LeetCode #2414 — MEDIUM

Length of the Longest Alphabetical Continuous Substring

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"abacaba"

Example 2

"abcde"

Core Insight

What unlocks the optimal approach

What is the longest possible continuous substring?
The size of the longest possible continuous substring is at most 26, so we can just brute force the answer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2414: Length of the Longest Alphabetical Continuous Substring
class Solution {
    public int longestContinuousSubstring(String s) {
        int ans = 1, cnt = 1;
        for (int i = 1; i < s.length(); ++i) {
            if (s.charAt(i) - s.charAt(i - 1) == 1) {
                ans = Math.max(ans, ++cnt);
            } else {
                cnt = 1;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2414: Length of the Longest Alphabetical Continuous Substring
func longestContinuousSubstring(s string) int {
	ans, cnt := 1, 1
	for i := range s[1:] {
		if s[i+1]-s[i] == 1 {
			cnt++
			ans = max(ans, cnt)
		} else {
			cnt = 1
		}
	}
	return ans
}

# Accepted solution for LeetCode #2414: Length of the Longest Alphabetical Continuous Substring
class Solution:
    def longestContinuousSubstring(self, s: str) -> int:
        ans = cnt = 1
        for x, y in pairwise(map(ord, s)):
            if y - x == 1:
                cnt += 1
                ans = max(ans, cnt)
            else:
                cnt = 1
        return ans

// Accepted solution for LeetCode #2414: Length of the Longest Alphabetical Continuous Substring
impl Solution {
    pub fn longest_continuous_substring(s: String) -> i32 {
        let mut ans = 1;
        let mut cnt = 1;
        let s = s.as_bytes();
        for i in 1..s.len() {
            if s[i] - s[i - 1] == 1 {
                cnt += 1;
                ans = ans.max(cnt);
            } else {
                cnt = 1;
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #2414: Length of the Longest Alphabetical Continuous Substring
function longestContinuousSubstring(s: string): number {
    let [ans, cnt] = [1, 1];
    for (let i = 1; i < s.length; ++i) {
        if (s.charCodeAt(i) - s.charCodeAt(i - 1) === 1) {
            ans = Math.max(ans, ++cnt);
        } else {
            cnt = 1;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.