Description

Given a string s, return the longest palindromic substring in s.

Example 1:

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Input: s = "babad"
Output: "bab"
Explanation: "aba" is also a valid answer.

Example 2:
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Input: s = "cbbd"
Output: "bb"

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Constraints:

1 <= s.length <= 1000
s consist of only digits and English letters.

Approach: Expand Around Center

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class Solution:
    def longestPalindrome(self, s: str) -> str:
        if not s or len(s) == 1:
            return s
        
        start, end = 0, 0

        def expand(l, r):
            while l >= 0 and r < len(s) and s[l] == s[r]:
                l -= 1
                r += 1
            return l + 1, r - 1

        for i in range(len(s)):
            l1, r1 = expand(i, i)
            l2, r2 = expand(i, i + 1)

            if r1 - l1 > end - start:
                start, end = l1, r1
            if r2 - l2 > end - start:
                start, end = l2, r2
        
        return s[start:end + 1]
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class Solution:
    def longestPalindrome(self, s: str) -> str:
        string = s
        longest = ''
        for i in range( len( string ) * 2 ):
            i = i / 2
            valid = True
            a = int( ( i - 0.5 ) // 1 )
            b = int( ( i + 1 ) // 1 )
            stringnew = i % 1 == 0 and string[ int( i ) ] or ''
            length = 0
            while valid:
                if a < 0 or b >= len( string ):
                    break
                if string[ a ] != string[ b ]:
                    valid = False
                    break
                stringnew = string[ a ] + stringnew + string[ b ]
                a -= 1
                b += 1
                length += 1
                
            if len( stringnew ) > len( longest ):
                longest = stringnew
        return longest
__import__("atexit").register(lambda: open("display_runtime.txt", 'w').write('0'))

Approach: Dynamic Programming

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class Solution:
    def longestPalindrome(self, s: str) -> str:
        n = len(s)
        if n < 2:
            return s
        
        dp = [[False] * n for _ in range(n)]
        start, max_len = 0, 1

        for j in range(1, n):
            for i in range(j):
                if s[i] == s[j]:
                    if j - i < 3:
                        dp[i][j] = True
                    else:
                        dp[i][j] = dp[i + 1][j - 1]
                
                if dp[i][j] and j - i + 1 > max_len:
                    max_len = j - i + 1
                    start = i
        
        return s[start:start + max_len]