Modifies String Transformation

Modified String Transformation 

Problem Statement

You are given a string s of lowercase English letters and an integer t representing the number of transformation steps.

Each transformation proceeds as follows:

  • If a character is 'z', it is transformed into "ab" (two characters).

  • Any other character ch is replaced with the next character in the English alphabet (e.g., 'a' becomes 'b', 'b' becomes 'c', and so on).

Your task is to calculate the length of the string after t transformations, modulo 10⁹ + 7.

Example

Input:

s = "abcyy"
t = 2

Output:

7

Explanation:

  • After 1st transformation: "a" → "b", "b" → "c", "c" → "d", "y" → "z", "y" → "z" → "bcdzz"

  • After 2nd transformation: "b" → "c", "c" → "d", "d" → "e", "z" → ab, "z" → ab → "cdeabab"

  • Final length: 7

Best Data Structure for the Job

We don't need complex data structures. A frequency array of 26 elements suffices to track how many of each character exists at each step. It prevents exponential memory usage and simulates the process mathematically.

Different Approaches

1. Brute Force (String Reconstruction)

Idea: Simulate the entire transformation using actual string concatenation for each character.

class Solution {
    public int lengthAfterTransformations(String s, int t) {
        final int MOD = 1_000_000_007;
        StringBuilder sb = new StringBuilder(s);

        for (int step = 0; step < t; step++) {
            StringBuilder next = new StringBuilder();
            for (char ch : sb.toString().toCharArray()) {
                if (ch == 'z') {
                    next.append("ab");
                } else {
                    next.append((char)(ch + 1));
                }
            }
            sb = next;
        }

        return sb.length() % MOD;
    }
}

2. Recursive Simulation (Inefficient)

Idea: Model each transformation recursively and return the total length instead of constructing strings.

class Solution {
    public int lengthAfterTransformations(String s, int t) {
        final int MOD = 1_000_000_007;
        return dfs(s, t, MOD);
    }

    private int dfs(String s, int t, int MOD) {
        if (t == 0) return s.length();
        int total = 0;

        for (char ch : s.toCharArray()) {
            if (ch == 'z') {
                total = (total + dfs("ab", t - 1, MOD)) % MOD;
            } else {
                total = (total + dfs("" + (char)(ch + 1), t - 1, MOD)) % MOD;
            }
        }

        return total;
    }
}

3. Optimized Dynamic Programming with Frequency Arrays (Efficient)

Idea: Track how many of each character exists without constructing the actual string. Simulate character transformations using a fixed-size frequency array for 26 letters.

class Solution {
    public int lengthAfterTransformations(String s, int t) {
        final int MOD = 1_000_000_007;
        long[] dp = new long[26];

        for (char ch : s.toCharArray()) {
            dp[ch - 'a']++;
        }

        for (int step = 0; step < t; step++) {
            long[] newDp = new long[26];

            for (int i = 0; i < 26; i++) {
                if (i == 25) { // 'z'
                    newDp[0] = (newDp[0] + dp[i]) % MOD;
                    newDp[1] = (newDp[1] + dp[i]) % MOD;
                } else {
                    newDp[i + 1] = (newDp[i + 1] + dp[i]) % MOD;
                }
            }

            dp = newDp;
        }

        long total = 0;
        for (long count : dp) {
            total = (total + count) % MOD;
        }

        return (int) total;
    }
}

⚠️ Common Mistakes and Pitfalls

  1. Ignoring the 'z' to "ab" transformation: A common mistake is treating it as a single character replacement.

  2. Incorrect Modulo Usage: Not applying modulo after each addition can lead to integer overflows.

  3. Using Strings to Simulate Large Inputs: Will result in Memory Limit Exceeded or timeout.

📊 Time and Space Complexity Comparison

Approach       Time ComplexitySpace Complexity
Brute Force        O(N * 2^t)            O(N * 2^t)
Recursive Simulation        O((2^t)^N)            High (stack + string copies)
Frequency-based Dynamic Programming        O(t * 26)            O(26)

🌍Real-world Applications

  • Language Evolution Engines: Simulating rules of transformation in string-based systems.

  • DNA Sequence Growth Models: Biological modeling where one base may become multiple others.

  • Automated Refactoring Engines: Tracking transformations in code or documents over multiple iterations.

  • Data Compression Analysis: Predicting post-processing size without actual decompression.


I hope you found this article insightful! Keep exploring, keep learning, and don’t forget to check back daily for more exciting problem breakdowns. If you know someone who would benefit from this, share it with them and help them grow too! That’s it for today—see you in the next post!

                                                                                                                                         Signing off!! 

                                                                                                Master DSA One Problem at a Time :)

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