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Formatted question description: https://leetcode.ca/all/2223.html
2223. Sum of Scores of Built Strings
 Difficulty: Hard.
 Related Topics: String, Binary Search, Rolling Hash, Suffix Array, String Matching, Hash Function.
 Similar Questions: Longest Happy Prefix.
Problem
You are building a string s
of length n
one character at a time, prepending each new character to the front of the string. The strings are labeled from 1
to n
, where the string with length i
is labeled si
.
 For example, for
s = "abaca"
,s1 == "a"
,s2 == "ca"
,s3 == "aca"
, etc.
The score of si
is the length of the longest common prefix between si
and sn
(Note that s == sn
).
Given the final string s
, return** the sum of the score of every **si
.
Example 1:
Input: s = "babab"
Output: 9
Explanation:
For s1 == "b", the longest common prefix is "b" which has a score of 1.
For s2 == "ab", there is no common prefix so the score is 0.
For s3 == "bab", the longest common prefix is "bab" which has a score of 3.
For s4 == "abab", there is no common prefix so the score is 0.
For s5 == "babab", the longest common prefix is "babab" which has a score of 5.
The sum of the scores is 1 + 0 + 3 + 0 + 5 = 9, so we return 9.
Example 2:
Input: s = "azbazbzaz"
Output: 14
Explanation:
For s2 == "az", the longest common prefix is "az" which has a score of 2.
For s6 == "azbzaz", the longest common prefix is "azb" which has a score of 3.
For s9 == "azbazbzaz", the longest common prefix is "azbazbzaz" which has a score of 9.
For all other si, the score is 0.
The sum of the scores is 2 + 3 + 9 = 14, so we return 14.
Constraints:

1 <= s.length <= 105

s
consists of lowercase English letters.
Solution

class Solution { public long sumScores(String s) { int n = s.length(); char[] ss = s.toCharArray(); int[] z = new int[n]; int l = 0; int r = 0; for (int i = 1; i < n; i++) { if (i <= r) { z[i] = Math.min(z[i  l], r  i + 1); } while (i + z[i] < n && ss[z[i]] == ss[i + z[i]]) { z[i]++; } if (i + z[i]  1 > r) { l = i; r = i + z[i]  1; } } long sum = n; for (int i = 0; i < n; i++) { sum += z[i]; } return sum; } }

Todo

print("Todo!")
Explain:
nope.
Complexity:
 Time complexity : O(n).
 Space complexity : O(n).