LeetCode #3303 — HARD

Find the Occurrence of First Almost Equal Substring

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

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?
Patterns Used
🔎String Matching→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

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

Pattern signal: String Matching

Example 1

"abcdefg"
"bcdffg"

Example 2

"ababbababa"
"bacaba"

Example 3

"abcd"
"dba"

Related Problems

  • Check Whether Two Strings are Almost Equivalent (check-whether-two-strings-are-almost-equivalent)
  • Count Almost Equal Pairs II (count-almost-equal-pairs-ii)
Step 02

Core Insight

What unlocks the optimal approach

  • Let <code>dp1[i]</code> represent the maximum length of a substring of <code>s</code> starting at index <code>i</code> that is also a prefix of <code>pattern</code>.
  • Let <code>dp2[i]</code> represent the maximum length of a substring of <code>s</code> ending at index <code>i</code> that is also a suffix of <code>pattern</code>.
  • Consider a window of size <code>pattern.length</code>. If <code>dp1[i] + i == i + pattern.length - 1 - dp2[i + pattern.length - 1]</code>, what does this signify?
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #3303: Find the Occurrence of First Almost Equal Substring
class Solution {
  public int minStartingIndex(String s, String pattern) {
    int[] z1 = zFunction(new StringBuilder(pattern).append(s).toString());
    int[] z2 = zFunction(new StringBuilder(pattern)
                             .reverse() //
                             .append(new StringBuilder(s).reverse())
                             .toString());

    // Match s[i..i + len(pattern) - 1] with `pattern` from both the prefix and
    // the suffix.
    for (int i = 0; i <= s.length() - pattern.length(); ++i)
      if (z1[pattern.length() + i] + z2[s.length() - i] >= pattern.length() - 1)
        return i;

    return -1;
  }

  // Returns the z array, where z[i] is the length of the longest prefix of
  // s[i..n) which is also a prefix of s.
  //
  // https://cp-algorithms.com/string/z-function.html#implementation
  private int[] zFunction(final String s) {
    final int n = s.length();
    int[] z = new int[n];
    int l = 0;
    int r = 0;
    for (int i = 1; i < n; ++i) {
      if (i < r)
        z[i] = Math.min(r - i, z[i - l]);
      while (i + z[i] < n && s.charAt(z[i]) == s.charAt(i + z[i]))
        ++z[i];
      if (i + z[i] > r) {
        l = i;
        r = i + z[i];
      }
    }
    return z;
  }

  private String reversed(final String s) {
    return new StringBuilder(s).reverse().toString();
  }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n + m)
Space
O(m)

Approach Breakdown

BRUTE FORCE
O(n × m) time
O(1) space

At each of the n starting positions in the text, compare up to m characters with the pattern. If a mismatch occurs, shift by one and restart. Worst case (e.g., searching "aab" in "aaaa...a") checks m characters at nearly every position: O(n × m).

KMP / Z-ALGO
O(n + m) time
O(m) space

KMP and Z-algorithm preprocess the pattern in O(m) to build a failure/Z-array, then scan the text in O(n) — never backtracking. Total: O(n + m). Rabin-Karp uses rolling hashes for O(n + m) expected time. All beat the O(n × m) brute force of checking every position.

Shortcut: Preprocessing avoids backtracking → O(n + m). The failure function is the key insight.
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.

Related Problems
#28Find the Index of the First Occurrenceeasy→#214Shortest Palindromehard→#459Repeated Substring Patterneasy→#572Subtree of Another Treeeasy→#686Repeated String Matchmedium→