[leetcode]Unique Substrings in Wraparound String

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: 1Explanation: Only the substring "a" of string "a" is in the string s.Example 2:Input: "cac"Output: 2Explanation: There are two substrings "a", "c" of string "cac" in the string s.Example 3:Input: "zab"Output: 6Explanation: There are six substrings "z", "a", "b", "za", "ab", "zab" of string "zab" in the string s.

Solution post: Concise Java solution using DP
It first get the length of the longest substring ending with each character: ‘a’, ‘b’, ‘c’ … ‘z’. Then the number of unique substrings is the sum of the lengths.
Example:
ababc:

variable a b c longest length 1 2 1

Thus the result is

    1  a+   2  b, ab+   1  c equals 6

code

int findSubstringInWraproundString(string p){    int count[26] = { 0 };    int len = 0;    for (int i = 0; i < p.length(); ++i)    {        if (i > 0 && (p[i] - p[i - 1] == 1 || p[i - 1] - p[i] == 25))            ++len;        else            len = 1;        int index = p[i] - 'a';        if (count[index] < len)            count[index] = len;    }    int ans = 0;    for (int i = 0; i < 26; ++i)        ans += count[i];    return ans;}