Is Find the Occurrence of First Almost Equal Substring easy or hard?

LeetCode classifies this problem as Hard because the naive solution is straightforward but inefficient for large inputs. The challenge is recognizing how to precompute matching segments so each substring check becomes constant time.

Find the Occurrence of First Almost Equal Substring Python/Java solution

Most implementations use a sliding window across the string and compare it with the pattern. The optimized solution adds prefix and suffix match arrays to avoid repeated comparisons. The algorithm can be implemented in Python, Java, C++, or JavaScript with O(n + m) time complexity.

How to solve Find the Occurrence of First Almost Equal Substring in O(n)?

Achieving near-linear time requires precomputing how many characters match between the pattern and each position of the string from both directions. Using prefix and suffix match arrays, you can verify whether a substring differs by at most one character in constant time per window. The total runtime becomes O(n + m).

What is the best approach for Find the Occurrence of First Almost Equal Substring?

The most efficient approach uses precomputation with prefix and suffix match lengths combined with a sliding window. By calculating how many characters match before and after a possible mismatch, you can check each window in constant time. This reduces the total complexity to O(n + m), where n is the length of the text and m is the pattern length.

Is Find the Occurrence of First Almost Equal Substring asked at Google/Amazon/Meta?

Problems involving approximate string matching and sliding window comparisons frequently appear in interviews at companies like Google, Amazon, and Meta. Variants that allow one mismatch or limited edits are common because they test knowledge of string algorithms and optimization techniques.

What data structure is used in Find the Occurrence of First Almost Equal Substring?

The problem primarily uses string processing techniques and arrays for prefix and suffix match lengths. The optimized solution relies on precomputed arrays and a sliding window over the string rather than complex data structures like trees or graphs.

What is the time complexity of Find the Occurrence of First Almost Equal Substring?

The brute-force sliding window approach runs in O(n * m) time because each window compares up to m characters. The optimized approach with precomputed prefix and suffix matches runs in O(n + m) time since each character of the text and pattern is processed a constant number of times.

Description Hints 3 Solution Video 18:37

3303. Find the Occurrence of First Almost Equal Substring

Hard16.0% acceptance1 company

You are given two strings s and pattern.

A string x is called almost equal to y if you can change at most one character in x to make it identical to y.

Return the smallest starting index of a substring in s that is almost equal to pattern. If no such index exists, return -1.

A substring is a contiguous non-empty sequence of characters within a string.

Example 1:

Input: s = "abcdefg", pattern = "bcdffg"

Output: 1

Explanation:

The substring s[1..6] == "bcdefg" can be converted to "bcdffg" by changing s[4] to "f".

Example 2:

Input: s = "ababbababa", pattern = "bacaba"

Output: 4

Explanation:

The substring s[4..9] == "bababa" can be converted to "bacaba" by changing s[6] to "c".

Example 3:

Input: s = "abcd", pattern = "dba"

Output: -1

Example 4:

Input: s = "dde", pattern = "d"

Output: 0

Constraints:

1 <= pattern.length < s.length <= 10⁵
s and pattern consist only of lowercase English letters.

Follow-up: Could you solve the problem if at most k consecutive characters can be changed?

Asked at 1 company

Google

Topics

String String Matching

Hints open one at a time

Hint 1

Let dp1[i] represent the maximum length of a substring of s starting at index i that is also a prefix of pattern.

Hint 2

Let dp2[i] represent the maximum length of a substring of s ending at index i that is also a suffix of pattern.

Hint 3

Consider a window of size pattern.length. If dp1[i] + i == i + pattern.length - 1 - dp2[i + pattern.length - 1], what does this signify?

18:37

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