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Link for the Problem – Longest Palindromic Substring– LeetCode Problem
Longest Palindromic Substring– LeetCode Problem
Problem:
Given a string s
, return the longest palindromic substring in s
.
Example 1:
Input: s = "babad" Output: "bab" Explanation: "aba" is also a valid answer.
Example 2:
Input: s = "cbbd" Output: "bb"
Constraints:
1 <= s.length <= 1000
s
consist of only digits and English letters.
Longest Palindromic Substring– LeetCode Solutions
class Solution { public: string longestPalindrome(string s) { // @ and $ signs are sentinels appended to each end to avoid bounds checking const string& t = join('@' + s + '$', '#'); const int n = t.length(); // t[i - maxExtends[i]..i) == // t[i + 1..i + maxExtends[i]] vector<int> maxExtends(n); int center = 0; for (int i = 1; i < n - 1; ++i) { const int rightBoundary = center + maxExtends[center]; const int mirrorIndex = center - (i - center); maxExtends[i] = rightBoundary > i && min(rightBoundary - i, maxExtends[mirrorIndex]); // Attempt to expand palindrome centered at i while (t[i + 1 + maxExtends[i]] == t[i - 1 - maxExtends[i]]) ++maxExtends[i]; // If palindrome centered at i expand past rightBoundary, // adjust center based on expanded palindrome. if (i + maxExtends[i] > rightBoundary) center = i; } // Find the maxExtend and bestCenter int maxExtend = 0; int bestCenter = -1; for (int i = 0; i < n; ++i) if (maxExtends[i] > maxExtend) { maxExtend = maxExtends[i]; bestCenter = i; } const int l = (bestCenter - maxExtend) / 2; const int r = (bestCenter + maxExtend) / 2; return s.substr(l, r - l); } private: string join(const string& s, char c) { string joined; for (int i = 0; i < s.length(); ++i) { joined += s[i]; if (i != s.length() - 1) joined += c; } return joined; } };
class Solution { public String longestPalindrome(String s) { final String t = join('@' + s + '$', '#'); final int n = t.length(); // t[i - maxExtends[i]..i) == // t[i + 1..i + maxExtends[i]] int[] maxExtends = new int[n]; int center = 0; for (int i = 1; i < n - 1; ++i) { final int rightBoundary = center + maxExtends[center]; final int mirrorIndex = center - (i - center); maxExtends[i] = rightBoundary > i && Math.min(rightBoundary - i, maxExtends[mirrorIndex]) > 0 ? 1 : 0; // Attempt to expand palindrome centered at i while (t.charAt(i + 1 + maxExtends[i]) == t.charAt(i - 1 - maxExtends[i])) ++maxExtends[i]; // If palindrome centered at i expand past rightBoundary, // adjust center based on expanded palindrome. if (i + maxExtends[i] > rightBoundary) center = i; } // Find the maxExtend and bestCenter int maxExtend = 0; int bestCenter = -1; for (int i = 0; i < n; ++i) if (maxExtends[i] > maxExtend) { maxExtend = maxExtends[i]; bestCenter = i; } return s.substring((bestCenter - maxExtend) / 2, (bestCenter + maxExtend) / 2); } private String join(final String s, char c) { StringBuilder sb = new StringBuilder(); for (int i = 0; i < s.length(); ++i) { sb.append(s.charAt(i)); if (i != s.length() - 1) sb.append(c); } return sb.toString(); } }
class Solution: def longestPalindrome(self, s: str) -> str: # @ and $ signs are sentinels appended to each end to avoid bounds checking t = '#'.join('@' + s + '$') n = len(t) # t[i - maxExtends[i]..i) == # t[i + 1..i + maxExtends[i]] maxExtends = [0] * n center = 0 for i in range(1, n - 1): rightBoundary = center + maxExtends[center] mirrorIndex = center - (i - center) maxExtends[i] = rightBoundary > i and \ min(rightBoundary - i, maxExtends[mirrorIndex]) # Attempt to expand palindrome centered at i while t[i + 1 + maxExtends[i]] == t[i - 1 - maxExtends[i]]: maxExtends[i] += 1 # If palindrome centered at i expand past rightBoundary, # adjust center based on expanded palindrome. if i + maxExtends[i] > rightBoundary: center = i # Find the maxExtend and bestCenter maxExtend, bestCenter = max((extend, i) for i, extend in enumerate(maxExtends)) return s[(bestCenter - maxExtend) // 2:(bestCenter + maxExtend) // 2]
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