Description#
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 <= 105s consists of only English lowercase letters.
Solutions#
Solution 1: Two Pointers#
We use two pointers $i$ and $j$ to point to the start and end of the current consecutive substring respectively. Traverse the string $s$, if the current character $s[j]$ is greater than $s[j-1]$, then move $j$ one step to the right, otherwise update $i$ to $j$, and update the length of the longest consecutive substring.
The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.
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| 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
|
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| 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;
}
}
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| 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;
}
};
|
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| 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
}
|
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| 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);
}
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| 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
}
}
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| #define max(a, b) (((a) > (b)) ? (a) : (b))
int longestContinuousSubstring(char* s) {
int n = strlen(s);
int i = 0;
int res = 1;
for (int j = 1; j < n; j++) {
if (s[j] - s[j - 1] != 1) {
res = max(res, j - i);
i = j;
}
}
return max(res, n - i);
}
|
