What is the time complexity of Scramble String?

The optimal solution has a time complexity of O(n^5).

What is the space complexity of Scramble String?

The space complexity is O(n^3).

What data structures are used to solve Scramble String?

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

What are the key insights for solving Scramble String?

Insight 1: The core idea is to use recursion with memoization. A string `s1` can be scrambled into `s2` if either `s1` and `s2` are identical, or there exists an index `k` where the substrings `s1[0:k]` is scrambled into `s2[0:k]` and `s1[k:]` is scrambled into `s2[k:]`, or `s1[0:k]` is scrambled into `s2[n-k:]` and `s1[k:]` is scrambled into `s2[0:n-k]` where n is the length of the strings. Insight 2: Using memoization drastically reduces the computational overhead by storing the results of previously computed subproblems.

Which companies ask LeetCode 87. Scramble String in interviews?

This problem is frequently asked at Amazon, Darwinbox, Google, Media.net, Rubrik.

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Scramble String - LeetCode 87 Solution

Scramble String - Complete Solution Guide

Scramble String is LeetCode problem 87, 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

We can scramble a string s to get a string t using the following algorithm: If the length of the string is 1, stop. If the length of the string is > 1, do the following: Split the string into two non-empty substrings at a random index, i.e., if the string is s , divide it to x and y where s = x + y . Randomly decide to swap the two substrings or to keep them in the same order. i.e., after this step, s may become s = x + y or s = y + x . Apply step 1 recursively on each of the two substrings x an

Detailed Explanation

The problem asks whether a string `s2` is a scrambled version of another string `s1`. A string is scrambled if it can be obtained from the original string through a series of operations involving splitting the string into two non-empty substrings and optionally swapping them, then recursively applying this process to the substrings. The input consists of two strings, `s1` and `s2`, of the same length, containing lowercase English letters. The output should be a boolean value: `true` if `s2` is a scrambled string of `s1`, and `false` otherwise. The strings' length is between 1 and 30, inclusive.

Solution Approach

The solution uses a recursive approach with memoization to determine if a string `s2` is a scrambled version of `s1`. The recursion breaks down the problem into smaller subproblems, checking if substrings of `s1` and `s2` are scrambled versions of each other. The function first checks if the substrings are identical or if their character counts differ; if so, it returns immediately. If the basic checks pass, it iterates through all possible split points `k` and recursively checks if the substrings can be scrambled with and without swapping.

Step-by-Step Algorithm

Step 1: Check if the result for the current `(i1, i2, length)` combination is already stored in the `memo`. If yes, return the stored value.
Step 2: Extract the substrings `sub1` from `s1` starting at `i1` with `length` and `sub2` from `s2` starting at `i2` with `length`.
Step 3: If `sub1` and `sub2` are equal, store `true` in the `memo` for the key `(i1, i2, length)` and return `true`.
Step 4: If the character frequencies of `sub1` and `sub2` are not equal, store `false` in the `memo` and return `false`.
Step 5: Iterate through all possible split points `k` from 1 to `length - 1`.
Step 6: For each `k`, recursively check two scenarios:
a. No Swap: Check if `s1[i1:i1+k]` is scrambled to `s2[i2:i2+k]` AND `s1[i1+k:i1+length]` is scrambled to `s2[i2+k:i2+length]`
b. Swap: Check if `s1[i1:i1+k]` is scrambled to `s2[i2+length-k:i2+length]` AND `s1[i1+k:i1+length]` is scrambled to `s2[i2:i2+length-k]`
Step 7: If either of the above scenarios returns `true`, store `true` in the `memo` and return `true`.
Step 8: If all the split points are checked and neither scenario returns `true`, store `false` in the `memo` and return `false`.

Key Insights

Insight 1: The core idea is to use recursion with memoization. A string `s1` can be scrambled into `s2` if either `s1` and `s2` are identical, or there exists an index `k` where the substrings `s1[0:k]` is scrambled into `s2[0:k]` and `s1[k:]` is scrambled into `s2[k:]`, or `s1[0:k]` is scrambled into `s2[n-k:]` and `s1[k:]` is scrambled into `s2[0:n-k]` where n is the length of the strings.
Insight 2: Using memoization drastically reduces the computational overhead by storing the results of previously computed subproblems.
Insight 3: Before recursion, checking if the character counts of the substrings are equal provides a quick way to prune unnecessary recursive calls. If the character counts are different, the strings cannot be scrambled versions of each other.

Complexity Analysis

Time Complexity: O(n^5)

Space Complexity: O(n^3)

Topics

This problem involves: String, Dynamic Programming.

Companies

Asked at: Amazon, Darwinbox, Google, Media.net, Rubrik.

Scramble String | LeetCode 87

Hard

String Dynamic Programming

Time: O(n^5)Space: O(n^3)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Arrays;

class Solution {
    private HashMap<String, Boolean> memo = new HashMap<>();

    public boolean isScramble(String s1, String s2) {
        int n = s1.length();
        return solve(s1, s2, 0, 0, n);
    }

    private boolean solve(String s1, String s2, int i1, int i2, int length) {
        String key = i1 + "," + i2 + "," + length;
        if (memo.containsKey(key)) {
            return memo.get(key);
        }

        String sub1 = s1.substring(i1, i1 + length);
        String sub2 = s2.substring(i2, i2 + length);

        if (sub1.equals(sub2)) {
            memo.put(key, true);
            return true;
        }

        if (!areAnagrams(sub1, sub2)) {
            memo.put(key, false);
            return false;
        }

        for (int k = 1; k < length; k++) {
            // Case 1: No swap
            if (solve(s1, s2, i1, i2, k) && solve(s1, s2, i1 + k, i2 + k, length - k)) {
                memo.put(key, true);
                return true;
            }

            // Case 2: Swap
            if (solve(s1, s2, i1, i2 + length - k, k) && solve(s1, s2, i1 + k, i2, length - k)) {
                memo.put(key, true);
                return true;
            }
        }

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

    private boolean areAnagrams(String s1, String s2) {
        if (s1.length() != s2.length()) {
            return false;
        }
        char[] arr1 = s1.toCharArray();
        char[] arr2 = s2.toCharArray();
        Arrays.sort(arr1);
        Arrays.sort(arr2);
        return Arrays.equals(arr1, arr2);
    }
}

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Problem Info

Description

Problem 87: Scramble String - Detailed explanation and solution approaches.

Topics

String Dynamic Programming

Companies

AmazonDarwinboxGoogleMedia.netRubrik

Stats

Difficulty:Hard

Languages:4

Companies:5

Problem #:87

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: Check if the result for the current `(i1, i2, length)` combination is already stored in the `memo`. If yes, return the stored value.
2
Step 2: Extract the substrings `sub1` from `s1` starting at `i1` with `length` and `sub2` from `s2` starting at `i2` with `length`.
3
Step 3: If `sub1` and `sub2` are equal, store `true` in the `memo` for the key `(i1, i2, length)` and return `true`.
4
Step 4: If the character frequencies of `sub1` and `sub2` are not equal, store `false` in the `memo` and return `false`.
5
Step 5: Iterate through all possible split points `k` from 1 to `length - 1`.
6
Step 6: For each `k`, recursively check two scenarios:
7
a. No Swap: Check if `s1[i1:i1+k]` is scrambled to `s2[i2:i2+k]` AND `s1[i1+k:i1+length]` is scrambled to `s2[i2+k:i2+length]`
8
b. Swap: Check if `s1[i1:i1+k]` is scrambled to `s2[i2+length-k:i2+length]` AND `s1[i1+k:i1+length]` is scrambled to `s2[i2:i2+length-k]`
9
Step 7: If either of the above scenarios returns `true`, store `true` in the `memo` and return `true`.
10
Step 8: If all the split points are checked and neither scenario returns `true`, store `false` in the `memo` and return `false`.

💡

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(n^5)

💾

Space Complexity

O(n^3)

📝

Why?

The time complexity arises from the recursive calls and the iterations within the solve function. The nested loops contribute to the O(n^5) factor. The space complexity comes from the memoization table, which stores boolean results for subproblems defined by the starting indices and lengths of the substrings.

📖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

Dynamic Programming (Bottom-up)

A bottom-up dynamic programming approach can also be used. A 3D table `dp[i][j][len]` stores whether `s1[i:i+len]` is a scrambled version of `s2[j:j+len]`. The table is filled starting with len=1 and iteratively increasing `len`.

Time: O(n^5)Space: O(n^3)

Trade-offs: The bottom-up approach avoids recursion overhead, but the logic might be slightly more complex to implement than the recursive memoization approach.

✨

Key Insights

1Insight 1: The core idea is to use recursion with memoization. A string `s1` can be scrambled into `s2` if either `s1` and `s2` are identical, or there exists an index `k` where the substrings `s1[0:k]` is scrambled into `s2[0:k]` and `s1[k:]` is scrambled into `s2[k:]`, or `s1[0:k]` is scrambled into `s2[n-k:]` and `s1[k:]` is scrambled into `s2[0:n-k]` where n is the length of the strings.
2Insight 2: Using memoization drastically reduces the computational overhead by storing the results of previously computed subproblems.
3Insight 3: Before recursion, checking if the character counts of the substrings are equal provides a quick way to prune unnecessary recursive calls. If the character counts are different, the strings cannot be scrambled versions of each other.

📚

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 or strings of different lengths. The problem states that strings must have the same length, and at least length 1. The initial check in `isScramble` implicitly handles the case where strings have different length.
Edge case 2: Strings that are equal. The code directly compares the substrings and returns true if they match without further recursion.
Edge case 3: Strings that are anagrams but not scrambled. The code checks if the substrings are anagrams before diving into recursion. However, it's possible for two strings to be anagrams but not scrambled versions of each other. The recursive checks handle this scenario.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using memoization, leading to exponential time complexity and TLE (Time Limit Exceeded).

✗Common mistake 2: Forgetting to check if the character counts of the substrings are equal before making recursive calls, which can significantly reduce the number of recursive calls.

✗Common mistake 3: Incorrectly defining the base cases for the recursion. The base cases should include the condition where the substrings are equal, and where the lengths are 1.

✅

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: This algorithm can be used in bioinformatics to compare and analyze DNA sequences that may have undergone rearrangements.

🏦Real-world scenario 2: It can be applied in data compression techniques where strings may be rearranged for efficient storage or transmission. Determining if a re-arranged string is derived from an original string can be useful for decompressing the data.

🎮Real-world scenario 3: Used in Natural language processing to determine sentence similarity.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

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

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 27, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Scramble String"?

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?