Count Largest Group
The problem is asking for the number of groups that have the most numbers, where each group represents numbers that have the same sum of digits.
Here’s a step by step explanation:
We start with an integer
n
. Our task is to consider all the integers from 1 ton
.For each of these numbers, we calculate the sum of its digits. For example, the sum of digits for the number 13 is 1 + 3 = 4. The sum of digits for the number 10 is 1 + 0 = 1.
We then group these numbers based on their digit sums. So all numbers that have a digit sum of 1 are in one group, all numbers that have a digit sum of 2 are in another group, and so on.
We look at the sizes of these groups. The size of a group is simply the number of elements in the group.
Our goal is to determine how many groups are the largest size. For example, if there are 3 groups of size 4 and no groups are larger than that, our answer would be 3, because there are 3 groups of the largest size.
Solution in Python:


The basic idea behind this solution is to create a map group_counts
that maps the sum of digits to the number of integers that have that sum of digits. We generate this map by looping through all the numbers from 1 to n
and for each number, we calculate the sum of its digits and increment the corresponding count in group_counts
.
Finally, we count how many of the sums have a count equal to the maximum count and return this count. This is done by first finding the maximum count using max(group_counts.values())
and then counting how many values in group_counts
are equal to the maximum count.
Solution in Java:


The Java function countLargestGroup
is designed to count the number of groups in a range of integers (from 1 to n
) that have the largest size, where a group is defined as a set of numbers having the same sum of digits.
Here’s how the function works:
It starts by initializing an ArrayList
cnt
of size 37 with all elements set to 0. The size 37 is chosen because the maximum sum of digits for any number up to 10,000 (the maximum possible value ofn
) is 36 (four digits of 9 each). This list will be used to store the size of each group, where the index in the list represents the sum of digits.Then, it iterates over all numbers from 1 to
n
. For each numberi
, it calculates the sum of digits by calling the helper functiondsum
.The
dsum
function calculates the sum of digits of a number using recursion. It takes a numbern
and returns the sum of its digits. It does so by adding the last digit ofn
(n % 10
) to the sum of digits of the rest of the number (n / 10
), recursively callingdsum
untiln
becomes 0.After calculating the sum of digits
c
for a numberi
, the function increments the count of the group associated withc
in thecnt
list.After counting the size of all groups, the function uses
Collections.frequency
to find the frequency of the largest group size in thecnt
list. This is the number of groups that have the largest size.Finally, it returns the count of groups that have the largest size.
This function effectively uses the properties of numbers and an ArrayList to group numbers based on their sum of digits and find the count of largest groups. The time complexity of this function is O(n), where n is the given number, as it iterates over all numbers from 1 to n
. The space complexity is O(1) as the size of the cnt
list is fixed.
This solution uses a Frequency Counter to count the sizes of groups of numbers that share the same sum of digits. The frequency counter is a common strategy for solving such problems.
Here are some additional highlevel concepts:
Digit Sum Calculation: This solution uses a helper function
dsum
to calculate the sum of the digits of a number. This function uses recursion and the modulo operator%
to separate the digits of the number.Grouping based on Characteristics: The numbers are grouped based on their digit sum. This grouping is essentially a categorization based on a shared characteristic, a common concept in many programming problems.
Usage of ArrayList and Collections: The code demonstrates usage of Java’s ArrayList and Collections classes for storing and manipulating a collection of elements.
Collections.nCopies
is used to initialize the ArrayList,ArrayList.set
is used to update values, andCollections.frequency
andCollections.max
are used to find the frequency of the maximum group size.Value as Index: This solution uses a value (the sum of digits) as an index into an array or ArrayList to count the frequency of that value. This is a common pattern in problems involving counting or grouping.
Building Blocks
Here are the building blocks for this problem:
Iteration: We loop through the numbers from 1 to
n
. This is done using a for loop:for i in range(1, n + 1)
.Sum of Digits: For each number, we calculate the sum of its digits. This is done using
key = sum(int(x) for x in str(i))
. This statement first converts the integer to a string, which allows us to iterate over each digit. Then it converts each digit back to an integer and sums them.Group Counting: We use a dictionary to count the number of integers for each possible sum of digits.
group_counts[key] += 1
increments the count of the group represented by the key.Maximum Group Size Calculation: After all numbers have been processed, we calculate the maximum group size:
max_count = max(group_counts.values())
.Count of Maximum Size Groups: We count how many groups have the maximum size and return this count:
return list(group_counts.values()).count(max_count)
. This line converts the dictionary values (group sizes) to a list and counts the number of times the maximum size appears.
We combine basic iteration, digit manipulation, dictionary usage for counting, and simple statistics (finding max and counting max values) to solve this problem.