Question

2414. Length of the Longest Alphabetical Continuous Substring

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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 <= 105
• s 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 $O(1)$.

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

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 {
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
}

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
  }
}

#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);
}