What is the brute force solution for Total Characters in String After Transformations II?

The brute force approach for Total Characters in String After Transformations II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves applying the transformation to each character in the string for each of the t transformations. This is straightforward but inefficient, as it repeatedly processes the entire string. The optimal Optimal Solution approach improves this to O(n + 26^3 log t).

What algorithm or data structure is used to solve Total Characters in String After Transformations II?

Total Characters in String After Transformations II uses the following concepts: Hash Table, Math, String, Dynamic Programming, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Total Characters in String After Transformations II?

Explain your thought process clearly when transitioning from brute force to optimal solutions. Practice matrix operations as they can often appear in advanced problems. Always check edge cases, such as empty strings or maximum transformation counts.

What are common mistakes when solving Total Characters in String After Transformations II?

Not considering the modulo operation, leading to overflow errors. Failing to account for wrapping around the alphabet when transforming characters.

#3337

Total Characters in String After Transformations II

Hard

Hash TableMathStringDynamic ProgrammingCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + 26^3 log t)
Space	O(1)	O(26^2)

💡

Intuition

Time O(n + 26^3 log t)Space O(26^2)

Instead of transforming the string character by character for each transformation, we can model the transformations as a mathematical operation using matrix exponentiation. This allows us to compute the final transformation in logarithmic time with respect to t.

⚙️

Algorithm

3 steps

1Step 1: Create a transformation matrix where each row represents the effect of transforming a character based on the nums array.
2Step 2: Use matrix exponentiation to compute the transformation matrix raised to the power of t.
3Step 3: Multiply the original character counts by the resulting matrix to get the final counts after t transformations.

solution.py27 lines

1# Full working Python code
2
3def matrix_mult(A, B):
4    return [[sum(x * y for x, y in zip(A_row, B_col)) % (10**9 + 7) for B_col in zip(*B)] for A_row in A]
5
6def matrix_pow(mat, p):
7    res = [[1 if i == j else 0 for j in range(26)] for i in range(26)]
8    while p:
9        if p % 2:
10            res = matrix_mult(res, mat)
11        mat = matrix_mult(mat, mat)
12        p //= 2
13    return res
14
15def total_characters(s, t, nums):
16    MOD = 10**9 + 7
17    trans_matrix = [[0] * 26 for _ in range(26)]
18    for i in range(26):
19        trans_matrix[i][(i + nums[i]) % 26] += 1
20    final_matrix = matrix_pow(trans_matrix, t)
21    count = [0] * 26
22    for char in s:
23        count[ord(char) - ord('a')] += 1
24    result_length = 0
25    for i in range(26):
26        result_length = (result_length + final_matrix[i][i] * count[i]) % MOD
27    return result_length

ℹ

Complexity note: The time complexity is O(n) for counting characters plus O(26^3 log t) for matrix exponentiation, where 26 is the number of letters in the alphabet. The space complexity is O(26^2) for storing the transformation matrix.

1Matrix exponentiation can drastically reduce the time complexity for repeated transformations.
2Understanding how to model problems mathematically can lead to more efficient solutions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.