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Count Largest Group

Easy

Asked by:

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Topics:

ArraysDynamic Programming

You are given an integer n.

We need to group the numbers from 1 to n according to the sum of its digits. For example, the numbers 14 and 5 belong to the same group, whereas 13 and 3 belong to different groups.

Return the number of groups that have the largest size, i.e. the maximum number of elements.

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

Try coding it in LeetCode

Count Largest Group

Easy

Asked by:

35 views

Topics:

ArraysDynamic Programming

You are given an integer n.

We need to group the numbers from 1 to n according to the sum of its digits. For example, the numbers 14 and 5 belong to the same group, whereas 13 and 3 belong to different groups.

Return the number of groups that have the largest size, i.e. the maximum number of elements.

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum value of `n`? Is `n` always a positive integer?
If multiple groups have the same largest size, should I return the count of such groups, or just one of them?
Can you provide a few examples of `n` and the expected output?
Are there any specific memory constraints I should be aware of given the potential size of `n`?
Is `n` guaranteed to be a valid integer, or should I handle potential non-integer or invalid input?

Brute Force Solution

Approach

The brute force approach to counting the largest group involves checking every possible number from 1 up to the input number. For each number, we determine its digit sum and assign it to a group. Finally, we identify the group with the most numbers and return its size.

Here's how the algorithm would work step-by-step:

Start by examining each number, one at a time, beginning with 1 and continuing up to the input number.
For each number, calculate the sum of its digits.
Based on the digit sum we just calculated, assign the current number to a corresponding group. Imagine having containers, each labeled with a specific digit sum, and placing the numbers into the appropriate container.
Once we've assigned all numbers up to the input, count how many numbers are in each group (container).
Find the group (container) that has the most numbers in it.
Return the count (size) of that largest group.

Code Implementation

def count_largest_group_brute_force(number):
    groups = {}

    for current_number in range(1, number + 1):
        digit_sum = 0
        temporary_number = current_number

        # Calculate the digit sum of the current number
        while temporary_number > 0:
            digit_sum += temporary_number % 10
            temporary_number //= 10

        # Assign the number to its respective group based on digit sum
        if digit_sum not in groups:
            groups[digit_sum] = []

        groups[digit_sum].append(current_number)

    largest_group_size = 0

    # Iterate through the groups to find the largest one
    for digit_sum in groups:
        group_size = len(groups[digit_sum])

        if group_size > largest_group_size:
            largest_group_size = group_size

    return largest_group_size

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each number from 1 to n. For each number, it calculates the digit sum. The number of digits in n is at most log10(n) + 1, so the digit sum calculation takes O(log n) time. The hashmap operations (inserting into and updating counts in a map) take O(1) on average. Finding the maximum count among the groups then takes O(k) where k is the number of distinct digit sums. Because k is at most 81 (the maximum digit sum for n = 10^4 is 36), it can be considered a constant. Therefore, the dominant factor is the iteration from 1 to n, making the overall time complexity O(n).

Space Complexity

O(N) – The solution uses a data structure (implicitly described as 'containers' or groups) to store the numbers based on their digit sums. In the worst-case scenario, each number from 1 to N could have a unique digit sum, leading to N distinct groups. Therefore, the space required to store these groups and their associated numbers scales linearly with the input number N. Thus, the auxiliary space complexity is O(N).

Optimal Solution

Approach

We need to find the group of numbers, based on their digit sums, that occurs most frequently within a given range. The efficient solution involves grouping the numbers by their digit sums and then simply counting which group is the largest.

Here's how the algorithm would work step-by-step:

For each number within the specified range, calculate the sum of its digits.
Group all the numbers based on their digit sums. This means all numbers with the same digit sum will belong to the same group.
Count the number of elements in each group.
Identify the group with the maximum number of elements. If multiple groups have the same maximum count, any of them will do.
Return the size of the largest group.

Code Implementation

def count_largest_group(limit):
    group_sizes = {}

    for number in range(1, limit + 1):
        digit_sum = 0
        temporary_number = number

        while temporary_number > 0:
            digit_sum += temporary_number % 10
            temporary_number //= 10

        # Group numbers by their digit sum
        if digit_sum not in group_sizes:
            group_sizes[digit_sum] = 0
        
        group_sizes[digit_sum] += 1

    # Find the largest group size
    max_group_size = 0
    for digit_sum in group_sizes:
        if group_sizes[digit_sum] > max_group_size:
            max_group_size = group_sizes[digit_sum]

    # Count how many groups have the max size
    largest_group_count = 0
    for digit_sum in group_sizes:
        if group_sizes[digit_sum] == max_group_size:

            #Need to find groups matching the max size
            largest_group_count += 1

    return largest_group_count

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each number from 1 to n to calculate its digit sum. The digit sum calculation for each number is proportional to the number of digits in n, which is logarithmic with base 10 to n, or log₁₀(n). However, for large n, log₁₀(n) will be smaller than the growth of n and can be seen as effectively constant. Grouping and counting the elements by digit sum takes time proportional to the number of elements, which is n. Therefore, the overall time complexity is dominated by the initial loop through the numbers 1 to n, making it O(n).

Space Complexity

O(1) – The provided algorithm primarily uses a hash map (or similar data structure) to group numbers by their digit sums. In the worst-case scenario, each number from 1 to n could have a unique digit sum. However, the maximum digit sum for any number 'n' is bounded by 9 * (number of digits in n), which grows much slower than n itself. Moreover, the number of distinct digit sums encountered is limited by the range of numbers from 1 to n, and the auxiliary space grows logarithmically with n (as the number of digits in n grows logarithmically). Therefore, a hash map with a size much smaller than n is sufficient to store these groups. Since the maximum possible digit sum increases at a logarithmic rate with respect to the input 'n', the space complexity can be considered O(1).

Edge Cases

Case	How to Handle
Input n is 0	Return 0, since there are no groups if n is zero.
Input n is a single-digit number (1-9)	Return 1, because each number forms a group of size 1 and all groups have same digit sum, making the number of largest groups 1.
Input n is a power of 10 (e.g., 10, 100, 1000)	Handle efficiently; the size of the largest group may not be 1.
Input n is a large number close to the maximum integer limit	Ensure the sum of digits calculation and group counting does not cause an integer overflow by using a larger data type if necessary.
Multiple groups have the same largest size.	The algorithm should correctly count the number of such groups with the largest size.
Input n produces a digit sum that is very large (e.g. when n is 99999)	Account for the potential for larger digit sums to require more space to store group sizes.
The calculation of digit sums is inefficient and leads to time limit exceeded for large n	Optimize digit sum calculation, perhaps caching results to avoid recalculating frequently, and minimize redundant loop traversals.
Memory constraints are exceeded when counting for large n	Consider more space-efficient data structures or algorithms for counting group sizes, like bucketing instead of a full hashmap if digit sum range is constrained.

Input n is 0

How to Handle:

Return 0, since there are no groups if n is zero.

Input n is a single-digit number (1-9)

How to Handle:

Return 1, because each number forms a group of size 1 and all groups have same digit sum, making the number of largest groups 1.

Input n is a power of 10 (e.g., 10, 100, 1000)

How to Handle:

Handle efficiently; the size of the largest group may not be 1.

Input n is a large number close to the maximum integer limit

How to Handle:

Ensure the sum of digits calculation and group counting does not cause an integer overflow by using a larger data type if necessary.

Multiple groups have the same largest size.

How to Handle:

The algorithm should correctly count the number of such groups with the largest size.

Input n produces a digit sum that is very large (e.g. when n is 99999)

How to Handle:

Account for the potential for larger digit sums to require more space to store group sizes.

The calculation of digit sums is inefficient and leads to time limit exceeded for large n

How to Handle:

Optimize digit sum calculation, perhaps caching results to avoid recalculating frequently, and minimize redundant loop traversals.

Memory constraints are exceeded when counting for large n

How to Handle:

Consider more space-efficient data structures or algorithms for counting group sizes, like bucketing instead of a full hashmap if digit sum range is constrained.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum value of `n`? Is `n` always a positive integer?
If multiple groups have the same largest size, should I return the count of such groups, or just one of them?
Can you provide a few examples of `n` and the expected output?
Are there any specific memory constraints I should be aware of given the potential size of `n`?
Is `n` guaranteed to be a valid integer, or should I handle potential non-integer or invalid input?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start by examining each number, one at a time, beginning with 1 and continuing up to the input number.
For each number, calculate the sum of its digits.
Based on the digit sum we just calculated, assign the current number to a corresponding group. Imagine having containers, each labeled with a specific digit sum, and placing the numbers into the appropriate container.
Once we've assigned all numbers up to the input, count how many numbers are in each group (container).
Find the group (container) that has the most numbers in it.
Return the count (size) of that largest group.

Code Implementation

def count_largest_group_brute_force(number):
    groups = {}

    for current_number in range(1, number + 1):
        digit_sum = 0
        temporary_number = current_number

        # Calculate the digit sum of the current number
        while temporary_number > 0:
            digit_sum += temporary_number % 10
            temporary_number //= 10

        # Assign the number to its respective group based on digit sum
        if digit_sum not in groups:
            groups[digit_sum] = []

        groups[digit_sum].append(current_number)

    largest_group_size = 0

    # Iterate through the groups to find the largest one
    for digit_sum in groups:
        group_size = len(groups[digit_sum])

        if group_size > largest_group_size:
            largest_group_size = group_size

    return largest_group_size

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

For each number within the specified range, calculate the sum of its digits.
Group all the numbers based on their digit sums. This means all numbers with the same digit sum will belong to the same group.
Count the number of elements in each group.
Identify the group with the maximum number of elements. If multiple groups have the same maximum count, any of them will do.
Return the size of the largest group.

Code Implementation

def count_largest_group(limit):
    group_sizes = {}

    for number in range(1, limit + 1):
        digit_sum = 0
        temporary_number = number

        while temporary_number > 0:
            digit_sum += temporary_number % 10
            temporary_number //= 10

        # Group numbers by their digit sum
        if digit_sum not in group_sizes:
            group_sizes[digit_sum] = 0
        
        group_sizes[digit_sum] += 1

    # Find the largest group size
    max_group_size = 0
    for digit_sum in group_sizes:
        if group_sizes[digit_sum] > max_group_size:
            max_group_size = group_sizes[digit_sum]

    # Count how many groups have the max size
    largest_group_count = 0
    for digit_sum in group_sizes:
        if group_sizes[digit_sum] == max_group_size:

            #Need to find groups matching the max size
            largest_group_count += 1

    return largest_group_count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input n is 0	Return 0, since there are no groups if n is zero.
Input n is a single-digit number (1-9)	Return 1, because each number forms a group of size 1 and all groups have same digit sum, making the number of largest groups 1.
Input n is a power of 10 (e.g., 10, 100, 1000)	Handle efficiently; the size of the largest group may not be 1.
Input n is a large number close to the maximum integer limit	Ensure the sum of digits calculation and group counting does not cause an integer overflow by using a larger data type if necessary.
Multiple groups have the same largest size.	The algorithm should correctly count the number of such groups with the largest size.
Input n produces a digit sum that is very large (e.g. when n is 99999)	Account for the potential for larger digit sums to require more space to store group sizes.
The calculation of digit sums is inefficient and leads to time limit exceeded for large n	Optimize digit sum calculation, perhaps caching results to avoid recalculating frequently, and minimize redundant loop traversals.
Memory constraints are exceeded when counting for large n	Consider more space-efficient data structures or algorithms for counting group sizes, like bucketing instead of a full hashmap if digit sum range is constrained.

Input n is 0

How to Handle:

Return 0, since there are no groups if n is zero.

Input n is a single-digit number (1-9)

How to Handle:

Return 1, because each number forms a group of size 1 and all groups have same digit sum, making the number of largest groups 1.

Input n is a power of 10 (e.g., 10, 100, 1000)

How to Handle:

Handle efficiently; the size of the largest group may not be 1.

Input n is a large number close to the maximum integer limit

How to Handle:

Ensure the sum of digits calculation and group counting does not cause an integer overflow by using a larger data type if necessary.

Multiple groups have the same largest size.

How to Handle:

The algorithm should correctly count the number of such groups with the largest size.

Input n produces a digit sum that is very large (e.g. when n is 99999)

How to Handle:

Account for the potential for larger digit sums to require more space to store group sizes.

The calculation of digit sums is inefficient and leads to time limit exceeded for large n

How to Handle:

Optimize digit sum calculation, perhaps caching results to avoid recalculating frequently, and minimize redundant loop traversals.

Memory constraints are exceeded when counting for large n

How to Handle:

Consider more space-efficient data structures or algorithms for counting group sizes, like bucketing instead of a full hashmap if digit sum range is constrained.