Problem Description

You are given a positive integer array nums.

The problem asks you to calculate two different sums:

1. Element sum: This is simply the sum of all elements in the array. For example, if nums = [15, 23, 8], the element sum would be 15 + 23 + 8 = 46.

2. Digit sum: This is the sum of all individual digits that appear in the numbers within the array. For the same example nums = [15, 23, 8], we would extract all digits:

   • From 15: digits are 1 and 5
   • From 23: digits are 2 and 3
   • From 8: digit is 8
   • The digit sum would be 1 + 5 + 2 + 3 + 8 = 19

Your task is to return the absolute difference between these two sums, which is |element_sum - digit_sum|.

The solution iterates through each number in the array. For each number v, it adds v to the element sum x, and then extracts each digit of v using modulo and integer division operations (v % 10 gives the last digit, v //= 10 removes the last digit) to add them to the digit sum y. Since the element sum is always greater than or equal to the digit sum (a number is always greater than or equal to the sum of its digits), the code directly returns x - y without needing the absolute value operation.

Intuition

The key observation is that we need to compute two separate sums from the same array - one treating each element as a whole number, and another breaking down each element into its individual digits.

The straightforward approach is to traverse the array once and calculate both sums simultaneously. For each number, we can:

1. Add it directly to the element sum
2. Extract its digits and add them to the digit sum

To extract digits from a number, we can use the mathematical property that n % 10 gives us the last digit of n, and n // 10 removes the last digit. By repeatedly applying these operations until the number becomes 0, we can extract all digits.

An important insight is that the element sum will always be greater than or equal to the digit sum. This is because any multi-digit number is always greater than the sum of its digits (for example, 23 > 2 + 3). Only single-digit numbers have equality. Therefore, we know that element_sum - digit_sum >= 0, which means we don't actually need to use the absolute value function - we can directly return x - y.

This single-pass approach is optimal because we only need to visit each element once, and for each element, we perform a constant amount of work (proportional to the number of digits, which is at most log10(n) for a number n).

Learn more about Math patterns.

Solution Approach

The solution uses a simulation approach where we traverse the array once and calculate both required sums simultaneously.

We initialize two variables:

• x to store the element sum
• y to store the digit sum

For each number v in the array nums:

1. Add to element sum: We directly add v to x using x += v

2. Extract and sum digits: We use a while loop to extract each digit from v:

   • v % 10 gives us the rightmost (last) digit
   • We add this digit to our digit sum y
   • v //= 10 removes the last digit by integer division
   • We repeat until v becomes 0

The digit extraction process works because:

• For a number like 234:
  • First iteration: 234 % 10 = 4, add to sum, then 234 // 10 = 23
  • Second iteration: 23 % 10 = 3, add to sum, then 23 // 10 = 2
  • Third iteration: 2 % 10 = 2, add to sum, then 2 // 10 = 0
  • Loop ends when v becomes 0

After processing all numbers, we return x - y. Since the element sum is always greater than or equal to the digit sum (a number is always >= the sum of its digits), we can return x - y directly without needing the absolute value operation.

Time complexity: O(n * d) where n is the length of the array and d is the average number of digits in each element. Space complexity: O(1) as we only use two variables regardless of input size.

Example Walkthrough

Let's walk through the solution with nums = [15, 23, 8].

Initialization:

• x = 0 (element sum)
• y = 0 (digit sum)

Processing first element (15):

• Add to element sum: x = 0 + 15 = 15
• Extract digits from 15:
  • 15 % 10 = 5, add to digit sum: y = 0 + 5 = 5
  • Update: 15 // 10 = 1
  • 1 % 10 = 1, add to digit sum: y = 5 + 1 = 6
  • Update: 1 // 10 = 0
  • Loop ends as number becomes 0

Processing second element (23):

• Add to element sum: x = 15 + 23 = 38
• Extract digits from 23:
  • 23 % 10 = 3, add to digit sum: y = 6 + 3 = 9
  • Update: 23 // 10 = 2
  • 2 % 10 = 2, add to digit sum: y = 9 + 2 = 11
  • Update: 2 // 10 = 0
  • Loop ends

Processing third element (8):

• Add to element sum: x = 38 + 8 = 46
• Extract digits from 8:
  • 8 % 10 = 8, add to digit sum: y = 11 + 8 = 19
  • Update: 8 // 10 = 0
  • Loop ends

Final calculation:

• Element sum: x = 46
• Digit sum: y = 19
• Return: x - y = 46 - 19 = 27

The absolute difference between the element sum and digit sum is 27.

Solution Implementation

Time and Space Complexity

The time complexity is O(n × log₁₀ M), where n is the length of the array nums and M is the maximum value of the elements in the array. This is because:

• The outer loop iterates through all n elements in the array
• For each element v, the inner while loop extracts digits by repeatedly dividing by 10 until v becomes 0
• The number of digits in a number v is ⌊log₁₀ v⌋ + 1, which is O(log₁₀ M) in the worst case when v = M
• Therefore, the total time complexity is O(n) × O(log₁₀ M) = O(n × log₁₀ M)

The space complexity is O(1) because the algorithm only uses a constant amount of extra space with variables x, y, and v, regardless of the input size.

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

Common Pitfalls

1. Modifying the Original Value During Digit Extraction

A common mistake is directly modifying the loop variable when extracting digits, which causes the element sum calculation to be incorrect.

Incorrect Implementation:

for num in nums:
    element_sum += num
    while num > 0:  # This modifies num!
        digit_sum += num % 10
        num //= 10  # num is being changed here

This seems correct at first glance, but if you need to use num again after the digit extraction (for debugging, logging, or future modifications), the value is lost.

Correct Solution:

for num in nums:
    element_sum += num
    temp = num  # Create a copy for digit extraction
    while temp > 0:
        digit_sum += temp % 10
        temp //= 10

2. Incorrect Handling of Single-Digit Numbers

Some might think single-digit numbers need special handling and add unnecessary complexity:

Overcomplicated Implementation:

for num in nums:
    element_sum += num
    if num < 10:
        digit_sum += num
    else:
        while num > 0:
            digit_sum += num % 10
            num //= 10

Why It's Unnecessary: The while loop naturally handles single-digit numbers correctly. For a single digit like 8:

• First iteration: 8 % 10 = 8, add to sum, then 8 // 10 = 0
• Loop ends, having correctly added the digit

3. Using String Conversion (Less Efficient)

While not incorrect, converting numbers to strings for digit extraction is less efficient:

Less Efficient Approach:

for num in nums:
    element_sum += num
    for digit_char in str(num):
        digit_sum += int(digit_char)

Why Mathematical Approach is Better:

• String conversion has overhead for memory allocation and type conversion
• Mathematical operations (%, //) are more direct and faster
• The mathematical approach uses O(1) space vs O(d) space for string conversion where d is the number of digits

4. Forgetting the Absolute Value (Problem-Specific)

While the current implementation correctly assumes element_sum >= digit_sum, if the problem were modified or if you're not certain about this property:

Safer Implementation:

return abs(element_sum - digit_sum)

This ensures correctness even if the assumption doesn't hold in edge cases or problem variations.