What is the time complexity of Check if an Original String Exists Given Two Encoded Strings?

The optimal solution has a time complexity of O(45^n).

What is the space complexity of Check if an Original String Exists Given Two Encoded Strings?

The space complexity is O(n^3).

What data structures are used to solve Check if an Original String Exists Given Two Encoded Strings?

This problem uses String, Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Check if an Original String Exists Given Two Encoded Strings?

Insight 1: The core idea is to use dynamic programming (with memoization) to explore all possible encodings for both strings simultaneously. Insight 2: The `diff` parameter in the recursive function keeps track of the difference in length between the portions of the original string matched by s1 and s2 up to the current indices. A positive `diff` means s1 is 'ahead' in terms of original string length, while a negative `diff` means s2 is ahead.

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Check if an Original String Exists Given Two Encoded Strings - LeetCode 2060 Solution

Q: Which companies ask LeetCode 2060. Check if an Original String Exists Given Two Encoded Strings in interviews?

This problem is frequently asked at BitGo.

Check if an Original String Exists Given Two Encoded Strings - Complete Solution Guide

Check if an Original String Exists Given Two Encoded Strings is LeetCode problem 2060, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 seco

Detailed Explanation

The problem asks us to determine if two encoded strings, `s1` and `s2`, could have originated from the same original string. The original string consists of lowercase English letters and can be encoded by splitting it into substrings, replacing some substrings with their lengths (as numeric strings), and concatenating the result. We need to return `true` if there exists at least one possible original string that can be encoded as both `s1` and `s2`, and `false` otherwise. The encoded strings consist of letters and digits (1-9), and the number of consecutive digits is limited to 3.

Solution Approach

The solution uses a recursive, memoized approach to explore all possible ways to decode `s1` and `s2` and check if there is a common original string. The `solve` function takes the current indices `i` and `j` for `s1` and `s2` respectively, and a `diff` parameter representing the difference in length of decoded strings so far. The function considers three possible cases: 1) `s1` provides a number (i.e., a substring representing the length of a part of the original string), 2) `s2` provides a number, and 3) both `s1` and `s2` provide characters that can be matched if their length difference is 0. Memoization is used to avoid recomputation of overlapping subproblems.

Step-by-Step Algorithm

Step 1: Define a recursive function `solve(i, j, diff)` that takes the indices `i` and `j` of `s1` and `s2` and the difference `diff` in length of decoded strings up to those indices.
Step 2: Base case: If both `i` and `j` have reached the end of their respective strings, return `true` if `diff` is 0 (meaning the decoded lengths are equal). Otherwise, return `false`.
Step 3: If the current character in `s1` is a digit, extract the number represented by consecutive digits, and recursively call `solve` with incremented index `k` of `s1` and updated `diff` (diff + num).
Step 4: If the current character in `s2` is a digit, extract the number represented by consecutive digits, and recursively call `solve` with incremented index `k` of `s2` and updated `diff` (diff - num).
Step 5: If `diff` is positive, it means `s1` is ahead. In this case, if the current character in `s2` is not a digit, try to match a length from s1 with character in s2 by decrementing the `diff`.
Step 6: If `diff` is negative, it means `s2` is ahead. In this case, if the current character in `s1` is not a digit, try to match a length from s2 with character in s1 by incrementing the `diff`.
Step 7: If `diff` is zero, both the strings match character by character. If the current characters in both strings are not digits and are equal, recursively call `solve` with `i+1`, `j+1` and same `diff`.
Step 8: Memoize the result of `solve(i, j, diff)` to avoid recomputation.
Step 9: Return the result of calling `solve(0, 0, 0)`.
Step 10: For the C code, explicitly allocate a 3D array for memoization because the input size is limited, and hash table would be an overhead. Note the shift and offset of the `diff` parameter since it can be negative.

Key Insights

Insight 1: The core idea is to use dynamic programming (with memoization) to explore all possible encodings for both strings simultaneously.
Insight 2: The `diff` parameter in the recursive function keeps track of the difference in length between the portions of the original string matched by s1 and s2 up to the current indices. A positive `diff` means s1 is 'ahead' in terms of original string length, while a negative `diff` means s2 is ahead.
Insight 3: Handling digits requires converting them to their numerical values and exploring all possible lengths that each digit string can represent. The digits can represent lengths in the original string which will then be matched against the letters of the other string.

Complexity Analysis

Time Complexity: O(45^n)

Space Complexity: O(n^3)

Topics

This problem involves: String, Dynamic Programming.

Companies

Asked at: BitGo.

Check if an Original String Exists Given Two Encoded Strings | LeetCode 2060

Hard

String Dynamic Programming

Time: O(45^n)Space: O(n^3)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public boolean possiblyEquals(String s1, String s2) {
        int n1 = s1.length();
        int n2 = s2.length();
        Map<String, Boolean> memo = new HashMap<>();

        return solve(s1, s2, 0, 0, 0, memo);
    }

    private boolean solve(String s1, String s2, int i, int j, int diff, Map<String, Boolean> memo) {
        String state = i + "," + j + "," + diff;
        if (memo.containsKey(state)) {
            return memo.get(state);
        }

        if (i == s1.length() && j == s2.length()) {
            return diff == 0;
        }

        // Option 1: s1 provides a number.
        if (i < s1.length() && Character.isDigit(s1.charAt(i))) {
            int num = 0;
            int k = i;
            while (k < s1.length() && Character.isDigit(s1.charAt(k))) {
                num = num * 10 + (s1.charAt(k) - '0');
                if (solve(s1, s2, k + 1, j, diff + num, memo)) {
                    memo.put(state, true);
                    return true;
                }
                k++;
            }
        }

        // Option 2: s2 provides a number.
        if (j < s2.length() && Character.isDigit(s2.charAt(j))) {
            int num = 0;
            int k = j;
            while (k < s2.length() && Character.isDigit(s2.charAt(k))) {
                num = num * 10 + (s2.charAt(k) - '0');
                if (solve(s1, s2, i, k + 1, diff - num, memo)) {
                    memo.put(state, true);
                    return true;
                }
                k++;
            }
        }

        // Option 3: Match characters or use wildcards.
        if (diff > 0) {  // s1 is ahead, s2 needs to provide a character to catch up.
            if (j < s2.length() && !Character.isDigit(s2.charAt(j))) {
                if (solve(s1, s2, i, j + 1, diff - 1, memo)) {
                    memo.put(state, true);
                    return true;
                }
            }
        } else if (diff < 0) {  // s2 is ahead, s1 needs to provide a character.
            if (i < s1.length() && !Character.isDigit(s1.charAt(i))) {
                if (solve(s1, s2, i + 1, j, diff + 1, memo)) {
                    memo.put(state, true);
                    return true;
                }
            }
        } else {  // diff == 0, balanced state.
            if (i < s1.length() && !Character.isDigit(s1.charAt(i)) && j < s2.length() && !Character.isDigit(s2.charAt(j))) {
                if (s1.charAt(i) == s2.charAt(j) && solve(s1, s2, i + 1, j + 1, 0, memo)) {
                    memo.put(state, true);
                    return true;
                }
            }
        }

        memo.put(state, false);
        return false;
    }
}

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Problem Info

Description

Problem 2060: Check if an Original String Exists Given Two Encoded Strings - Detailed explanation and solution approaches.

Topics

String Dynamic Programming

Companies

BitGo

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2060

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Define a recursive function `solve(i, j, diff)` that takes the indices `i` and `j` of `s1` and `s2` and the difference `diff` in length of decoded strings up to those indices.
2
Step 2: Base case: If both `i` and `j` have reached the end of their respective strings, return `true` if `diff` is 0 (meaning the decoded lengths are equal). Otherwise, return `false`.
3
Step 3: If the current character in `s1` is a digit, extract the number represented by consecutive digits, and recursively call `solve` with incremented index `k` of `s1` and updated `diff` (diff + num).
4
Step 4: If the current character in `s2` is a digit, extract the number represented by consecutive digits, and recursively call `solve` with incremented index `k` of `s2` and updated `diff` (diff - num).
5
Step 5: If `diff` is positive, it means `s1` is ahead. In this case, if the current character in `s2` is not a digit, try to match a length from s1 with character in s2 by decrementing the `diff`.
6
Step 6: If `diff` is negative, it means `s2` is ahead. In this case, if the current character in `s1` is not a digit, try to match a length from s2 with character in s1 by incrementing the `diff`.
7
Step 7: If `diff` is zero, both the strings match character by character. If the current characters in both strings are not digits and are equal, recursively call `solve` with `i+1`, `j+1` and same `diff`.
8
Step 8: Memoize the result of `solve(i, j, diff)` to avoid recomputation.
9
Step 9: Return the result of calling `solve(0, 0, 0)`.
10
Step 10: For the C code, explicitly allocate a 3D array for memoization because the input size is limited, and hash table would be an overhead. Note the shift and offset of the `diff` parameter since it can be negative.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(45^n)

💾

Space Complexity

O(n^3)

📝

Why?

The time complexity is determined by the number of possible states (i, j, diff) multiplied by the cost of the transitions from each state. The space complexity is dominated by the size of the memoization table, which stores the results for each possible state.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Iterative DP

Instead of recursion with memoization, an iterative dynamic programming approach can be used. A 3D table `dp[i][j][diff]` would store whether the substrings `s1[0...i]` and `s2[0...j]` can be encoded to have a length difference of `diff`. The table would be filled bottom-up.

Time: O(N1 * N2 * D * 3^N1) - Similar to the recursive approachSpace: O(N1 * N2 * D) - Similar to the recursive approach

Trade-offs: Iterative DP avoids the recursion overhead, but it might be slightly harder to implement and understand the dependencies.

✨

Key Insights

1Insight 1: The core idea is to use dynamic programming (with memoization) to explore all possible encodings for both strings simultaneously.
2Insight 2: The `diff` parameter in the recursive function keeps track of the difference in length between the portions of the original string matched by s1 and s2 up to the current indices. A positive `diff` means s1 is 'ahead' in terms of original string length, while a negative `diff` means s2 is ahead.
3Insight 3: Handling digits requires converting them to their numerical values and exploring all possible lengths that each digit string can represent. The digits can represent lengths in the original string which will then be matched against the letters of the other string.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty strings. The base case `i == n1 && j == n2` handles the scenario where both strings are fully consumed. The function returns `true` only if the length difference `diff` is 0.
Edge case 2: Strings containing only numbers. The algorithm correctly parses and matches the number of characters against the other string, and will return true or false.
Edge case 3: Strings with consecutive digits. The algorithm iterates all valid number combinations of consecutive digits, and will try to match with other possibilities of splitting. Note that problem constraints limit consecutive digits to 3.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the `diff` parameter correctly. The `diff` is crucial for determining if the decoded strings can potentially match.

✗Common mistake 2: Incorrectly parsing digits. It's important to consider all possible numbers that can be formed from consecutive digits.

✗Common mistake 3: Not using memoization. Without memoization, the algorithm would have exponential time complexity and would time out for larger inputs.

✗Common mistake 4: Failing to account for cases where letters from either string compensate for lengths from number in the other string, e.g., 'a1' matching to 'bb'.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data compression. This algorithm can be used to analyze different encodings of strings to check which one uses fewer bytes without loss of data.

🏦Real-world scenario 2: Pattern matching with variable-length placeholders. In scenarios where you want to check if a string matches a pattern allowing replacements of subsequences with numbers indicating a wildcard length, the approach can be adapted.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 20, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Check if an Original String Exists Given Two Encoded Strings"?

For this hard-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?