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Author: NinePiece2

Solutions/0647-palindromic-substrings/README.md

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+<p>Given a string <code>s</code>, return <em>the number of <strong>palindromic substrings</strong> in it</em>.</p>
+
+<p>A string is a <strong>palindrome</strong> when it reads the same backward as forward.</p>
+
+<p>A <strong>substring</strong> is a contiguous sequence of characters within the string.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;abc&quot;
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> Three palindromic strings: &quot;a&quot;, &quot;b&quot;, &quot;c&quot;.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;aaa&quot;
+<strong>Output:</strong> 6
+<strong>Explanation:</strong> Six palindromic strings: &quot;a&quot;, &quot;a&quot;, &quot;a&quot;, &quot;aa&quot;, &quot;aa&quot;, &quot;aaa&quot;.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
+	<li><code>s</code> consists of lowercase English letters.</li>
+</ul>

Solutions/0647-palindromic-substrings/solution.py

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+class Solution:
+    def countSubstrings(self, s: str) -> int:
+        result, n = 0, len(s)
+        for k in range(n * 2 - 1):
+            i, j = k // 2, (k + 1) // 2
+            while ~i and j < n and s[i] == s[j]:
+                result += 1
+                i, j = i - 1, j + 1
+        
+        return result