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.
struct Solution;
impl Solution {
fn find_substring_in_wrapround_string(p: String) -> i32 {
let mut count: [usize; 26] = [0; 26];
let p: Vec<u8> = p.bytes().collect();
let n = p.len();
for i in 0..n {
let j = (p[i] - b'a') as usize;
count[j] = 1;
}
let mut l = 1;
for i in 1..n {
let j = (p[i] - b'a') as usize;
let k = (p[i - 1] - b'a') as usize;
l = if (k + 1) % 26 == j { l + 1 } else { 1 };
count[j] = count[j].max(l);
}
let res: usize = count.iter().sum();
res as i32
}
}
#[test]
fn test() {
let p = "a".to_string();
let res = 1;
assert_eq!(Solution::find_substring_in_wrapround_string(p), res);
let p = "cac".to_string();
let res = 2;
assert_eq!(Solution::find_substring_in_wrapround_string(p), res);
let p = "zab".to_string();
let res = 6;
assert_eq!(Solution::find_substring_in_wrapround_string(p), res);
}