You are given a positive integer array nums.
nums.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 <= 20001 <= nums[i] <= 2000Problem Overview: You are given an integer array nums. Compute two values: the sum of all elements in the array and the sum of every digit from those elements. The task is to return the absolute difference between these two sums.
Approach 1: Iterative Digit Extraction to Compute Sums (Time: O(n * d), Space: O(1))
This approach processes each number in the array and extracts its digits one by one. First, iterate through the array and accumulate the element sum. Then for each number, repeatedly apply num % 10 to get the last digit and add it to the digit sum, followed by num /= 10 to remove the digit. This continues until the number becomes zero. The key idea is that digit extraction can be done using simple arithmetic operations without converting numbers to strings. If the array has n elements and each number contains up to d digits, the time complexity becomes O(n * d). Space complexity remains O(1) since only running totals are stored. This is the most straightforward implementation and works well in any language.
Approach 2: Mathematical Handling by Modulo (Time: O(n * d), Space: O(1))
This version focuses on computing the difference directly using arithmetic properties instead of maintaining two separate totals. For each element, add the full number to the element sum while simultaneously subtracting its digits from a running difference using modulo operations. Each iteration extracts digits with % 10 and reduces the number using integer division by 10. The insight is that the difference between element sum and digit sum can be accumulated incrementally rather than calculated at the end. This keeps the logic tight and avoids maintaining multiple aggregates. Complexity remains O(n * d) because every digit of every number must still be processed, while space usage stays O(1). The algorithm relies purely on arithmetic operations commonly used in math problems.
Both approaches rely on simple iteration over the input array, a common pattern in array problems. Digit extraction using modulo is also a fundamental technique in many math and number manipulation questions.
Recommended for interviews: Interviewers typically expect the iterative digit extraction approach. It clearly demonstrates that you understand how to manipulate numbers using modulo and division. The mathematical accumulation variant is slightly cleaner but conceptually the same. Showing the basic digit extraction first proves your understanding, while writing it in a compact form shows implementation maturity.
This approach uses iteration to extract each digit for calculating the digit sum. The element sum is simply the sum of all numbers. For the digit sum, convert each number to its constituent digits by using modulo and integer division.
This C solution iterates over the array to compute both the element sum and the digit sum. The digit extraction uses modulo and division to break down each integer into its digits.
Time Complexity: O(n * log(max(nums[i]))), where n is the number of elements and max(nums[i]) is the maximum number in nums. Space Complexity: O(1).
This approach is the same iterative technique but focuses on optimizing the handling by utilizing mathematical operations directly to extract each digit, avoiding string conversions.
This C solution iteratively calculates the digit sum for each number using modulo and integer division, ensuring efficient digit extraction without string conversion.
Time Complexity: O(n * log(max(nums[i]))), where n is the number of elements and max(nums[i]) is the maximum number in nums. Space Complexity: O(1).
We traverse the array nums, calculate the sum of the elements x and the sum of the digits y, and finally return |x - y|. Since x is always greater than or equal to y, we can directly return x - y.
The time complexity is O(n times log_{10} M), where n and M are the length of the array nums and the maximum value of the elements in the array, respectively. The space complexity is O(1).
| Approach | Complexity |
|---|---|
| Iterative Digit Extraction to Compute Sums | Time Complexity: O(n * log(max(nums[i]))), where n is the number of elements and max(nums[i]) is the maximum number in nums. Space Complexity: O(1). |
| Mathematical Handling by Modulo | Time Complexity: O(n * log(max(nums[i]))), where n is the number of elements and max(nums[i]) is the maximum number in nums. Space Complexity: O(1). |
| Simulation | — |
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Iterative Digit Extraction to Compute Sums | O(n * d) | O(1) | General case. Clear and easy to implement during interviews. |
| Mathematical Handling by Modulo | O(n * d) | O(1) | When you want a concise solution that accumulates the difference directly. |
2535. Difference Between Element Sum and Digit Sum of an Array | Weekly Contest 328 | LeetCode 2535 • Bro Coders • 1,690 views views
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