467. Unique Substrings in Wraparound String

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.

Solution：

思路：
ex: zabcd
z = 1
za = 2 + 1 = 3
zab = 3 + 2 + 1 = 6
zabc = 4 + 3 + 2 + 1 = 10
zabcd = 5 + 4 + 3 + 2 + 1 = 15
但需要考虑到是要考虑到重复，比如说以c为结尾子串当前找到，后续也有可能重新再找到以c为结尾，其中的组合不能重复累加，
所以利用
count[index] = Math.max(count[index], maxLengthCur);
来取max

Time Complexity: O(N) Space Complexity: O(1)

Solution Code：

public class Solution {
    public int findSubstringInWraproundString(String p) {
        // count[i] is the maximum unique substring end with ith letter.
        // 0 - 'a', 1 - 'b', ..., 25 - 'z'.
        int[] count = new int[26];
        
        // store longest contiguous substring ends at current position.
        int maxLengthCur = 0; 

        for (int i = 0; i < p.length(); i++) {
            if (i > 0 && (p.charAt(i) - p.charAt(i - 1) == 1 || (p.charAt(i - 1) - p.charAt(i) == 25))) {
                maxLengthCur++;
            }
            else {
                maxLengthCur = 1;
            }
            
            int index = p.charAt(i) - 'a';
            count[index] = Math.max(count[index], maxLengthCur);
        }
        
        // Sum to get result
        int sum = 0;
        for (int i = 0; i < 26; i++) {
            sum += count[i];
        }
        return sum;
    }
}