What is the time complexity of String Transformation?

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

What data structures are used to solve String Transformation?

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

What are the key insights for solving String Transformation?

Insight 1: If `t` cannot be obtained from `s` through rotations, the answer is 0. This can be checked by verifying that `t` is a substring of `s + s`. Insight 2: The rotations can be represented as cyclic shifts. The key idea is to find the period `p` of the string `s`. If `s` is a periodic string, i.e., `s = (sub) * m`, where `sub` is some substring of length `p` and `m` is some integer, performing a cyclic shift by `p` will yield the same string. Knowing `p` helps determine the number of rotations that result in the same string. We find period `p` using KMP's LPS array.

Which companies ask LeetCode 2851. String Transformation in interviews?

This problem is frequently asked at Atlassian, MathWorks, Snowflake.

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String Transformation - LeetCode 2851 Solution

Q: How to solve String Transformation?

The solution involves the following steps: first, check if `t` can even be obtained by rotating `s` (i.e., `t` is a substring of `s + s`). If not, return 0. Then, calculate the LPS (Longest Proper Prefix which is also a Suffix) array for `s` using the KMP algorithm to determine the smallest period `p` of `s`. If such period exists (s is periodic), then some rotations will not change the string. Next, compute the result based on whether `s` equals `t` or not. The formula is derived based on the observation that the number of operations which effectively changes the string and which keeps it unchanged are related, and it can be simplified and computed using modular arithmetic operations and inverse modulo.

String Transformation - Complete Solution Guide

String Transformation is LeetCode problem 2851, 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

You are given two strings s and t of equal length n . You can perform the following operation on the string s : Remove a suffix of s of length l where 0 < l < n and append it at the start of s . For example, let s = 'abcd' then in one operation you can remove the suffix 'cd' and append it in front of s making s = 'cdab' . You are also given an integer k . Return the number of ways in which s can be transformed into t in exactly k operations. Since the answer can be large, return it modulo 10 9 +

Detailed Explanation

The problem asks us to find the number of ways to transform a string `s` into another string `t` using exactly `k` operations. Each operation involves taking a suffix of `s` (of length greater than 0 and less than `n`, where `n` is the length of `s`) and moving it to the beginning of the string. The goal is to count how many distinct sequences of `k` operations will result in `s` becoming equal to `t`, and the answer should be returned modulo 10^9 + 7.

Solution Approach

The solution involves the following steps: first, check if `t` can even be obtained by rotating `s` (i.e., `t` is a substring of `s + s`). If not, return 0. Then, calculate the LPS (Longest Proper Prefix which is also a Suffix) array for `s` using the KMP algorithm to determine the smallest period `p` of `s`. If such period exists (s is periodic), then some rotations will not change the string. Next, compute the result based on whether `s` equals `t` or not. The formula is derived based on the observation that the number of operations which effectively changes the string and which keeps it unchanged are related, and it can be simplified and computed using modular arithmetic operations and inverse modulo.

Step-by-Step Algorithm

Step 1: Check if `t` is a rotation of `s`. If `t` is not a substring of `s + s`, return 0.
Step 2: Compute the LPS (Longest Proper Prefix which is also a Suffix) array `pi` for string `s` using the KMP algorithm.
Step 3: Calculate the period `p` of `s`. If `s` is periodic, `p = n - pi[n-1]`, else `p=n`.
Step 4: Calculate `m = n / p`, which is the number of times the period repeats in the string.
Step 5: Calculate `n_inv = pow(n, MOD - 2, MOD)`, which is the modular inverse of `n` modulo `MOD` using Fermat's Little Theorem.
Step 6: Calculate `term_n_minus_1_pow_k = pow(n - 1, k, MOD)`, which is (n-1)^k modulo MOD.
Step 7: Calculate `term_neg_1_pow_k = (k % 2 == 0) ? 1 : (MOD - 1)`, which is (-1)^k modulo MOD.
Step 8: Calculate `m_mod = m % MOD`.
Step 9: Calculate `term_m_mul = (m_mod * term_n_minus_1_pow_k) % MOD`.
Step 10: If `s == t`, calculate the numerator as `(n_minus_m_mod * term_neg_1_pow_k + term_m_mul) % MOD`. Otherwise, calculate the numerator as `((MOD - m_mod) * term_neg_1_pow_k + term_m_mul) % MOD`.
Step 11: Calculate the final result as `(numerator * n_inv) % MOD` and return it.

Key Insights

Insight 1: If `t` cannot be obtained from `s` through rotations, the answer is 0. This can be checked by verifying that `t` is a substring of `s + s`.
Insight 2: The rotations can be represented as cyclic shifts. The key idea is to find the period `p` of the string `s`. If `s` is a periodic string, i.e., `s = (sub) * m`, where `sub` is some substring of length `p` and `m` is some integer, performing a cyclic shift by `p` will yield the same string. Knowing `p` helps determine the number of rotations that result in the same string. We find period `p` using KMP's LPS array.
Insight 3: The problem can be modeled as a linear equation modulo `MOD`. We need to calculate the number of ways to achieve `t` by a certain combination of rotations that change the string and rotations that do not. The core calculation is simplified by using modular arithmetic properties and Fermat's Little Theorem for calculating modular inverses.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Math, String, Dynamic Programming, String Matching.

Companies

Asked at: Atlassian, MathWorks, Snowflake.

String Transformation | LeetCode 2851

Hard

Math String Dynamic Programming String Matching

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int numberOfWays(String s, String t, long k) {
        int MOD = (int)1e9 + 7;
        int n = s.length();

        if ((s + s).indexOf(t) == -1) {
            return 0;
        }

        int[] pi = _compute_pi(s);
        int lps = pi[n - 1];
        int p = n;
        if (lps > 0 && n % (n - lps) == 0) {
            p = n - lps;
        }
        
        int m = n / p;

        long n_inv = power(n, MOD - 2, MOD);
        long term_n_minus_1_pow_k = power(n - 1, k, MOD);
        
        long term_neg_1_pow_k;
        if (k % 2 == 0) {
            term_neg_1_pow_k = 1;
        } else {
            term_neg_1_pow_k = MOD - 1;
        }

        long m_mod = m % MOD;
        
        long term_m_mul = (m_mod * term_n_minus_1_pow_k) % MOD;
        
        if (s.equals(t)) {
            long n_minus_m_mod = (n - m) % MOD;
            long term_nm_mul = (n_minus_m_mod * term_neg_1_pow_k) % MOD;
            long numerator = (term_nm_mul + term_m_mul) % MOD;
            long result = (numerator * n_inv) % MOD;
            return (int)result;
        } else {
            long term_neg_m_mul = (((long)MOD - m_mod) * term_neg_1_pow_k) % MOD;
            long numerator = (term_neg_m_mul + term_m_mul) % MOD;
            long result = (numerator * n_inv) % MOD;
            return (int)result;
        }
    }

    private int[] _compute_pi(String pattern) {
        int p_len = pattern.length();
        int[] pi = new int[p_len];
        for (int i = 1; i < p_len; i++) {
            int j = pi[i - 1];
            while (j > 0 && pattern.charAt(i) != pattern.charAt(j)) {
                j = pi[j - 1];
            }
            if (pattern.charAt(i) == pattern.charAt(j)) {
                j += 1;
            }
            pi[i] = j;
        }
        return pi;
    }
    
    private long power(long base, long exp, int mod) {
        long res = 1;
        base %= mod;
        while (exp > 0) {
            if (exp % 2 == 1) res = (res * base) % mod;
            base = (base * base) % mod;
            exp /= 2;
        }
        return res;
    }
}

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Problem Info

Description

Problem 2851: String Transformation - Detailed explanation and solution approaches.

Topics

Math String Dynamic Programming String Matching

Companies

AtlassianMathWorksSnowflake

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:2851

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

11 steps

1
Step 1: Check if `t` is a rotation of `s`. If `t` is not a substring of `s + s`, return 0.
2
Step 2: Compute the LPS (Longest Proper Prefix which is also a Suffix) array `pi` for string `s` using the KMP algorithm.
3
Step 3: Calculate the period `p` of `s`. If `s` is periodic, `p = n - pi[n-1]`, else `p=n`.
4
Step 4: Calculate `m = n / p`, which is the number of times the period repeats in the string.
5
Step 5: Calculate `n_inv = pow(n, MOD - 2, MOD)`, which is the modular inverse of `n` modulo `MOD` using Fermat's Little Theorem.
6
Step 6: Calculate `term_n_minus_1_pow_k = pow(n - 1, k, MOD)`, which is (n-1)^k modulo MOD.
7
Step 7: Calculate `term_neg_1_pow_k = (k % 2 == 0) ? 1 : (MOD - 1)`, which is (-1)^k modulo MOD.
8
Step 8: Calculate `m_mod = m % MOD`.
9
Step 9: Calculate `term_m_mul = (m_mod * term_n_minus_1_pow_k) % MOD`.
10
Step 10: If `s == t`, calculate the numerator as `(n_minus_m_mod * term_neg_1_pow_k + term_m_mul) % MOD`. Otherwise, calculate the numerator as `((MOD - m_mod) * term_neg_1_pow_k + term_m_mul) % MOD`.
11
Step 11: Calculate the final result as `(numerator * n_inv) % MOD` and return it.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the KMP algorithm (O(n)) and the string comparison operation (O(n)). Calculating power modulo is O(log k) but can be considered smaller than linear time. The space complexity is dominated by the LPS array, which needs to store prefix lengths for each position in the string.

📖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

Brute Force

Try all possible combinations of `k` rotations. This would involve recursively exploring different suffix lengths for each operation and keeping track of the number of ways to reach `t`. However, this is highly inefficient.

Time: O(n^k), where n is string length and k is the number of operationsSpace: O(k) (for recursion depth)

Trade-offs: This is a simple but highly inefficient approach. It's not practical for larger values of `k` and `n` because it exceeds time limit.

✨

Key Insights

1Insight 1: If `t` cannot be obtained from `s` through rotations, the answer is 0. This can be checked by verifying that `t` is a substring of `s + s`.
2Insight 2: The rotations can be represented as cyclic shifts. The key idea is to find the period `p` of the string `s`. If `s` is a periodic string, i.e., `s = (sub) * m`, where `sub` is some substring of length `p` and `m` is some integer, performing a cyclic shift by `p` will yield the same string. Knowing `p` helps determine the number of rotations that result in the same string. We find period `p` using KMP's LPS array.
3Insight 3: The problem can be modeled as a linear equation modulo `MOD`. We need to calculate the number of ways to achieve `t` by a certain combination of rotations that change the string and rotations that do not. The core calculation is simplified by using modular arithmetic properties and Fermat's Little Theorem for calculating modular inverses.

📚

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: When `t` cannot be obtained by rotating `s`. The solution correctly returns 0 in this case by checking if `t` is a substring of `s + s`.
Edge case 2: When `s` is equal to `t`. The solution handles this case separately, correctly calculating the result based on the formula described above.
Edge case 3: When `k` is very large. The solution uses modular arithmetic to avoid integer overflow.
Edge case 4: when the string 's' is aperiodic (i.e., does not repeat itself). The LPS array can correctly determine this.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the modulo operation correctly. It's crucial to apply the modulo operation after every multiplication and addition to prevent integer overflow.

✗Common mistake 2: Not considering the case where `t` cannot be obtained from `s` by rotations. The problem statement specifies that we should return 0 in this case.

✗Common mistake 3: Failing to correctly calculate the period `p` of the string `s` using the LPS array.

✗Common mistake 4: Incorrectly calculating modular inverse. Using the property power(a, MOD - 2, MOD) to find the modular inverse

✅

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: Cryptography. Cyclic shifts and modular arithmetic are fundamental concepts in cryptography for encryption and decryption algorithms.

🏦Real-world scenario 2: Data compression. The KMP algorithm and LPS arrays are used in data compression algorithms for pattern matching and finding repeated sequences.

🎮Real-world scenario 3: Bioinformatics. String matching algorithms like KMP are used in DNA sequencing and analyzing genetic patterns.

📱Real-world scenario 4: Rolling hash functions often utilize modular arithmetic.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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: October 30, 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

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

@AlgoMasterJul 13

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

@InterviewProJul 6

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

@CodeNinjaJun 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.

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 "String Transformation"?

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

🤖 Need help with this problem?