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.
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.
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Example 1:
\n\n\nInput: s = "babab"\nOutput: 9\nExplanation:\nFor s1 == "b", the longest common prefix is "b" which has a score of 1.\nFor s2 == "ab", there is no common prefix so the score is 0.\nFor s3 == "bab", the longest common prefix is "bab" which has a score of 3.\nFor s4 == "abab", there is no common prefix so the score is 0.\nFor s5 == "babab", the longest common prefix is "babab" which has a score of 5.\nThe sum of the scores is 1 + 0 + 3 + 0 + 5 = 9, so we return 9.\n\n
Example 2:
\n\n\nInput: s = "azbazbzaz"\nOutput: 14\nExplanation: \nFor s2 == "az", the longest common prefix is "az" which has a score of 2.\nFor s6 == "azbzaz", the longest common prefix is "azb" which has a score of 3.\nFor s9 == "azbazbzaz", the longest common prefix is "azbazbzaz" which has a score of 9.\nFor all other si, the score is 0.\nThe sum of the scores is 2 + 3 + 9 = 14, so we return 14.\n\n\n
\n
Constraints:
\n\n1 <= s.length <= 105s consists of lowercase English letters.Compiled with clang 11 using the latest C++ 20 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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