How do you solve Count Largest Group on LeetCode?

Count Largest Group (LeetCode #1399) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The sum of digits can be calculated efficiently using string manipulation.

What is the brute force solution for Count Largest Group?

The brute force approach for Count Largest Group is Brute Force, with O(n²) time complexity and O(1) space complexity. We can group numbers by calculating the sum of their digits and counting how many numbers fall into each group. This is straightforward but can be inefficient as we check each number multiple times. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Largest Group?

Count Largest Group uses the following concepts: Hash Table, Math, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Largest Group?

Always consider edge cases, such as the smallest values of n. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice writing clean and efficient code, as readability can be just as important as functionality.

What are common mistakes when solving Count Largest Group?

Not resetting the count for each new maximum size found. Overcomplicating the digit sum calculation instead of using simple arithmetic or string methods.

#1399

Count Largest Group

Easy

Hash TableMathCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of recalculating the digit sum multiple times, we can directly count the occurrences of each digit sum in a single pass, making the process more efficient.

⚙️

Algorithm

3 steps

1Step 1: Initialize a dictionary to count groups based on the sum of digits.
2Step 2: For each number from 1 to n, calculate the sum of its digits and update the count in the dictionary.
3Step 3: Find the maximum group size and count how many groups have that size.

solution.py9 lines

1def countLargestGroup(n):
2    groups = defaultdict(int)
3    
4    for i in range(1, n + 1):
5        digit_sum = sum(int(d) for d in str(i))
6        groups[digit_sum] += 1
7    
8    max_size = max(groups.values())
9    return sum(1 for size in groups.values() if size == max_size)

ℹ

Complexity note: The time complexity is O(n) because we only pass through the numbers once, calculating the digit sum in constant time for each number. The space complexity is O(n) due to storing the counts of groups.

1The sum of digits can be calculated efficiently using string manipulation.
2Using a dictionary (or hash map) allows us to count occurrences without worrying about the size of the groups in advance.

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