Total Characters in String After Transformations II - Video Solutions
Total Characters in String After Transformations II | Matrix Exponentiation | Leetcode 3337
Total Characters in String After Transformations II - Video Solution
Watch 10 video solutions for Total Characters in String After Transformations II, a hard level problem involving Hash Table, Math, String. This walkthrough by Techdose has 6,424 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
Problem Statement
You are given a string s consisting of lowercase English letters, an integer t representing the number of transformations to perform, and an array nums of size 26. In one transformation, every character in s is replaced according to the following rules:
- Replace
s[i]with the nextnums[s[i] - 'a']consecutive characters in the alphabet. For example, ifs[i] = 'a'andnums[0] = 3, the character'a'transforms into the next 3 consecutive characters ahead of it, which results in"bcd". - The transformation wraps around the alphabet if it exceeds
'z'. For example, ifs[i] = 'y'andnums[24] = 3, the character'y'transforms into the next 3 consecutive characters ahead of it, which results in"zab".
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, nums = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2]
Output: 7
Explanation:
-
First Transformation (t = 1):
'a'becomes'b'asnums[0] == 1'b'becomes'c'asnums[1] == 1'c'becomes'd'asnums[2] == 1'y'becomes'z'asnums[24] == 1'y'becomes'z'asnums[24] == 1- String after the first transformation:
"bcdzz"
-
Second Transformation (t = 2):
'b'becomes'c'asnums[1] == 1'c'becomes'd'asnums[2] == 1'd'becomes'e'asnums[3] == 1'z'becomes'ab'asnums[25] == 2'z'becomes'ab'asnums[25] == 2- 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, nums = [2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2]
Output: 8
Explanation:
-
First Transformation (t = 1):
'a'becomes'bc'asnums[0] == 2'z'becomes'ab'asnums[25] == 2'b'becomes'cd'asnums[1] == 2'k'becomes'lm'asnums[10] == 2- String after the first transformation:
"bcabcdlm"
-
Final Length of the string: The string is
"bcabcdlm", which has 8 characters.
Constraints:
1 <= s.length <= 105sconsists only of lowercase English letters.1 <= t <= 109nums.length == 261 <= nums[i] <= 25
Approach Overview
Problem Overview: You start with a string s. Each transformation step changes every character according to predefined rules, potentially expanding one character into multiple new characters. After applying the transformation t times, return the total number of characters in the final string. Directly building the string quickly becomes impossible because its size grows exponentially.
Approach 1: Direct Simulation of Transformations (Counting DP) (Time: O(t * 26), Space: O(26))
Instead of constructing the full string, track how many times each character appears. Use a frequency array of size 26 where freq[i] represents the count of the i-th letter. For each transformation step, compute a new frequency array based on how each character expands according to the rules. This avoids large string allocations and turns the process into simple counting updates. The algorithm iterates through all 26 characters per step and accumulates counts. This approach works well when t is relatively small and is easy to implement using arrays or a hash table style frequency map.
Approach 2: Mathematical Calculation Without Simulation (Matrix / Transition DP) (Time: O(26^3 log t), Space: O(26^2))
When t is very large, even iterating t times becomes too slow. Model the transformation rules as a transition matrix where M[i][j] represents how many characters of type j are produced from character i in one step. The state vector contains counts of each letter. Applying a transformation becomes a matrix multiplication: next = current × M. Repeating the transformation t times becomes matrix exponentiation: M^t. Fast exponentiation reduces the runtime to logarithmic in t. This approach relies on mathematical modeling and dynamic programming style state transitions.
Recommended for interviews: Start with the counting simulation. It shows that you recognized the exponential growth problem and replaced string construction with frequency tracking. For large constraints, interviewers expect the mathematical transition approach using matrix exponentiation. That solution demonstrates deeper understanding of state transitions, repeated transformations, and optimization techniques used in advanced dynamic programming problems.
Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Direct Simulation with Character Counting | O(t * 26) | O(26) | When transformation count t is moderate and a straightforward DP approach is sufficient |
| Mathematical Transition Matrix (Matrix Exponentiation) | O(26^3 log t) | O(26^2) | When t is extremely large and repeated transformations must be computed efficiently |