LeetCode #2060 — HARD

Check if an Original String Exists Given Two Encoded Strings

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

Solve on LeetCode
The Problem

Problem Statement

An original string, consisting of lowercase English letters, can be encoded by the following steps:

  • Arbitrarily split it into a sequence of some number of non-empty substrings.
  • Arbitrarily choose some elements (possibly none) of the sequence, and replace each with its length (as a numeric string).
  • Concatenate the sequence as the encoded string.

For example, one way to encode an original string "abcdefghijklmnop" might be:

  • Split it as a sequence: ["ab", "cdefghijklmn", "o", "p"].
  • Choose the second and third elements to be replaced by their lengths, respectively. The sequence becomes ["ab", "12", "1", "p"].
  • Concatenate the elements of the sequence to get the encoded string: "ab121p".

Given two encoded strings s1 and s2, consisting of lowercase English letters and digits 1-9 (inclusive), return true if there exists an original string that could be encoded as both s1 and s2. Otherwise, return false.

Note: The test cases are generated such that the number of consecutive digits in s1 and s2 does not exceed 3.

Example 1:

Input: s1 = "internationalization", s2 = "i18n"
Output: true
Explanation: It is possible that "internationalization" was the original string.
- "internationalization" 
  -> Split:       ["internationalization"]
  -> Do not replace any element
  -> Concatenate:  "internationalization", which is s1.
- "internationalization"
  -> Split:       ["i", "nternationalizatio", "n"]
  -> Replace:     ["i", "18",                 "n"]
  -> Concatenate:  "i18n", which is s2

Example 2:

Input: s1 = "l123e", s2 = "44"
Output: true
Explanation: It is possible that "leetcode" was the original string.
- "leetcode" 
  -> Split:      ["l", "e", "et", "cod", "e"]
  -> Replace:    ["l", "1", "2",  "3",   "e"]
  -> Concatenate: "l123e", which is s1.
- "leetcode" 
  -> Split:      ["leet", "code"]
  -> Replace:    ["4",    "4"]
  -> Concatenate: "44", which is s2.

Example 3:

Input: s1 = "a5b", s2 = "c5b"
Output: false
Explanation: It is impossible.
- The original string encoded as s1 must start with the letter 'a'.
- The original string encoded as s2 must start with the letter 'c'.

Constraints:

  • 1 <= s1.length, s2.length <= 40
  • s1 and s2 consist of digits 1-9 (inclusive), and lowercase English letters only.
  • The number of consecutive digits in s1 and s2 does not exceed 3.
Patterns Used
📊Dynamic Programming→

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: An original string, consisting of lowercase English letters, can be encoded by the following steps: Arbitrarily split it into a sequence of some number of non-empty substrings. Arbitrarily choose some elements (possibly none) of the sequence, and replace each with its length (as a numeric string). Concatenate the sequence as the encoded string. For example, one way to encode an original string "abcdefghijklmnop" might be: Split it as a sequence: ["ab", "cdefghijklmn", "o", "p"]. Choose the second and third elements to be replaced by their lengths, respectively. The sequence becomes ["ab", "12", "1", "p"]. Concatenate the elements of the sequence to get the encoded string: "ab121p". Given two encoded strings s1 and s2, consisting of lowercase English letters and digits 1-9 (inclusive), return true if there exists an original string that could be encoded as both s1 and s2. Otherwise,

Baseline thinking

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

Pattern signal: Dynamic Programming

Example 1

"internationalization"
"i18n"

Example 2

"l123e"
"44"

Example 3

"a5b"
"c5b"

Related Problems

  • Valid Word Abbreviation (valid-word-abbreviation)
  • Check If Two String Arrays are Equivalent (check-if-two-string-arrays-are-equivalent)
Step 02

Core Insight

What unlocks the optimal approach

  • For s1 and s2, divide each into a sequence of single alphabet strings and digital strings. The problem now becomes comparing if two sequences are equal.
  • A single alphabet string has no variation, but a digital string has variations. For example: "124" can be interpreted as 1+2+4, 12+4, 1+24, and 124 wildcard characters.
  • There are four kinds of comparisons: a single alphabet vs another; a single alphabet vs a number, a number vs a single alphabet, and a number vs another number. In the case of a number vs another (a single alphabet or a number), can you decrease the number by the min length of both?
  • There is a recurrence relation in the search which ends when either a single alphabet != another, or one sequence ran out, or both sequences ran out.
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 #2060: Check if an Original String Exists Given Two Encoded Strings
class Solution {
  public boolean possiblyEquals(String s1, String s2) {
    Map<Integer, Boolean>[][] mem = new Map[s1.length() + 1][s2.length() + 1];
    for (int i = 0; i <= s1.length(); ++i)
      for (int j = 0; j <= s2.length(); ++j)
        mem[i][j] = new HashMap<>();
    return f(s1, s2, 0, 0, 0, mem);
  }

  // Returns true if s1[i..n) matches s2[j..n), accounting for the padding
  // difference. Here, `paddingDiff` represents the signed padding. A positive
  // `paddingDiff` indicates that s1 has an additional number of offset bytes
  // compared to s2.
  private boolean f(final String s1, final String s2, int i, int j, int paddingDiff,
                    Map<Integer, Boolean>[][] mem) {
    if (mem[i][j].containsKey(paddingDiff))
      return mem[i][j].get(paddingDiff);
    if (i == s1.length() && j == s2.length())
      return paddingDiff == 0;
    if (i < s1.length() && Character.isDigit(s1.charAt(i))) {
      // Add padding on s1.
      final int nextLetterIndex = getNextLetterIndex(s1, i);
      for (final int num : getNums(s1.substring(i, nextLetterIndex)))
        if (f(s1, s2, nextLetterIndex, j, paddingDiff + num, mem))
          return true;
    } else if (j < s2.length() && Character.isDigit(s2.charAt(j))) {
      // Add padding on s2.
      final int nextLetterIndex = getNextLetterIndex(s2, j);
      for (final int num : getNums(s2.substring(j, nextLetterIndex)))
        if (f(s1, s2, i, nextLetterIndex, paddingDiff - num, mem))
          return true;
    } else if (paddingDiff > 0) {
      // `s1` has more padding, so j needs to catch up.
      if (j < s2.length())
        return f(s1, s2, i, j + 1, paddingDiff - 1, mem);
    } else if (paddingDiff < 0) {
      // `s2` has more padding, so i needs to catch up.
      if (i < s1.length())
        return f(s1, s2, i + 1, j, paddingDiff + 1, mem);
    } else { // paddingDiff == 0
      // There's no padding difference, so consume the next letter.
      if (i < s1.length() && j < s2.length() && s1.charAt(i) == s2.charAt(j))
        return f(s1, s2, i + 1, j + 1, 0, mem);
    }
    mem[i][j].put(paddingDiff, false);
    return false;
  }

  private int getNextLetterIndex(final String s, int i) {
    int j = i;
    while (j < s.length() && Character.isDigit(s.charAt(j)))
      ++j;
    return j;
  }

  private List<Integer> getNums(final String s) {
    List<Integer> nums = new ArrayList<>(List.of(Integer.parseInt(s)));
    if (s.length() == 2) {
      nums.add(Integer.parseInt(s.substring(0, 1)) + Integer.parseInt(s.substring(1, 2)));
    } else if (s.length() == 3) {
      nums.add(Integer.parseInt(s.substring(0, 1)) + Integer.parseInt(s.substring(1, 3)));
      nums.add(Integer.parseInt(s.substring(0, 2)) + Integer.parseInt(s.substring(2, 3)));
      nums.add(Integer.parseInt(s.substring(0, 1)) + Integer.parseInt(s.substring(1, 2)) +
               Integer.parseInt(s.substring(2, 3)));
    }
    return nums;
  }
}
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(n × m)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

Related Problems
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