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Difference Between Element Sum and Digit Sum of an Array

Easy

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Topics:

Arrays

You are given a positive integer array nums.

The element sum is the sum of all the elements in nums.
The digit sum is the sum of all the digits (not necessarily distinct) that appear in nums.

Return the absolute difference between the element sum and digit sum of nums.

Note that the absolute difference between two integers x and y is defined as |x - y|.

Example 1:

Input: nums = [1,15,6,3]
Output: 9
Explanation: 
The element sum of nums is 1 + 15 + 6 + 3 = 25.
The digit sum of nums is 1 + 1 + 5 + 6 + 3 = 16.
The absolute difference between the element sum and digit sum is |25 - 16| = 9.

Example 2:

Input: nums = [1,2,3,4]
Output: 0
Explanation:
The element sum of nums is 1 + 2 + 3 + 4 = 10.
The digit sum of nums is 1 + 2 + 3 + 4 = 10.
The absolute difference between the element sum and digit sum is |10 - 10| = 0.

Constraints:

1 <= nums.length <= 2000
1 <= nums[i] <= 2000

Try coding it in LeetCode

Difference Between Element Sum and Digit Sum of an Array

Easy

Asked by:

11 views

Topics:

Arrays

You are given a positive integer array nums.

The element sum is the sum of all the elements in nums.
The digit sum is the sum of all the digits (not necessarily distinct) that appear in nums.

Return the absolute difference between the element sum and digit sum of nums.

Note that the absolute difference between two integers x and y is defined as |x - y|.

Example 1:

Input: nums = [1,15,6,3]
Output: 9
Explanation: 
The element sum of nums is 1 + 15 + 6 + 3 = 25.
The digit sum of nums is 1 + 1 + 5 + 6 + 3 = 16.
The absolute difference between the element sum and digit sum is |25 - 16| = 9.

Example 2:

Input: nums = [1,2,3,4]
Output: 0
Explanation:
The element sum of nums is 1 + 2 + 3 + 4 = 10.
The digit sum of nums is 1 + 2 + 3 + 4 = 10.
The absolute difference between the element sum and digit sum is |10 - 10| = 0.

Constraints:

1 <= nums.length <= 2000
1 <= nums[i] <= 2000

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 possible value for each integer in the input array `nums`?
Can the input array `nums` be empty or null?
Are all the numbers in `nums` guaranteed to be non-negative?
Should the return value be an integer, and if the difference is too large to fit in a standard integer, what should I do?
Could you provide an example of an input `nums` and the expected output to ensure my understanding of the problem is correct?

Brute Force Solution

Approach

We're given a collection of numbers. We need to find two things: the sum of all the numbers directly, and the sum of all the individual digits that make up those numbers. The brute force method simply calculates each sum separately and then finds the difference.

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

First, calculate the total sum of all the numbers in the collection.
Next, take each number individually and break it down into its individual digits.
Add up all of these individual digits to get the total digit sum.
Finally, subtract the total digit sum from the total sum of the numbers. The result is the answer.

Code Implementation

def difference_between_element_sum_and_digit_sum(numbers):
    element_sum = 0
    digit_sum = 0

    # Calculate the sum of all elements in the array.
    for number in numbers:
        element_sum += number

    # Iterate through the numbers to calculate digit sum.
    for number in numbers:
        string_number = str(number)

        # Convert each number to string to iterate over digits
        for digit in string_number:
            digit_sum += int(digit)

    return element_sum - digit_sum

Big(O) Analysis

Time Complexity

O(n log(k)) – The first step iterates through the input array of n numbers to calculate the element sum, which takes O(n) time. The second step iterates through the same array of n numbers. For each number, we extract its digits, which takes time proportional to the number of digits in that number. Assuming k is the maximum number of digits in any number in the input array, extracting digits takes O(log(k)) time on average (where k represents the maximum value in the input). Therefore, the digit sum calculation takes O(n log(k)) time. Finally, the difference calculation is O(1). The dominant term is O(n log(k)), so the overall time complexity is O(n log(k)).

Space Complexity

O(1) – The algorithm calculates the sum of numbers and the sum of digits iteratively without using any auxiliary data structures that scale with the input size. The intermediate sums and the final difference are stored in a constant number of variables. Therefore, the space complexity is constant, regardless of the number of input numbers N.

Optimal Solution

Approach

The goal is to find the difference between two things: the total when you add up all the numbers in a group, and the total when you add up all the digits that make up those numbers. The best way is to calculate each sum separately and then find the difference.

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

First, add up all the numbers in the group to get the element sum.
Next, for each number in the group, break it down into its individual digits and add them together. Do this for every number, and then add up all the digit sums to get the total digit sum.
Finally, subtract the total digit sum from the element sum. The result is the difference between the element sum and digit sum.

Code Implementation

def difference_between_sums(numbers):
    element_sum = sum(numbers)

    digit_sum = 0
    # Iterate through each number to calculate its digit sum
    for number in numbers:
        temporary_number = number
        while temporary_number > 0:
            digit = temporary_number % 10
            digit_sum += digit
            temporary_number //= 10

    # Calculate the difference between the element sum and digit sum
    difference = element_sum - digit_sum
    
    return difference

Big(O) Analysis

Time Complexity

O(n*m) – Let n be the number of elements in the input array and m be the maximum number of digits in any of the numbers in the array. Calculating the element sum takes O(n) time. Calculating the digit sum involves iterating through each element in the array (n elements). For each element, we iterate through its digits, which in the worst case is m digits. Thus, calculating the digit sum takes O(n*m) time. Since O(n*m) dominates O(n), the overall time complexity is O(n*m).

Space Complexity

O(1) – The algorithm calculates the element sum and digit sum iteratively. It primarily uses a few integer variables to store intermediate calculations like the element sum, digit sum, and potentially the current number being processed. The space used by these variables remains constant regardless of the size of the input array (N), where N is the number of elements in the array. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array	Return 0 since the sum of elements and digit sum would both be 0.
Array with a single element	The solution should work correctly, calculating element sum and digit sum, then return the absolute difference.
Array containing large numbers that may cause integer overflow when summed	Use a data type like long or long long to store the element sum and digit sum to prevent overflow.
Array with a large number of elements to check for performance implications	The solution has O(n) complexity so it should scale linearly and handle it efficiently, where n is the number of elements.
Array where all numbers are single-digit numbers	The digit sum and element sum will be identical so the absolute difference will be 0.
Array containing numbers with leading zeros (e.g., 007).	The problem states positive integers, so leading zeros shouldn't naturally occur, but if they do, treat '007' as 7 during digit sum calculation, effectively ignoring them.
Array containing only zeros.	The element sum and digit sum will both be 0, resulting in a return value of 0.
Array containing a very large number with many digits	The digit sum calculation should handle numbers with a large number of digits without causing performance issues; the calculation is still linear to the digits of each number.

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 possible value for each integer in the input array `nums`?
Can the input array `nums` be empty or null?
Are all the numbers in `nums` guaranteed to be non-negative?
Should the return value be an integer, and if the difference is too large to fit in a standard integer, what should I do?
Could you provide an example of an input `nums` and the expected output to ensure my understanding of the problem is correct?

Brute Force Solution

Approach

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

First, calculate the total sum of all the numbers in the collection.
Next, take each number individually and break it down into its individual digits.
Add up all of these individual digits to get the total digit sum.
Finally, subtract the total digit sum from the total sum of the numbers. The result is the answer.

Code Implementation

def difference_between_element_sum_and_digit_sum(numbers):
    element_sum = 0
    digit_sum = 0

    # Calculate the sum of all elements in the array.
    for number in numbers:
        element_sum += number

    # Iterate through the numbers to calculate digit sum.
    for number in numbers:
        string_number = str(number)

        # Convert each number to string to iterate over digits
        for digit in string_number:
            digit_sum += int(digit)

    return element_sum - digit_sum

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, add up all the numbers in the group to get the element sum.
Next, for each number in the group, break it down into its individual digits and add them together. Do this for every number, and then add up all the digit sums to get the total digit sum.
Finally, subtract the total digit sum from the element sum. The result is the difference between the element sum and digit sum.

Code Implementation

def difference_between_sums(numbers):
    element_sum = sum(numbers)

    digit_sum = 0
    # Iterate through each number to calculate its digit sum
    for number in numbers:
        temporary_number = number
        while temporary_number > 0:
            digit = temporary_number % 10
            digit_sum += digit
            temporary_number //= 10

    # Calculate the difference between the element sum and digit sum
    difference = element_sum - digit_sum
    
    return difference

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0 since the sum of elements and digit sum would both be 0.
Array with a single element	The solution should work correctly, calculating element sum and digit sum, then return the absolute difference.
Array containing large numbers that may cause integer overflow when summed	Use a data type like long or long long to store the element sum and digit sum to prevent overflow.
Array with a large number of elements to check for performance implications	The solution has O(n) complexity so it should scale linearly and handle it efficiently, where n is the number of elements.
Array where all numbers are single-digit numbers	The digit sum and element sum will be identical so the absolute difference will be 0.
Array containing numbers with leading zeros (e.g., 007).	The problem states positive integers, so leading zeros shouldn't naturally occur, but if they do, treat '007' as 7 during digit sum calculation, effectively ignoring them.
Array containing only zeros.	The element sum and digit sum will both be 0, resulting in a return value of 0.
Array containing a very large number with many digits	The digit sum calculation should handle numbers with a large number of digits without causing performance issues; the calculation is still linear to the digits of each number.