Introduction

In this blog post, we will explore how to solve the LeetCode problem "1399" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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. Example 1: Input: n = 13 Output: 4 Explanation: There are 9 groups in total, they are grouped according sum of its digits of numbers from 1 to 13: [1,10], [2,11], [3,12], [4,13], [5], [6], [7], [8], [9]. There are 4 groups with largest size. Example 2: Input: n = 2 Output: 2 Explanation: There are 2 groups [1], [2] of size 1. Constraints: 1 <= n <= 104

Explanation

Here's the solution approach, complexity analysis, and Python code:

• Digit Sum Calculation: Calculate the sum of digits for each number from 1 to n.
• Group Counting: Use a dictionary (or array if the digit sum range is known and small) to count the frequency of each digit sum.

• Maximum Frequency Analysis: Determine the maximum frequency among all digit sum counts and return the number of digit sums that have this maximum frequency.

• Runtime Complexity: O(n), where n is the input number.

• Storage Complexity: O(1) - The number of unique digit sums is bounded by the constraints. (Technically O(log n) because the digit sum grows logarithmically with n, but because the problem constrains n to be <= 10^4, we can say that the storage is constant since the max digit sum is 36. )

Code

    def countLargestGroup(n: int) -> int:
    """
    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.
    """
    digit_sums = {}
    for i in range(1, n + 1):
        digit_sum = 0
        num = i
        while num > 0:
            digit_sum += num % 10
            num //= 10

        if digit_sum in digit_sums:
            digit_sums[digit_sum] += 1
        else:
            digit_sums[digit_sum] = 1

    max_frequency = 0
    for digit_sum in digit_sums:
        max_frequency = max(max_frequency, digit_sums[digit_sum])

    count = 0
    for digit_sum in digit_sums:
        if digit_sums[digit_sum] == max_frequency:
            count += 1

    return count