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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:

1. naive method to get all the substring from p and judge whether it is in the wrapround string 2. use DP for naive method, there are lots of repetition 1) for string p: "abcd" a => ends with "a", there are total 1 here ab => ends with b, there are "ab", "b", total 2 abc => ends with c, there are "abc", "bc" , "c", total 3 abcd =>ends with d, there are "abcd", "bcd", "cd", "d" , total 4 the answer of the substring is 1+2+3+4 so the number of unique substirng of p in s is for every letter in p which has the maximum unique continuous substring end in each character and then sum, but it needs to get rid of the duplicated one 2) if string p : "abcdabc" the latter "abc" will be duplicated for ending character a, b, c. therefore, we need to get the maximum of number of substring in same ending character. 3) if string p: ade ad is not continous substring, so the maximum of number of substring ending in "d" will be 1 again

 1
 
   if p is None or len(p) == 0: 
 2             return 0
 3         dp= [0] * 26              # 26 letters
 4         maxCnt = 0
 5         
 6         for i in range(0, len(p)):
 7             if i > 0 and (ord(p[i]) - ord(p[i-1]) == 1 or ord(p[i-1]) - ord(p[i]) == 25):
 8                 maxCnt += 1
 9             else:
10                 maxCnt = 1
11             index = ord(p[i]) - 97
12             
13             dp[index] = max(dp[index], maxCnt)
14             
15             #print ("directly: ", p[i], dp[index])
16         
17         ans = 0
18         for ele in dp:
19             ans += ele
20         return ans
21     

[Leetcode] DP-- 467. Unique Substrings in Wraparound String