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 <= 10^5sconsists 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;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).