Find the Occurrence of First Almost Equal Substring

To solve this problem, scan s while comparing substrings of length equal to pattern. Track mismatches and return the first index where at most one character differs. Using dynamic programming or direct comparison ensures efficient detection of almost equal substrings without checking all possible substrings naively.

Problem Statement

You are given two strings s and pattern. A substring of s is considered almost equal to pattern if it can be converted into pattern by changing at most one character. Your task is to find the starting index of the first such substring in s.

If no substring satisfies this condition, return -1. For example, with s = "abcdefg" and pattern = "bcdffg", the substring starting at index 1 matches after changing one character. Constraints: 1 <= pattern.length < s.length <= 105 and both strings consist of lowercase letters only.

Examples

Example 1

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

Output: 1

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

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 details omitted.

Constraints

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

Solution Approach

Sliding Window with Mismatch Counting

Use a window of size pattern.length sliding over s. Count character mismatches for each window and return the first index with at most one mismatch. This directly targets the almost equal substring requirement and avoids unnecessary substring generation.

Dynamic Programming Prefix Matching

Let dp1[i] store the length of the longest prefix of pattern matching the substring of s starting at i. Iterate through s, adjusting for one mismatch allowance. This ensures you quickly identify positions where a single character change makes the substring match pattern.

Early Termination Optimization

Stop scanning once a valid almost equal substring is found. Skipping further checks avoids unnecessary computation and aligns with the problem's goal of returning the first valid occurrence.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n * m) in the naive approach where n = s.length and m = pattern.length, but using prefix DP or optimized sliding window can reduce redundant comparisons. Space complexity is O(n) if storing prefix match lengths or O(1) for direct sliding window counting.

What Interviewers Usually Probe

• Check if candidates understand handling at most one mismatch correctly.
• Look for efficient substring scanning instead of generating all possibilities.
• See if dynamic programming or optimized sliding window is applied for string matching patterns.

Common Pitfalls or Variants

Common pitfalls

• Forgetting to allow exactly one mismatch, returning only exact matches.
• Over-scanning s or re-comparing the same substrings, increasing time complexity.
• Misaligning substring indices when calculating dp1 or mismatch counts.

Follow-up variants

• Allow k mismatches instead of just one, generalizing almost equal substrings.
• Find all occurrences of almost equal substrings rather than the first.
• Consider case-insensitive almost equality for string matching problems.

FAQ

What does almost equal mean in this problem?

A substring is almost equal to pattern if changing at most one character can make it identical to pattern.

Can the solution handle patterns near the length of s efficiently?

Yes, using sliding window or DP approaches minimizes unnecessary comparisons even when pattern is close to s in length.

Why not check every substring of s directly?

Checking every substring naively increases time complexity to O(n*m); optimized methods reduce redundant comparisons.

Is dynamic programming necessary?

DP is not required but helps efficiently track prefix matches and quickly detect valid almost equal substrings.

How does this relate to String plus String Matching pattern?

It directly uses string matching techniques while accounting for a single allowed mismatch, a key variant of standard pattern matching.