Scale Engineer

Total Characters in String After Transformations I

Med
#2962Time: O(t * L_avg)Space: O(L_max)
Patterns
MathDynamic ProgrammingCounting
Data structures
Hash TableString

Prompt

You are given a string s and an integer t, representing the number of transformations to perform. In one transformation, every character in s is replaced according to the following rules:

  • If the character is 'z', replace it with the string "ab".
  • Otherwise, replace it with the next character in the alphabet. For example, 'a' is replaced with 'b', 'b' is replaced with 'c', and so on.

Return the length of the resulting string after exactly t transformations.

Since the answer may be very large, return it modulo 109 + 7.

 

Example 1:

Input: s = "abcyy", t = 2

Output: 7

Explanation:

  • First Transformation (t = 1):
    • 'a' becomes 'b'
    • 'b' becomes 'c'
    • 'c' becomes 'd'
    • 'y' becomes 'z'
    • 'y' becomes 'z'
    • String after the first transformation: "bcdzz"
  • Second Transformation (t = 2):
    • 'b' becomes 'c'
    • 'c' becomes 'd'
    • 'd' becomes 'e'
    • 'z' becomes "ab"
    • 'z' becomes "ab"
    • String after the second transformation: "cdeabab"
  • Final Length of the string: The string is "cdeabab", which has 7 characters.

Example 2:

Input: s = "azbk", t = 1

Output: 5

Explanation:

  • First Transformation (t = 1):
    • 'a' becomes 'b'
    • 'z' becomes "ab"
    • 'b' becomes 'c'
    • 'k' becomes 'l'
    • String after the first transformation: "babcl"
  • Final Length of the string: The string is "babcl", which has 5 characters.

 

Constraints:

  • 1 <= s.length <= 105
  • s consists only of lowercase English letters.
  • 1 <= t <= 105

Approaches

3 approaches with complexity analysis and trade-offs.

The most straightforward approach is to directly simulate the transformation process. We can loop t times, and in each iteration, we build a new string by applying the transformation rules to every character of the current string. This method is easy to conceptualize but is highly impractical due to the potential for exponential growth in the string's length.

Algorithm

  • Start with the initial string s.
  • Loop t times, representing each transformation.
  • Inside the loop, create a new StringBuilder to build the string for the next state.
  • Iterate through each character of the current string.
  • If the character is 'z', append the string "ab" to the StringBuilder.
  • Otherwise, append the next character in the alphabet (e.g., 'a' becomes 'b').
  • After iterating through all characters, replace the current string with the string generated by the StringBuilder.
  • After t transformations, the length of the final string is the answer. Since the length can be very large, this approach is not practical.

Walkthrough

This method involves a step-by-step simulation of the process described in the problem. We maintain the string as it transforms. For each of the t steps, we construct a new string based on the transformations of the characters in the previous string.

For example, if s = "az" and t = 1:

  1. We start with "az".
  2. We create a new empty string.
  3. The first character 'a' transforms to 'b'. The new string is now "b".
  4. The second character 'z' transforms to "ab". The new string is now "bab".
  5. After one transformation, the result is "bab" and its length is 3.

While simple, this approach is not feasible for the given constraints because the string length can become enormous. If s contains many 'z's, the length can grow very rapidly. For instance, a single 'z' becomes "ab" (length 2), which then becomes "bc" (length 2), then "cd" (length 2), and so on, until it becomes "yz" which transforms to "z(ab)", resulting in "zab" (length 3). The growth becomes complex and leads to performance issues.

// NOTE: This code is for illustrative purposes only and will fail due to TLE/MLE.class Solution {    public int totalCharacters(String s, int t) {        String currentString = s;        for (int i = 0; i < t; i++) {            StringBuilder nextString = new StringBuilder();            for (char c : currentString.toCharArray()) {                if (c == 'z') {                    nextString.append("ab");                } else {                    nextString.append((char)(c + 1));                }            }            currentString = nextString.toString();            // The string length can exceed memory limits here.        }        // The length can also exceed standard integer types.        long length = currentString.length();        int MOD = 1_000_000_007;        return (int)(length % MOD);    }}

Complexity

Time

O(t * L_avg)

Space

O(L_max)

Trade-offs

Pros

  • Very simple to understand and implement.

Cons

  • Extremely inefficient in terms of both time and space.

  • The length of the string can grow exponentially, quickly exceeding memory limits (Memory Limit Exceeded).

  • Processing these very long strings at each step is very slow, leading to a timeout (Time Limit Exceeded).

Solutions

class Solution {public  int lengthAfterTransformations(String s, int t) {    final int mod = (int)1 e9 + 7;    int[][] f = new int[t + 1][26];    for (char c : s.toCharArray()) {      f[0][c - 'a']++;    }    for (int i = 1; i <= t; ++i) {      f[i][0] = f[i - 1][25] % mod;      f[i][1] = (f[i - 1][0] + f[i - 1][25]) % mod;      for (int j = 2; j < 26; j++) {        f[i][j] = f[i - 1][j - 1] % mod;      }    }    int ans = 0;    for (int j = 0; j < 26; ++j) {      ans = (ans + f[t][j]) % mod;    }    return ans;  }}

Video walkthrough

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