[LeetCode] Problem 5 - Longest Palindromic Substring

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Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example

No.1

Input: “babad”

Output: “bab”

Note: “aba” is also a valid answer.

No.2

Input: “cbbd”

Output: “bb”

O(n^2) runtime, O(n) space – Dynamic programming

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public String longestPalindrome(String s) {
    String str = "";
    int n = s.length();

    if (n > 1000)
        return str;

    boolean[][] dp = new boolean[n][n];

    // P[i, i] ← true
    // P[i, i+1] ← (Si = Si+1)
    // P[i, j] ← (P[i+1, j-1] and Si = Sj)
    for (int i = n - 1; i >= 0; i--) {
        for (int j = i; j < n; j++) {
            dp[i][j] = (s.charAt(i) == s.charAt(j)) && (j - i + 1 < 3 || dp[i+1][j-1]);

            if (dp[i][j] && (j - i + 1) > str.length())
                str = s.substring(i, j + 1);
        }
    }

    return str;
}

O(n^2) runtime, O(1) space – Simpler solution

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public String longestPalindrome(String s) {
    String str = "";
    int n = s.length();

    if (n > 1000)
        return str;

    for (int i = 0; i < n; i++) {
        int len1 = expandAroundCenter(s, i, i);
        int len2 = expandAroundCenter(s, i, i + 1);
        int max = Math.max(len1, len2);

        if (max > str.length())
            str = s.substring(i - (max - 1) / 2, i + 1 + max / 2);
    }

    return str;
}

private int expandAroundCenter(String s, int i, int j) {
    while (i >= 0 && j < s.length() && s.charAt(i) == s.charAt(j)) {
        i--;
        j++;
    }

    return j - i - 1;
}