From 138e4c5156c2f8fb77aec01d172f81e170a06842 Mon Sep 17 00:00:00 2001
From: forest000014 <forest000014@gmail.com>
Date: Thu, 2 Jan 2025 23:11:17 +0900
Subject: [PATCH] Palindromic Substrings

---
 palindromic-substrings/forest000014.java | 38 ++++++++++++++++++++++++
 1 file changed, 38 insertions(+)
 create mode 100644 palindromic-substrings/forest000014.java

diff --git a/palindromic-substrings/forest000014.java b/palindromic-substrings/forest000014.java
new file mode 100644
index 000000000..42e69387d
--- /dev/null
+++ b/palindromic-substrings/forest000014.java
@@ -0,0 +1,38 @@
+/*
+time complexity: O(n^2)
+space complexity: O(1)
+
+모든 가능한 조합(O(n^2))에 대해 palindrome 여부를 검사(O(n))하는 brute force는 O(n^3).
+i번째 문자에서 시작하여, 앞/뒤로 한 글자씩 늘려가면서 탐색하되, 한번이라도 앞/뒤 글자가 서로 다르다면 그 이후는 탐색하지 않을 수 있음. 이 경우는 O(n^2).
+*/
+class Solution {
+    public int countSubstrings(String s) {
+        int ans = 0;
+        for (int i = 0; i < s.length(); i++) {
+            int head = i, tail = i;
+            while (head >= 0 && tail < s.length()) {
+                if (s.charAt(head) == s.charAt(tail)) {
+                    ans++;
+                } else {
+                    break;
+                }
+                head--;
+                tail++;
+            }
+
+            head = i;
+            tail = i + 1;
+            while (head >= 0 && tail < s.length()) {
+                if (s.charAt(head) == s.charAt(tail)) {
+                    ans++;
+                } else {
+                    break;
+                }
+                head--;
+                tail++;
+            }
+        }
+
+        return ans;
+    }
+}