# 467. Unique Substrings in Wraparound String

## Description

We define the string base to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so base will look like this:

• "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....".

Given a string s, return the number of unique non-empty substrings of s are present in base.

Example 1:

Input: s = "a"
Output: 1
Explanation: Only the substring "a" of s is in base.


Example 2:

Input: s = "cac"
Output: 2
Explanation: There are two substrings ("a", "c") of s in base.


Example 3:

Input: s = "zab"
Output: 6
Explanation: There are six substrings ("z", "a", "b", "za", "ab", and "zab") of s in base.


Constraints:

• 1 <= s.length <= 105
• s consists of lowercase English letters.

## Solutions

• class Solution {
public int findSubstringInWraproundString(String p) {
int[] dp = new int[26];
int k = 0;
for (int i = 0; i < p.length(); ++i) {
char c = p.charAt(i);
if (i > 0 && (c - p.charAt(i - 1) + 26) % 26 == 1) {
++k;
} else {
k = 1;
}
dp[c - 'a'] = Math.max(dp[c - 'a'], k);
}
int ans = 0;
for (int v : dp) {
ans += v;
}
return ans;
}
}

• class Solution {
public:
int findSubstringInWraproundString(string p) {
vector<int> dp(26);
int k = 0;
for (int i = 0; i < p.size(); ++i) {
char c = p[i];
if (i && (c - p[i - 1] + 26) % 26 == 1)
++k;
else
k = 1;
dp[c - 'a'] = max(dp[c - 'a'], k);
}
int ans = 0;
for (int& v : dp) ans += v;
return ans;
}
};

• class Solution:
def findSubstringInWraproundString(self, p: str) -> int:
dp = [0] * 26
k = 0
for i, c in enumerate(p):
if i and (ord(c) - ord(p[i - 1])) % 26 == 1:
k += 1
else:
k = 1
idx = ord(c) - ord('a')
dp[idx] = max(dp[idx], k)
return sum(dp)


• func findSubstringInWraproundString(p string) int {
dp := make([]int, 26)
k := 0
for i := range p {
c := p[i]
if i > 0 && (c-p[i-1]+26)%26 == 1 {
k++
} else {
k = 1
}
dp[c-'a'] = max(dp[c-'a'], k)
}
ans := 0
for _, v := range dp {
ans += v
}
return ans
}

• function findSubstringInWraproundString(p: string): number {
const n = p.length;
const dp = new Array(26).fill(0);
let cur = 1;
dp[p.charCodeAt(0) - 'a'.charCodeAt(0)] = 1;
for (let i = 1; i < n; i++) {
if ((p.charCodeAt(i) - p.charCodeAt(i - 1) + 25) % 26 == 0) {
cur++;
} else {
cur = 1;
}
const index = p.charCodeAt(i) - 'a'.charCodeAt(0);
dp[index] = Math.max(dp[index], cur);
}
return dp.reduce((r, v) => r + v);
}


• impl Solution {
pub fn find_substring_in_wrapround_string(p: String) -> i32 {
let n = p.len();
let p = p.as_bytes();
let mut dp = [0; 26];
let mut cur = 1;
dp[(p[0] - b'a') as usize] = 1;
for i in 1..n {
if (p[i] - p[i - 1] + 25) % 26 == 0 {
cur += 1;
} else {
cur = 1;
}
let index = (p[i] - b'a') as usize;
dp[index] = dp[index].max(cur);
}
dp.into_iter().sum()
}
}