LeetCode 3931: Check Adjacent Digit Differences ·

Table of Contents

Meta Data
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Difficulty: easy First Attempt: 2026-05-17

Submissions: Accepted (first try)

Approach
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Convert characters to digits and compare each adjacent pair once (String Template 4). Any pair differing by more than 2 means False.

My Solution
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class Solution:
    def isAdjacentDiffAtMostTwo(self, s: str) -> bool:
        return all(abs(int(a) - int(b)) <= 2 for a, b in zip(s, s[1:]))

Complexity
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Time O(n), space O(n) (the s[1:] copy).

Mistakes to Avoid
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Convert characters to digits before subtracting.

Interview Explanation
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I pair the string with itself shifted by one and return True when every adjacent pair differs by at most 2. Time O(n).

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2026-03-29·2 mins

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LeetCode 3838: Weighted Word Mapping

2026-02-14·1 min

LeetCode Easy Array String Simulation

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LeetCode 3823: Reverse Letters Then Special Characters in a String

2026-01-31·2 mins

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LeetCode 3823 note: two converging-pointer passes, swapping only letters the first time and only special characters the second

LeetCode 2273: Find Resultant Array After Removing Anagrams

2026-01-29·2 mins

LeetCode Easy Hash-Table String Sorting

LeetCode 2273 note: repeatedly deleting a word that is an anagram of the previous one keeps exactly the words that are not anagrams of the last kept word; compare sorted strings

LeetCode 2697: Lexicographically Smallest Palindrome

2026-01-24·2 mins

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LeetCode 2697 note: converging pointers, setting both ends to the smaller character whenever they differ

LeetCode 3258: Count Substrings That Satisfy K-Constraint I

2026-01-18·2 mins

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LeetCode 3258 note: valid means zeros <= k or ones <= k; it is invalid only when both exceed k, and after shrinking there are right - left + 1 valid substrings ending at right

Meta Data#

Approach#

My Solution#

Complexity#

Mistakes to Avoid#

Interview Explanation#

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