Total Characters in String After Transformations I - Video Solutions

Problem 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.