_467.java

package com.fishercoder.solutions;

/**Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so s will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....".

 Now we have another string p. Your job is to find out how many unique non-empty substrings of p are present in s. In particular, your input is the string p and you need to output the number of different non-empty substrings of p in the string s.

 Note: p consists of only lowercase English letters and the size of p might be over 10000.

 Example 1:
 Input: "a"
 Output: 1

 Explanation: Only the substring "a" of string "a" is in the string s.
 Example 2:
 Input: "cac"
 Output: 2
 Explanation: There are two substrings "a", "c" of string "cac" in the string s.
 Example 3:
 Input: "zab"
 Output: 6
 Explanation: There are six substrings "z", "a", "b", "za", "ab", "zab" of string "zab" in the string s.*/
public class _467 {
    /**A naive solution would definitely result in TLE.
     * Since the pattern keeps repeating, DP is the way to go.
     * Because the input consists merely of lowercase English letters, we could maintain an array of size 26,
     * keep updating this array, counting the substrings that end with this letter, keep updating it with the largest one possible.
     * 
     * Inspired by this: https://discuss.leetcode.com/topic/70658/concise-java-solution-using-dp*/
    public static int findSubstringInWraproundString(String p) {
        if (p == null || p.isEmpty()) {
            return 0;
        }
        if (p.length() < 2) {
            return p.length();
        }
        int count = 0;
        int[] dp = new int[26];
        dp[p.charAt(0) - 'a'] = 1;
        int len = 1;
        for (int i = 1; i < p.length(); i++) {
            if (p.charAt(i) - 1 == p.charAt(i - 1) || (p.charAt(i) == 'a' && p.charAt(i - 1) == 'z')) {
                len++;
            } else {
                len = 1;
            }
            dp[p.charAt(i) - 'a'] = Math.max(len, dp[p.charAt(i) - 'a']);
        }

        for (int i : dp) {
            count += i;
        }
        return count;
    }

    public static void main(String... args) {
//        String p = "a";
//        String p = "abcgha";
        String p = "zab";
        System.out.println(findSubstringInWraproundString(p));
    }
}