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Count Largest Group

Difficulty: Easy

Problem Description

You are given an integer n. Each number from 1 to n is grouped according to the sum of its digits. Return the number of groups that have the largest size.

Key Insights

  • Each number can be represented by the sum of its digits.
  • Numbers with the same digit sum form a group.
  • We need to determine how many groups have the maximum size.
  • A hash table (or dictionary) can be utilized to count occurrences of each digit sum.

Space and Time Complexity

Time Complexity: O(n) - We iterate through numbers from 1 to n, calculating their digit sums. Space Complexity: O(n) - In the worst case, we could have n different digit sums stored in the hash table.

Solution

To solve the problem, we can follow these steps:

  1. Initialize a hash table (or dictionary) to count the occurrences of each digit sum.
  2. Iterate through each number from 1 to n, compute its digit sum, and update the count in the hash table.
  3. Determine the maximum size of the groups by finding the highest value in the hash table.
  4. Count how many groups have this maximum size and return that count.

In this approach, we use a hash table to efficiently track and count the sizes of groups formed by each unique digit sum.

Code Solutions

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

    # Dictionary to count groups by digit sum
    digit_sum_count = defaultdict(int)

    # Function to calculate digit sum
    def digit_sum(x):
        return sum(int(d) for d in str(x))

    # Count occurrences of each digit sum
    for i in range(1, n + 1):
        s = digit_sum(i)
        digit_sum_count[s] += 1

    # Find the maximum group size
    max_size = max(digit_sum_count.values())

    # Count how many groups have the maximum size
    return sum(1 for size in digit_sum_count.values() if size == max_size)
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