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:
'z', replace it with the string "ab".'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:
'a' becomes 'b''b' becomes 'c''c' becomes 'd''y' becomes 'z''y' becomes 'z'"bcdzz"'b' becomes 'c''c' becomes 'd''d' becomes 'e''z' becomes "ab"'z' becomes "ab""cdeabab""cdeabab", which has 7 characters.Example 2:
Input: s = "azbk", t = 1
Output: 5
Explanation:
'a' becomes 'b''z' becomes "ab"'b' becomes 'c''k' becomes 'l'"babcl""babcl", which has 5 characters.
Constraints:
1 <= s.length <= 105s consists only of lowercase English letters.1 <= t <= 105Problem Overview: You start with a string s. During each transformation, every character moves to the next letter in the alphabet. When a character reaches 'z', it expands into the two-character string "ab". After performing t transformations, return the total number of characters in the resulting string.
Approach 1: Simulate Transformations with Length Calculation (Time: O(n + 26t), Space: O(26))
Instead of rebuilding the string every round, track how many times each character appears. Use a frequency array of size 26 where count[i] stores the number of occurrences of the i-th letter. For each transformation, shift counts forward: characters 'a' through 'y' move to the next index, while 'z' produces two characters, increasing the counts of 'a' and 'b'. Repeat this update for t rounds. The total length at any step is simply the sum of all frequencies. This approach avoids string concatenation and keeps the state compact using a small fixed array.
The key insight is that the transformation only depends on character counts, not the order of characters. Using counting makes each step constant time with respect to alphabet size. This pattern appears frequently in problems involving repeated string updates and can be viewed as a small-state dynamic programming process combined with simple counting.
Approach 2: Mathematical Simplification Using Exponents (Time: O(n + log t) or O(n + 26t) precompute, Space: O(26))
Each character contributes to the final length depending on how many times it reaches 'z' within t steps. When that happens, the character splits into two new characters which continue transforming independently. Instead of simulating every step, compute how a single character grows after t transformations. Precompute the growth for each letter by observing when the chain hits 'z' and creates additional characters.
This reduces the problem to summing contributions from the initial characters. For example, if a letter becomes 'z' after k steps, the remaining transformations generate extra characters starting from 'a' and 'b'. Using precomputed growth values or exponent-style recurrence avoids repeatedly shifting arrays and scales better when t becomes large. The reasoning relies on simple alphabet arithmetic and ideas commonly used in math-driven DP transitions.
Recommended for interviews: The counting simulation is the expected solution. It shows you recognize that the string order is irrelevant and that only frequencies matter. Brute string simulation would explode in size, while the 26-slot counting DP keeps updates predictable and efficient. The mathematical simplification demonstrates deeper optimization thinking but usually comes after you explain the counting model.
This approach focuses on simulating each transformation step by step. Begin with the initial length of the string. For each character in the string, calculate the change in length based on the transformation rules. If the character is not 'z', it contributes 1 to the final length; if it is 'z', it contributes 2 due to its transformation to 'ab'. Iterate this process for each transformation up to t times, updating the total length accordingly.
The function total_length_after_transformations simulates each transformation for a given number of t, adjusting the length by accounting for each rule's effect on each character within the string s for t times.
Python
JavaScript
Time Complexity: O(t * n), where n is the length of the string because it processes every character for each transformation.
Space Complexity: O(n), needing to store the potentially transformed string.
This mathematical method avoids simulating every transformation by leveraging the predictable pattern that the transformations create. Consider that each non-'z' character simply advances through the alphabet. Importantly, 'z' contributes significantly due to its transformation to 'ab'. Devise a formula or recurring relation that describes the total length after t transformations. Solve this by setting up a simulation using mathematical functions that model the string length changes without modifying string values.
The total_length_formula uses the property of exponential growth when 'z' is present, doubling contributions each time transformation occurs. This avoids actually constructing the string, hence limiting memory use and improving performance.
Time Complexity: O(n + log t).
Space Complexity: O(1). This approach eliminates the need to store intermediate string states fully.
We define f[i][j] to represent the count of the j-th letter in the alphabet after i transformations. Initially, f[0][j] is the count of the j-th letter in the string s.
After each transformation, the count of the j-th letter in the alphabet can be calculated as follows:
$
\begin{align}
f[i][0] &= f[i - 1][25] \
f[i][1] &= f[i - 1][0] + f[i - 1][25] \
f[i][2] &= f[i - 1][1] \
f[i][3] &= f[i - 1][2] \
&\vdots \
f[i][25] &= f[i - 1][24]
\end{align}
The answer is f[t][0] + f[t][1] + ldots + f[t][25].
Since the answer can be very large, we take the result modulo 10^9 + 7.
The time complexity is O(t times |\Sigma|), and the space complexity is O(t times |\Sigma|), where |\Sigma|$ is the size of the alphabet.
Python
Java
C++
Go
TypeScript
| Approach | Complexity |
|---|---|
| Simulate Transformations with Length Calculation | Time Complexity: O(t * n), where n is the length of the string because it processes every character for each transformation. |
| Mathematical Simplification Using Exponents | Time Complexity: O(n + log t). |
| Recurrence | — |
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Naive String Simulation | Exponential | O(n * growth) | Conceptual understanding only; impractical because the string grows rapidly |
| Simulate with Character Counting | O(n + 26t) | O(26) | Standard solution when t is moderate; avoids constructing the string |
| Mathematical Contribution / Precomputed Growth | O(n + log t) or O(n + 26t) | O(26) | Large t values where repeated simulation becomes expensive |
Total Characters in String After Transformations I | Made Easy | Leetcode 3335 | codestorywithMIK • codestorywithMIK • 13,153 views views
Watch 9 more video solutions →Practice Total Characters in String After Transformations I with our built-in code editor and test cases.
Practice on FleetCodePractice this problem
Open in Editor