#2851Hard~60 min

String Transformation

Time O(n + logk) · Space O(n) · Official statement on LeetCode

string dynamic programming kmp algorithm

Solutions

// Time:  O(n + logk)
// Space: O(n)

// dp, math, kmp
class Solution {
public:
    int numberOfWays(string s, string t, long long k) {
        const int n = size(s);
        vector<int> dp(2);
        dp[1] = (((pow(n - 1, k, MOD) - pow(-1, k, MOD)) * pow(n, MOD - 2, MOD)) % MOD + MOD) % MOD;
        dp[0] = (dp[1] + pow(-1, k, MOD)) % MOD;
        int result = 0;
        for (const auto& i : KMP(s + s.substr(0, size(s) - 1), t)) {
            result = (result + dp[static_cast<int>(i != 0)]) % MOD;
        }
        return result;
    }

private:
    int64_t pow(int64_t a, int64_t b, int64_t m) {
        a %= m;
        int64_t result = 1;
        while (b) {
            if (b & 1) {
                result = (result * a) % m;
            }
            a = (a * a) % m;
            b >>= 1;
        }
        return result;
    }

    vector<int> KMP(const string& text, const string& pattern) {
        vector<int> result;
        const vector<int> prefix = getPrefix(pattern);
        int j = -1;
        for (int i = 0; i < text.length(); ++i) {
            while (j > -1 && pattern[j + 1] != text[i]) {
                j = prefix[j];
            }
            if (pattern[j + 1] == text[i]) {
                ++j;
            }
            if (j == pattern.length() - 1) {
                result.emplace_back(i - j);
                j = prefix[j];
            }
        }
        return result;
    }

    vector<int> getPrefix(const string& pattern) {
        vector<int> prefix(pattern.length(), -1);
        int j = -1;
        for (int i = 1; i < pattern.length(); ++i) {
            while (j > -1 && pattern[j + 1] != pattern[i]) {
                j = prefix[j];
            }
            if (pattern[j + 1] == pattern[i]) {
                ++j;
            }
            prefix[i] = j;
        }
        return prefix;
    }

    const int MOD = 1e9 + 7;
};

// Time:  O(n + logk)
// Space: O(n)
// dp, matrix exponentiation, kmp
class Solution2 {
public:
    int numberOfWays(string s, string t, long long k) {
        const int n = size(s);
        vector<vector<int>> T = {{    0,           1},
                                 {n - 1, (n - 1) - 1}};
        const auto dp = matrixMult({{1, 0}}, matrixExpo(T, k))[0];  // [dp[0], dp[1]] * T^k
        int result = 0;
        for (const auto& i : KMP(s + s.substr(0, size(s) - 1), t)) {
            result = (result + dp[static_cast<int>(i != 0)]) % MOD;
        }
        return result;
    }

private:
    vector<vector<int>> matrixExpo(const vector<vector<int>>& A, int64_t pow) {
        vector<vector<int>> result(A.size(), vector<int>(A.size()));
        vector<vector<int>> A_exp(A);
        for (int i = 0; i < A.size(); ++i) {
            result[i][i] = 1;
        }
        while (pow) {
            if (pow % 2 == 1) {
                result = matrixMult(result, A_exp);
            }
            A_exp = matrixMult(A_exp, A_exp);
            pow /= 2;
        }
        return result;
    }

    vector<vector<int>> matrixMult(const vector<vector<int>>& A, const vector<vector<int>>& B) {
        vector<vector<int>> result(A.size(), vector<int>(B[0].size()));
        for (int i = 0; i < A.size(); ++i) {
            for (int j = 0; j < B[0].size(); ++j) {
                int64_t entry = 0;
                for (int k = 0; k < B.size(); ++k) {
                    entry = (static_cast<int64_t>(A[i][k]) * B[k][j] % MOD + entry) % MOD;
                }
                result[i][j] = static_cast<int>(entry);
            }
        }
        return result;
    }

    vector<int> KMP(const string& text, const string& pattern) {
        vector<int> result;
        const vector<int> prefix = getPrefix(pattern);
        int j = -1;
        for (int i = 0; i < text.length(); ++i) {
            while (j > -1 && pattern[j + 1] != text[i]) {
                j = prefix[j];
            }
            if (pattern[j + 1] == text[i]) {
                ++j;
            }
            if (j == pattern.length() - 1) {
                result.emplace_back(i - j);
                j = prefix[j];
            }
        }
        return result;
    }

    vector<int> getPrefix(const string& pattern) {
        vector<int> prefix(pattern.length(), -1);
        int j = -1;
        for (int i = 1; i < pattern.length(); ++i) {
            while (j > -1 && pattern[j + 1] != pattern[i]) {
                j = prefix[j];
            }
            if (pattern[j + 1] == pattern[i]) {
                ++j;
            }
            prefix[i] = j;
        }
        return prefix;
    }

    const int MOD = 1e9 + 7;
};

// Time:  O(n + logk)
// Space: O(n)
// dp, matrix exponentiation, z-function
class Solution3 {
public:
    int numberOfWays(string s, string t, long long k) {
        const int n = size(s);
        vector<vector<int>> T = {{    0,           1},
                                 {n - 1, (n - 1) - 1}};
        const auto dp = matrixMult({{1, 0}}, matrixExpo(T, k))[0];  // [dp[0], dp[1]] * T^k
        const auto& z = z_function(t + s + s.substr(0, size(s) - 1));
        int result = 0;
        for (int i = 0; i < n; ++i) {
            if (z[i + size(t)] >= size(t)) {
                result = (result + dp[static_cast<int>(i != 0)]) % MOD;
            }
        }
        return result;
    }

private:
    vector<vector<int>> matrixExpo(const vector<vector<int>>& A, int64_t pow) {
        vector<vector<int>> result(A.size(), vector<int>(A.size()));
        vector<vector<int>> A_exp(A);
        for (int i = 0; i < A.size(); ++i) {
            result[i][i] = 1;
        }
        while (pow) {
            if (pow % 2 == 1) {
                result = matrixMult(result, A_exp);
            }
            A_exp = matrixMult(A_exp, A_exp);
            pow /= 2;
        }
        return result;
    }

    vector<vector<int>> matrixMult(const vector<vector<int>>& A, const vector<vector<int>>& B) {
        vector<vector<int>> result(A.size(), vector<int>(B[0].size()));
        for (int i = 0; i < A.size(); ++i) {
            for (int j = 0; j < B[0].size(); ++j) {
                int64_t entry = 0;
                for (int k = 0; k < B.size(); ++k) {
                    entry = (static_cast<int64_t>(A[i][k]) * B[k][j] % MOD + entry) % MOD;
                }
                result[i][j] = static_cast<int>(entry);
            }
        }
        return result;
    }

    // Template: https://cp-algorithms.com/string/z-function.html
    vector<int> z_function(const string& s) {  // Time: O(n), Space: O(n)
        vector<int> z(size(s));
        for (int i = 1, l = 0, r = 0; i < size(z); ++i) {
            if (i <= r) {
                z[i] = min(r - i + 1, z[i - l]);
            }
            while (i + z[i] < size(z) && s[z[i]] == s[i + z[i]]) {
                ++z[i];
            }
            if (i + z[i] - 1 > r) {
                l = i, r = i + z[i] - 1;
            }
        }
        return z;
    }

    const int MOD = 1e9 + 7;
};

Beginner Explanation

What is String Transformation?

String Transformation (LeetCode #2851) is a Hard problem that primarily trains string.

How to think about it

Restate the goal in your own words before coding.
Work a tiny example by hand so the invariant becomes obvious.
Identify the pattern — this problem aligns with dynamic programming and kmp algorithm.
Only then translate the idea into code.

Why this problem matters

Hard problems force you to combine patterns and prove complexity carefully — interview gold. Official solution notes mention: DP, Matrix Exponentiation, Math, Z-Function.

AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.

Interview Walkthrough

Interview approach for String Transformation

Opening (30–60 seconds)

Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
State a brute force so the interviewer knows you can solve it naively.
Propose the optimal direction tied to dynamic programming and kmp algorithm.

Core solution narrative

Define the state you track (pointers, DP cell, set membership, stack top, etc.).
Explain the transition when you process the next element.
Call out time (O(n + logk)) and space (O(n)) before coding.
Code cleanly; narrate variable names.

What interviewers listen for

Correctness on edge cases
Complexity honesty
Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)

Follow-up questions they may ask

Can you solve it with less memory?
What if the input stream is infinite / doesn't fit in RAM?
How would tests look for adversarial inputs?

Optimized Approach

Optimized solution notes

The reference solutions on AlgoForge target O(n + logk) time and O(n) space.

Pattern focus: dynamic programming and kmp algorithm

Use the pattern as a checklist:

dynamic programming — confirm the invariant holds after each step
kmp algorithm — confirm the invariant holds after each step

Multiple methods appear in the source solutions — compare them and explain when each is preferable.

Implementation tips

Prefer readable names over micro-optimizations in interviews.
Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
After AC-level logic, re-scan for off-by-one and null checks.

Complexity Analysis

Complexity

Measure	Bound
Time	O(n + logk)
Space	O(n)

How to justify this in an interview

Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
Space: include hash maps, recursion stack, and output allocation when the problem asks for it.

If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.

Common Mistakes

Common mistakes on String Transformation

Skipping edge cases — empty collections, single-element inputs, max constraints.
Wrong invariant for dynamic programming and kmp algorithm — updating state too early or too late.
Mutating input unexpectedly when the problem forbids it.
Off-by-one in windows, ranges, or binary search bounds.
Ignoring overflow / precision for integer arithmetic problems.
Overengineering — jumping to an advanced structure when a simpler approach works.

Alternative Approaches

Alternatives

The source file includes more than one method. Compare:

Primary optimized path — best complexity for typical interviews.
Secondary approach — often brute force, sorting-based, or space-optimized variant.

Practice articulating when you would pick each (constraints, readability, follow-ups).

Edge Cases

Edge cases checklist

Minimum input size
Maximum input size / time limits
Duplicates and already-sorted input
Negative numbers / zeros (if applicable)
Disconnected structures (graphs/trees)
Single path vs branching recursion depth

Pattern Recognition

Spotting this pattern

Signal phrases that point to dynamic programming and kmp algorithm:

Sorted input or ability to sort without changing the answer class
Need for contiguous subarray / substring → consider sliding window
Need for O(1) membership → hash set/map
Optimal substructure + overlapping subproblems → DP
Connectivity / components → graph DFS/BFS or Union-Find

Primary topics: string.

Follow-up Interview Questions

Follow-ups

How does the solution change if the input is a stream?
Can you solve it in-place?
What if duplicates must be handled differently?
How would you parallelize the approach?
Design tests that would break a buggy implementation.

Practice Recommendations

What to practice next

Re-solve String Transformation in a second language (cpp, python).
Drill 3–5 more problems tagged string.
Teach the solution out loud in under 5 minutes.
Add this problem to your revision calendar in 3 days and 14 days.

Visualization

Conceptual diagram for String Transformation: show input structure (string), highlight the moving parts of the dynamic programming and kmp algorithm approach, and annotate each step with the maintained invariant and complexity.

Study checklist

Read the official problem statement on LeetCode
Solve on paper / whiteboard first
Implement the dynamic programming and kmp algorithm approach
Verify edge cases from the checklist
State time and space complexity aloud
Compare with the AlgoForge reference solution
Schedule a revision session

Revision notes

String Transformation (#2851) — Hard. Pattern: dynamic programming and kmp algorithm. Complexity: O(n + logk) time / O(n) space. Re-derive the invariant before coding.

FAQs

What is the time complexity of String Transformation?+

The reference solutions aim for O(n + logk) time and O(n) space. Always re-derive complexity from the code you write in the interview.

What pattern does String Transformation use?+

It primarily maps to dynamic programming and kmp algorithm, within the broader topic of string.

Is String Transformation good for interviews?+

Yes — as a Hard problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.

Where can I read the official statement?+

Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/string-transformation/

#	Title	Difficulty	Topics
3036	Number of Subarrays That Match a Pattern II	Hard	string
3037	Find Pattern in Infinite Stream II	Hard	string
28	Implement strStr()	Easy	string
459	Repeated Substring Pattern	Easy	string
1455	Check If a Word Occurs As a Prefix of Any Word in a Sentence	Easy	string
1668	Maximum Repeating Substring	Easy	string
1910	Remove All Occurrences of a Substring	Medium	string
2800	Shortest String That Contains Three Strings	Medium	string