# Leetcode – Count Largest Group Solution

The “Count Largest Group” problem from Leetcode serves as an interesting playground to explore algorithms, data structures, and Pythonic best practices. This problem is particularly useful for practicing hash tables, integer manipulations, and sorting algorithms. This exhaustive article aims to provide a deep dive into various ways to solve this problem, and their associated time and space complexities.

1. Understanding the Problem Statement
2. Preliminary Thoughts
3. Brute-Force Approach
4. Hash Table-Based Approach
5. Sorting and Iterating
6. Time and Space Complexity Analysis
7. Testing and Edge Cases
9. Conclusion

## 1. Understanding the Problem Statement

The problem is to count the number of groups having the largest size, given that the task is to group the integers from 1 to n based on the sum of their digits.

### Constraints and Assumptions

• 1 <= n <= 10^4

## 2. Preliminary Thoughts

At its core, the problem requires us to group numbers based on the sum of their digits and then find the largest group. Hence, the problem can be broken down into two parts:

1. Calculating the sum of digits for each number from 1 to n.
2. Grouping these numbers based on the sum of their digits and finding the largest group.

## 3. Brute-Force Approach

A naive way to approach this problem would be to loop through numbers 1 to n, compute the sum of their digits, and store them in groups. Finally, we can find the largest group size.

Python Code:

def countLargestGroup(n):
groups = {}
for i in range(1, n+1):
digit_sum = sum(int(digit) for digit in str(i))
if digit_sum not in groups:
groups[digit_sum] = []
groups[digit_sum].append(i)

max_size = max(len(group) for group in groups.values())
count = sum(1 for group in groups.values() if len(group) == max_size)

return count

## 4. Hash Table-Based Approach

We can optimize the above solution by using a hash table to store the frequency of each group size.

Python Code:

from collections import defaultdict

def countLargestGroup(n):
groups = defaultdict(list)
for i in range(1, n+1):
digit_sum = sum(int(digit) for digit in str(i))
groups[digit_sum].append(i)

group_sizes = defaultdict(int)
for group in groups.values():
size = len(group)
group_sizes[size] += 1

max_size = max(group_sizes.keys())
return group_sizes[max_size]

## 5. Sorting and Iterating

While not necessary for this problem, one could sort the group sizes before finding the largest one, which might be useful for variations of this problem.

## 6. Time and Space Complexity Analysis

• Brute-Force Approach: Time Complexity is O(n * d), where d is the number of digits, and Space Complexity is O(n).
• Hash Table-Based Approach: Time Complexity remains O(n * d), but it is slightly more optimized in terms of space with a complexity of O(d * m), where m is the number of unique digit sums.

## 7. Testing and Edge Cases

Important test cases to consider:

• n is at its lower limit (1).
• n is at its upper limit (10^4).
• n is a power of 10.