0005-longest-palindromic-substring.java

// Solution: Expanding Around Center

// Time Complexity: O(n^2)
// Extra Space Complexity: O(1)

class Solution1 {

    public String longestPalindrome(String s) {
        int strLength = s.length();

        if (strLength < 2) {
            return s;
        }

        int resultLength = 0;
        String result = "";

        for (int i = 0; i < s.length(); i++) {
            //Odd length
            int left = i, right = i;
            while (
                left >= 0 &&
                right < s.length() &&
                s.charAt(left) == s.charAt(right)
            ) {
                if ((right - left + 1) > resultLength) {
                    result = s.substring(left, right + 1);
                    resultLength = right - left + 1;
                }
                left -= 1;
                right += 1;
            }

            // even length
            left = i;
            right = i + 1;
            while (
                left >= 0 &&
                right < s.length() &&
                s.charAt(left) == s.charAt(right)
            ) {
                if ((right - left + 1) > resultLength) {
                    result = s.substring(left, right + 1);
                    resultLength = right - left + 1;
                }
                left -= 1;
                right += 1;
            }
        }

        return result;
    }
}

// Solution: A more Optimized Expand Around Center

// TIme Complexity: O(n^2)
// Extra Space Complexity: O(1)

class Solution2 {

    public String longestPalindrome(String s) {
        int best = 0;
        int start = 0, end = 0;

        for (int i = 0; i < s.length(); i++) {
            int left = i - 1;
            while (i < s.length() - 1 && s.charAt(i) == s.charAt(i + 1)) {
                ++i;
            }

            int right = i + 1;
            while (
                left >= 0 &&
                right < s.length() &&
                s.charAt(left) == s.charAt(right)
            ) {
                --left;
                ++right;
            }

            if (right - left > best) {
                best = right - left;
                start = left + 1;
                end = right;
            }
        }

        return s.substring(start, end);
    }
}

// Solution: Dynamic Programming

// Time Complexity: O(n^2)
// Extra Spce Complexity: O(n^2)

class Solution3 {

    public String longestPalindrome(String s) {
        int len = s.length();
        int left = 0, right = 1, max = 0;

        var isPalindrome = new boolean[len][len];

        for (int i = len - 1; i >= 0; i--) {
            for (int j = i; j < len; j++) {
                if (i == j) {
                    isPalindrome[i][j] = true;
                } else if (s.charAt(i) == s.charAt(j)) {
                    if (j - i == 1) {
                        isPalindrome[i][j] = true;
                    } else {
                        isPalindrome[i][j] = isPalindrome[i + 1][j - 1];
                    }
                }

                if (isPalindrome[i][j] && j - i + 1 > max) {
                    max = j - i + 1;
                    left = i;
                    right = j + 1;
                }
            }
        }

        return s.substring(left, right);
    }
}