LeetCode #2851 — HARD

String Transformation

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

Solve on LeetCode
The Problem

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 109 + 7.

Example 1:

Input: s = "abcd", t = "cdab", k = 2
Output: 2
Explanation: 
First way:
In first operation, choose suffix from index = 3, so resulting s = "dabc".
In second operation, choose suffix from index = 3, so resulting s = "cdab".

Second way:
In first operation, choose suffix from index = 1, so resulting s = "bcda".
In second operation, choose suffix from index = 1, so resulting s = "cdab".

Example 2:

Input: s = "ababab", t = "ababab", k = 1
Output: 2
Explanation: 
First way:
Choose suffix from index = 2, so resulting s = "ababab".

Second way:
Choose suffix from index = 4, so resulting s = "ababab".

Constraints:

  • 2 <= s.length <= 5 * 105
  • 1 <= k <= 1015
  • s.length == t.length
  • s and t consist of only lowercase English alphabets.
Patterns Used
📊Dynamic Programming→🔎String Matching→

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: 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 109 + 7.

Baseline thinking

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

Pattern signal: Math · Dynamic Programming · String Matching

Example 1

"abcd"
"cdab"
2

Example 2

"ababab"
"ababab"
1
Step 02

Core Insight

What unlocks the optimal approach

  • String <code>t</code> can be only constructed if it is a rotated version of string <code>s</code>.
  • Use KMP algorithm or Z algorithm to find the number of indices from where <code>s</code> is equal to <code>t</code>.
  • Use Dynamic Programming to count the number of ways.
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 #2851: String Transformation
class Solution {
    private static final int M = 1000000007;

    private int add(int x, int y) {
        if ((x += y) >= M) {
            x -= M;
        }
        return x;
    }

    private int mul(long x, long y) {
        return (int) (x * y % M);
    }

    private int[] getZ(String s) {
        int n = s.length();
        int[] z = new int[n];
        for (int i = 1, left = 0, right = 0; i < n; ++i) {
            if (i <= right && z[i - left] <= right - i) {
                z[i] = z[i - left];
            } else {
                int z_i = Math.max(0, right - i + 1);
                while (i + z_i < n && s.charAt(i + z_i) == s.charAt(z_i)) {
                    z_i++;
                }
                z[i] = z_i;
            }
            if (i + z[i] - 1 > right) {
                left = i;
                right = i + z[i] - 1;
            }
        }
        return z;
    }

    private int[][] matrixMultiply(int[][] a, int[][] b) {
        int m = a.length, n = a[0].length, p = b[0].length;
        int[][] r = new int[m][p];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < p; ++j) {
                for (int k = 0; k < n; ++k) {
                    r[i][j] = add(r[i][j], mul(a[i][k], b[k][j]));
                }
            }
        }
        return r;
    }

    private int[][] matrixPower(int[][] a, long y) {
        int n = a.length;
        int[][] r = new int[n][n];
        for (int i = 0; i < n; ++i) {
            r[i][i] = 1;
        }
        int[][] x = new int[n][n];
        for (int i = 0; i < n; ++i) {
            System.arraycopy(a[i], 0, x[i], 0, n);
        }
        while (y > 0) {
            if ((y & 1) == 1) {
                r = matrixMultiply(r, x);
            }
            x = matrixMultiply(x, x);
            y >>= 1;
        }
        return r;
    }

    public int numberOfWays(String s, String t, long k) {
        int n = s.length();
        int[] dp = matrixPower(new int[][] {{0, 1}, {n - 1, n - 2}}, k)[0];
        s += t + t;
        int[] z = getZ(s);
        int m = n + n;
        int result = 0;
        for (int i = n; i < m; ++i) {
            if (z[i] >= n) {
                result = add(result, dp[i - n == 0 ? 0 : 1]);
            }
        }
        return result;
    }
}
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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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.

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