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Formatted question description: https://leetcode.ca/all/2414.html

# 2414. Length of the Longest Alphabetical Continuous Substring

• Difficulty: Medium.
• Related Topics: .
• Similar Questions: Longest Consecutive Sequence, Arithmetic Slices, Max Consecutive Ones, Maximum Number of Vowels in a Substring of Given Length, Number of Zero-Filled Subarrays.

## Problem

An alphabetical continuous string is a string consisting of consecutive letters in the alphabet. In other words, it is any substring of the string "abcdefghijklmnopqrstuvwxyz".

• For example, "abc" is an alphabetical continuous string, while "acb" and "za" are not.

Given a string s consisting of lowercase letters only, return the length of the **longest alphabetical continuous substring.**

Example 1:

Input: s = "abacaba"
Output: 2
Explanation: There are 4 distinct continuous substrings: "a", "b", "c" and "ab".
"ab" is the longest continuous substring.


Example 2:

Input: s = "abcde"
Output: 5
Explanation: "abcde" is the longest continuous substring.


Constraints:

• 1 <= s.length <= 105

• s consists of only English lowercase letters.

## Solution (Java, C++, Python)

• class Solution {
public int longestContinuousSubstring(String s) {
int x = 0, len = 0, res = 0;
for(char ch : s.toCharArray()) {
len = ch-'a'-x==1 ? len+1 : 1;  // add to len or restart
res = Math.max(res, len);
x = ch-'a';
}
return res;
}
}

############

class Solution {
public int longestContinuousSubstring(String s) {
int ans = 0;
int i = 0, j = 1;
for (; j < s.length(); ++j) {
ans = Math.max(ans, j - i);
if (s.charAt(j) - s.charAt(j - 1) != 1) {
i = j;
}
}
ans = Math.max(ans, j - i);
return ans;
}
}

• class Solution:
def longestContinuousSubstring(self, s: str) -> int:
ans = 0
i, j = 0, 1
while j < len(s):
ans = max(ans, j - i)
if ord(s[j]) - ord(s[j - 1]) != 1:
i = j
j += 1
ans = max(ans, j - i)
return ans

############

# 2414. Length of the Longest Alphabetical Continuous Substring
# https://leetcode.com/problems/length-of-the-longest-alphabetical-continuous-substring/

class Solution:
def longestContinuousSubstring(self, s: str) -> int:
N = len(s)
res = length = 1

def f(x):
return ord(x) - ord("a")

prev = f(s[0])

for i in range(1, N):
curr = f(s[i])

if prev + 1 == curr:
length += 1
res = max(res, length)
else:
length = 1

prev = curr

return res


• class Solution {
public:
int longestContinuousSubstring(string s) {
int ans = 0;
int i = 0, j = 1;
for (; j < s.size(); ++j) {
ans = max(ans, j - i);
if (s[j] - s[j - 1] != 1) {
i = j;
}
}
ans = max(ans, j - i);
return ans;
}
};

• func longestContinuousSubstring(s string) int {
ans := 0
i, j := 0, 1
for ; j < len(s); j++ {
ans = max(ans, j-i)
if s[j]-s[j-1] != 1 {
i = j
}
}
ans = max(ans, j-i)
return ans
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

• function longestContinuousSubstring(s: string): number {
const n = s.length;
let res = 1;
let i = 0;
for (let j = 1; j < n; j++) {
if (s[j].charCodeAt(0) - s[j - 1].charCodeAt(0) !== 1) {
res = Math.max(res, j - i);
i = j;
}
}
return Math.max(res, n - i);
}


• impl Solution {
pub fn longest_continuous_substring(s: String) -> i32 {
let s = s.as_bytes();
let n = s.len();
let mut res = 1;
let mut i = 0;
for j in 1..n {
if s[j] - s[j - 1] != 1 {
res = res.max(j - i);
i = j;
}
}
res.max(n - i) as i32
}
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).