add LongestPalindromicSubstring.java

Java/Dynamic-Programming/LongestPalindromicSubstring.java

@@ -0,0 +1,56 @@
+/**
+ * Longest Palindromic Substring
+ * 
+ * Given a string s, return the longest palindromic substring in s.
+ * 
+ * Example 1
+ * Input: s = "cabbad"
+ * Output: "abba"
+ * 
+ * Example 2
+ * Input: s = "cbbd"
+ * Output: "bb"
+ * 
+ * Author: Frank Garcia (https://github.com/frank-xyz)
+ */
+
+public class LongestPalindromicSubstring {
+    public static void main(String[] args) {
+        System.out.println(new LongestPalindromicSubstring().longestPalindrome("cabbad"));
+    }
+
+    public String longestPalindrome(String s) {
+        if (s == null || s.length() == 0) return s; // base case: empty input string
+
+        int start = 0; // the start boundary for the longest palindromic substring in s 
+        int end = 0; // the end boundary for the longest palindromic substring in s
+
+        for (int i = 0; i < s.length(); i++) { // loop to iterate through the input string
+            int lenOdd = expandFromMiddle(s, i, i); // get the length of the palindrome with an odd number of characters
+            int lenEven = expandFromMiddle(s, i, i + 1); // get the length of the palindrome with an even number of characters
+            int lenMax = Math.max(lenOdd, lenEven); // the maximum length between lenOdd and lenEven
+
+            // end - start + 1 = length of current longest palindrome
+            if (lenMax > (end - start + 1)) { // if lenMax is greater than the length of the current longest palindrome
+                start = i - ((lenMax - 1) / 2); // update the start boundary of the new longest palindromic substring
+                end = i + (lenMax / 2); // update the end boundary of the new longest palindromic substring 
+            }
+        }
+
+        return s.substring(start, end + 1); // return the substring found between these boundaries
+    }
+
+    private int expandFromMiddle(String s, int left, int right) {
+        if (s == null || left > right) return 0; // boundary check
+
+        // loop while the substring in s between index left and index right is a palindrome 
+        while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {
+            left--; // decrement left to move left (i.e. expand from middle)
+            right++; // increment right to move right (i.e. expand from middle)
+        }
+
+        return right - left - 1; // return the length of the palindromic substring
+    }
+    // time: O(n^2), where n is the number of characters from the center of the string to the beginning AND to the end of the string
+    // space: O(n)
+}