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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).

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