Create 2025-05-06-Leetcode-5.md

Author: hyeonukim

_posts/Leetcode/DP/2025-05-06-Leetcode-5.md

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+---
+title: Leetcode 5. Longest Palindromic Substring
+description: Explanation for Leetcode 5 - Longest Palindromic Substring, and its solution in Python.
+date: 2025-05-06
+categories: [Leetcode, Dynamic Programming, Medium]
+tags: [Leetcode, Python, Dynamic Programming, Medium]
+math: true
+---
+
+## Problem
+[Leetcode 5 - Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/description/)
+
+Example:
+```
+Input: s = "babad"
+Output: "bab"
+Explanation: "aba" is also a valid answer.
+
+Input: s = "cbbd"
+Output: "bb"
+
+Input: s = "a"
+Output: "a"
+
+Input: s = "ac"
+Output: "a"
+```
+
+## Approach
+
+We can solve this problem using Dynamic Programming. The idea is to maintain a 2D DP table where `dp[i][j]` is `True` if the substring `s[i:j+1]` is a palindrome. We iterate over all possible substrings and update the DP table accordingly.
+
+Here is the Python code for the solution:
+```python
+class Solution:
+    def longestPalindrome(self, s: str) -> str:
+        n = len(s)
+        if n <= 1:
+            return s
+        
+        dp = [[False] * n for _ in range(n)]
+        start, max_length = 0, 1
+        
+        # Every single character is a palindrome
+        for i in range(n):
+            dp[i][i] = True
+        
+        # Check for substrings of length 2
+        for i in range(n - 1):
+            if s[i] == s[i + 1]:
+                dp[i][i + 1] = True
+                start = i
+                max_length = 2
+        
+        # Check for substrings of length 3 or more
+        for length in range(3, n + 1):
+            for i in range(n - length + 1):
+                j = i + length - 1
+                if s[i] == s[j] and dp[i + 1][j - 1]:
+                    dp[i][j] = True
+                    start = i
+                    max_length = length
+        
+        return s[start:start + max_length]
+```
+
+## Time Complexity and Space Complexity
+
+Time Complexity: $O(n^2)$  
+Space Complexity: $O(n^2)$