
 
 
+class Solution:
+    def countSubstrings(self, s: str) -> int:
+        ret = 0
+
+        for num in range(1, len(s) + 1):
+            for i in range(len(s) - num + 1):
+                ss = s[i:i + num]
+                if ss == ss[::-1]:
+                    ret += 1
+
+        return ret
+
+# 해설보고 스스로 풀이
+
+class Solution:
+    def countSubstrings(self, s: str) -> int:
+        dp = {}
+
+        for start in range(len(s)):
+            for end in range(start, -1, -1):
+                if start == end:
+                    dp[(start, end)] = True
+
+                elif start + 1 == end:
+                    dp[(start, end)] = s[start] == s[end]
+
+                else:
+                    dp[(start, end)] = dp[(start + 1, end - 1)] and s[start] == s[end]
+
+        return dp.values().count(True)
+
+# 기존 값 재활용하려면 end 부터 세야하는게 이해가 안감
+# -> 다시 풀이
+class Solution:
+    def countSubstrings(self, s: str) -> int:
+        dp = {}
+
+        for end in range(len(s)):
+            for start in range(end, -1, -1):
+                if start == end:
+                    dp[(start, end)] = True
+
+                elif start + 1 == end:
+                    dp[(start, end)] = s[start] == s[end]
+
+                else:
+                    dp[(start, end)] = dp[(start + 1, end - 1)] and s[start] == s[end]
+
+        return list(dp.values()).count(True)
+