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You are given an integer n. Let f(x) be the sum of digits of x for x in the range [1, n].

Return the number of groups that have the largest size, where a group is defined as a set of integers having the same sum of digits.

Clarifying Questions

Before implementing the solution, let’s clarify the problem with some questions:

  1. What is the range of n?
    • Let’s assume 1 <= n <= 10^4.
  2. What should be the output if there’s only one integer?
    • If there’s only one integer, the output should be 1.

Given these questions, we know that we will need to handle fairly large values of n efficiently.

Strategy

  1. Sum of Digits Calculation: Create a helper function to compute the sum of digits of any integer.
  2. Group Counting: Use a dictionary to count the frequency of the sum of digits for each number in the range [1, n].
  3. Find the Largest Group Size: Determine the maximum frequency in the dictionary.
  4. Count Largest Groups: Count how many groups have this maximum frequency.

Code

Here’s how we can implement this:

def count_largest_group(n: int) -> int:
    from collections import defaultdict

    def sum_of_digits(x: int) -> int:
        total = 0
        while x > 0:
            total += x % 10
            x //= 10
        return total

    digit_sum_count = defaultdict(int)
    
    # Calculate sum of digits for each number and count occurrences
    for i in range(1, n + 1):
        digit_sum = sum_of_digits(i)
        digit_sum_count[digit_sum] += 1

    # Find the size of the largest group
    max_size = max(digit_sum_count.values())
    
    # Count the number of groups that have the largest size
    largest_groups_count = sum(1 for size in digit_sum_count.values() if size == max_size)
    
    return largest_groups_count

Time Complexity

  1. Sum of Digits Calculation: For each number from 1 to n, we compute the sum of digits. This can be done in O(log k) time where k is the number itself.
  2. Iteration and Counting: We iterate through the numbers from 1 to n, which is O(n).
  3. Finding Maximum and Counting: These operations are O(m) where m is the number of unique sums of digits.

Since n can be as large as 10^4, the dominant term is O(n * log n). Thus, the overall time complexity is:

O(n log n)

This should be efficient enough for the given problem constraints.

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