LeetCode #3302 — MEDIUM

Find the Lexicographically Smallest Valid Sequence

Move from brute-force thinking to an efficient approach using two pointers strategy.

Solve on LeetCode
The Problem

Problem Statement

You are given two strings word1 and word2.

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.

A sequence of indices seq is called valid if:

  • The indices are sorted in ascending order.
  • Concatenating the characters at these indices in word1 in the same order results in a string that is almost equal to word2.

Return an array of size word2.length representing the lexicographically smallest valid sequence of indices. If no such sequence of indices exists, return an empty array.

Note that the answer must represent the lexicographically smallest array, not the corresponding string formed by those indices.

Example 1:

Input: word1 = "vbcca", word2 = "abc"

Output: [0,1,2]

Explanation:

The lexicographically smallest valid sequence of indices is [0, 1, 2]:

  • Change word1[0] to 'a'.
  • word1[1] is already 'b'.
  • word1[2] is already 'c'.

Example 2:

Input: word1 = "bacdc", word2 = "abc"

Output: [1,2,4]

Explanation:

The lexicographically smallest valid sequence of indices is [1, 2, 4]:

  • word1[1] is already 'a'.
  • Change word1[2] to 'b'.
  • word1[4] is already 'c'.

Example 3:

Input: word1 = "aaaaaa", word2 = "aaabc"

Output: []

Explanation:

There is no valid sequence of indices.

Example 4:

Input: word1 = "abc", word2 = "ab"

Output: [0,1]

Constraints:

  • 1 <= word2.length < word1.length <= 3 * 105
  • word1 and word2 consist only of lowercase English letters.
Patterns Used
👉Two Pointers→📊Dynamic Programming→🎯Greedy→

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 word1 and word2. 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. A sequence of indices seq is called valid if: The indices are sorted in ascending order. Concatenating the characters at these indices in word1 in the same order results in a string that is almost equal to word2. Return an array of size word2.length representing the lexicographically smallest valid sequence of indices. If no such sequence of indices exists, return an empty array. Note that the answer must represent the lexicographically smallest array, not the corresponding string formed by those indices.

Baseline thinking

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

Pattern signal: Two Pointers · Dynamic Programming · Greedy

Example 1

"vbcca"
"abc"

Example 2

"bacdc"
"abc"

Example 3

"aaaaaa"
"aaabc"

Related Problems

  • Smallest K-Length Subsequence With Occurrences of a Letter (smallest-k-length-subsequence-with-occurrences-of-a-letter)
Step 02

Core Insight

What unlocks the optimal approach

  • Let <code>dp[i]</code> be the longest suffix of <code>word2</code> that exists as a subsequence of suffix of the substring of <code>word1</code> starting at index <code>i</code>.
  • If <code>dp[i + 1] < m</code> and <code>word1[i] == word2[m - dp[i + 1] - 1]</code>,<code>dp[i] = dp[i + 1] + 1</code>. Otherwise, <code>dp[i] = dp[i + 1]</code>.
  • For each index <code>i</code>, greedily select characters using the <code>dp</code> array to know whether a solution exists.
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
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 #3302: Find the Lexicographically Smallest Valid Sequence
class Solution {
  public int[] validSequence(String word1, String word2) {
    int[] ans = new int[word2.length()];
    // last[j] := the index i of the last occurrence in word1, where
    // word1[i] == word2[j]
    int[] last = new int[word2.length()];
    Arrays.fill(last, -1);

    int i = word1.length() - 1;
    int j = word2.length() - 1;
    while (i >= 0 && j >= 0) {
      if (word1.charAt(i) == word2.charAt(j))
        last[j--] = i;
      --i;
    }

    boolean canSkip = true;
    j = 0;
    for (i = 0; i < word1.length(); ++i) {
      if (j == word2.length())
        break;
      if (word1.charAt(i) == word2.charAt(j)) {
        ans[j++] = i;
      } else if (canSkip && (j == word2.length() - 1 || i < last[j + 1])) {
        canSkip = false;
        ans[j++] = i;
      }
    }

    return j == word2.length() ? ans : new int[0];
  }
}
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)
Space
O(1)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS
O(n) time
O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.

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