Problem Description

You need to group all integers from 1 to n based on their digit sum. Two numbers belong to the same group if the sum of their digits is equal.

For example:

• The number 14 has digit sum 1 + 4 = 5

• The number 5 has digit sum 5

• So 14 and 5 belong to the same group (both have digit sum of 5)

• The number 13 has digit sum 1 + 3 = 4

• The number 3 has digit sum 3

• So 13 and 3 belong to different groups (digit sums of 4 and 3 respectively)

After grouping all numbers from 1 to n by their digit sums, you need to find the size of the largest group(s). Then return how many groups have this maximum size.

For instance, if after grouping you have:

• Group with digit sum 3: contains 2 numbers
• Group with digit sum 4: contains 3 numbers
• Group with digit sum 5: contains 3 numbers
• Group with digit sum 6: contains 1 number

The maximum group size is 3, and there are 2 groups with this size (digit sums 4 and 5), so the answer would be 2.

Intuition

The key insight is that we need to track two things: the size of each group and which group size is the largest.

Since we're grouping numbers by their digit sum, we can use the digit sum as a key to count how many numbers fall into each group. For any number from 1 to n, we can calculate its digit sum by repeatedly taking the remainder when divided by 10 (to get the last digit) and then dividing by 10 (to remove the last digit).

As we process each number from 1 to n, we calculate its digit sum and increment the count for that particular sum. This naturally builds up our groups.

The clever part is tracking the answer as we go. Instead of first building all groups and then finding the maximum size and counting how many groups have that size, we can do it in one pass:

• Whenever we encounter a group that becomes larger than our current maximum size, we know this is now the only group with the maximum size, so we reset our answer to 1
• Whenever we encounter a group that equals our current maximum size, we've found another group with the maximum size, so we increment our answer

This way, by the time we've processed all numbers from 1 to n, we already have our answer without needing a second pass through the data.

The constraint that n ≤ 10^4 tells us the maximum possible digit sum is 9 + 9 + 9 + 9 = 36 (for 9999), so we know our groups are limited to at most 36 different digit sums, making this approach very efficient.

Learn more about Math patterns.

Solution Approach

We implement the solution using a hash table (or array) to count occurrences of each digit sum.

Data Structure Setup:

• Create a counter cnt to track how many numbers have each digit sum
• Initialize variables mx = 0 to track the maximum group size and ans = 0 to count groups with maximum size

Main Algorithm:

For each number i from 1 to n:

1. Calculate digit sum:

   • Initialize s = 0 to store the sum
   • While i > 0:
     • Add the last digit to sum: s += i % 10
     • Remove the last digit: i //= 10

2. Update the count for this digit sum:

   • Increment cnt[s] by 1

3. Update the answer based on the new count:

   • If cnt[s] > mx: We found a new maximum group size
     • Update mx = cnt[s]
     • Reset ans = 1 (only one group has this new maximum size)
   • Else if cnt[s] == mx: We found another group with the maximum size
     • Increment ans += 1

Why this works:

• Since we know the maximum digit sum is 36 (from 9999), we could alternatively use an array of size 40 instead of a hash table
• By tracking mx and ans during the iteration, we avoid a second pass through the data
• The counter naturally groups numbers by their digit sum as we process them

Time Complexity: O(n × log n) where the log n factor comes from calculating the digit sum of each number (number of digits is proportional to log n)

Space Complexity: O(1) since we're storing at most 36 different digit sums

Example Walkthrough

Let's walk through the solution with n = 15.

Initial Setup:

• cnt = {} (empty counter for digit sums)
• mx = 0 (maximum group size)
• ans = 0 (count of groups with maximum size)

Processing each number from 1 to 15:

Final Groups:

• Digit sum 1: {1, 10} → size 2
• Digit sum 2: {2, 11} → size 2
• Digit sum 3: {3, 12} → size 2
• Digit sum 4: {4, 13} → size 2
• Digit sum 5: {5, 14} → size 2
• Digit sum 6: {6, 15} → size 2
• Digit sum 7: {7} → size 1
• Digit sum 8: {8} → size 1
• Digit sum 9: {9} → size 1

Result: Maximum group size is 2, and there are 6 groups with this size, so the answer is 6.

The key insight demonstrated here is how we dynamically track the answer:

• When we hit number 10, it creates the first group of size 2 (digit sum 1), making it the new maximum
• As we continue, numbers 11-15 each create groups of size 2, incrementing our answer each time
• We never need to look back at our groups - the answer is computed on the fly

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log n)

The algorithm iterates through all numbers from 1 to n, which takes O(n) iterations. For each number i, we calculate the sum of its digits by repeatedly dividing by 10 and taking the remainder. The number of digits in a number i is approximately log₁₀(i), so extracting all digits takes O(log i) time. In the worst case when i = n, this becomes O(log n). Therefore, the overall time complexity is O(n × log n).

Space Complexity: O(log n)

The space is primarily used by the Counter dictionary cnt. The key insight is that the dictionary stores digit sums as keys. The maximum possible digit sum occurs for the number n, and since n has at most O(log n) digits, and each digit can be at most 9, the maximum possible sum is 9 × O(log n) = O(log n). The number of unique digit sums is bounded by this maximum value, so the dictionary can have at most O(log n) entries. Therefore, the space complexity is O(log n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Modifying the Loop Variable While Calculating Digit Sum

A common mistake is directly modifying the loop variable i when calculating the digit sum:

Incorrect Code:

for i in range(1, n + 1):
    digit_sum = 0
    while i > 0:  # BUG: Modifying loop variable i
        digit_sum += i % 10
        i //= 10  # This destroys the original value of i
  
    # Now i is 0, not the original number!
    digit_sum_count[digit_sum] += 1

Why it's wrong: After calculating the digit sum, i becomes 0, so you lose track of which number you were processing. This breaks the loop iteration.

Solution: Always use a temporary variable to preserve the original value:

for i in range(1, n + 1):
    digit_sum = 0
    temp = i  # Create a copy
    while temp > 0:
        digit_sum += temp % 10
        temp //= 10
  
    # i is still intact
    digit_sum_count[digit_sum] += 1

2. Incorrect Answer Tracking Logic

Another pitfall is resetting the counter at the wrong time or not handling the first occurrence correctly:

Incorrect Code:

if digit_sum_count[digit_sum] >= max_group_size:  # BUG: Using >= instead of >
    max_group_size = digit_sum_count[digit_sum]
    num_largest_groups = 1  # This resets even when equal!

Why it's wrong: Using >= would reset the counter every time we see a group with the current maximum size, losing count of previous groups with the same size.

Solution: Use strict comparison operators correctly:

if digit_sum_count[digit_sum] > max_group_size:
    # New maximum found
    max_group_size = digit_sum_count[digit_sum]
    num_largest_groups = 1
elif digit_sum_count[digit_sum] == max_group_size:
    # Another group with same maximum size
    num_largest_groups += 1

3. Post-Processing Instead of Real-Time Tracking

Some might try to find the answer after building the complete frequency map:

Less Efficient Approach:

# Build complete frequency map first
for i in range(1, n + 1):
    digit_sum = calculate_digit_sum(i)
    digit_sum_count[digit_sum] += 1

# Then find the answer (requires second pass)
max_size = max(digit_sum_count.values())
return sum(1 for count in digit_sum_count.values() if count == max_size)

Why it's suboptimal: This requires two passes through the data - one to build the map and another to find the answer.

Solution: Track the answer while building the frequency map (as shown in the original solution), which requires only one pass and is more efficient.