Date: 2026-06-06
Time: 16:23
difference-between-element-sum-and-digit-sum-of-an-array/solution.pyThis file solves LeetCode 2535. It computes two sums over an input array — the element sum (sum of the numbers themselves) and the digit sum (sum of every individual digit across all numbers) — then returns their absolute difference.
It follows the repo's convention: one solution file per problem directory, exporting a single function that encapsulates the algorithm.
differencebetweenelementanddigit_sum(nums: list[int]) -> int — The sole public function. Contract:
abs(elementsum - digitsum), a non-negative integerSingle-pass accumulation. Both sums are computed in one loop over nums, avoiding a second traversal. The digit extraction happens inline via repeated % 10 / //= 10 — standard modular arithmetic rather than string conversion. This is the idiomatic numeric approach and avoids allocating intermediate string objects.
Destructive parameter reuse. The loop variable num is mutated in-place by the while num > 0 loop. This is safe because num is a rebinding of the loop variable, not a mutation of the input list.
Imports: None. The solution is self-contained with no stdlib or third-party dependencies.
Imported by: The test_solution.py in the same directory imports this function. The massive "Imported By" list in the prompt is misleading — those are unrelated test files that happen to share a test runner infrastructure, not actual consumers of this function.
1. Initialize elementsum and digitsum to 0.
2. For each num in nums:
num to element_sum (the whole number).num % 10, accumulate into digit_sum, then integer-divide num by 10. Repeat until num reaches 0.3. Return abs(elementsum - digitsum).
For input [1, 15, 6, 3]: element sum = 25, digit sum = 1+1+5+6+3 = 16, result = 9.
while num > 0 guard means num = 0 contributes zero digits (not even a single "0" digit). This is correct per the LeetCode constraint that 1 <= nums[i] <= 2000, but would silently skip zeros if the constraint were relaxed.abs() call. In practice, for positive integers, element sum >= digit sum always holds (since a multi-digit number is always larger than the sum of its digits), so abs() is technically redundant but makes the contract explicit.num variable is a local copy of each element.None. The function trusts its caller to provide a valid list[int] of positive integers, consistent with LeetCode's constraint-based model. An empty list returns 0 (both sums start at 0).
difference-between-element-sum-and-digit-sum-of-an-array/test_solution.py — See what edge cases the test suite covers (empty list, single-digit numbers, large values)subtract-the-product-and-sum-of-digits-of-an-integer/solution.py:subtractProductAndSum — A closely related digit-decomposition problem using the same % 10 / //= 10 patternadd-digits/solution.py:addDigits — Another digit-sum problem, likely using the digital root formula instead of explicit extractiondifference-between-element-sum-and-digit-sum-of-an-array/review.md — The code review notes for this solution, may flag the abs() redundancy or alternative approacheselement-sum-gte-digit-sum — For positive integers, element sum is always >= digit sum, making abs() a no-op guard rather than a functional requirementzero-input-skipped — A 0 element contributes nothing to digit_sum because the while num > 0 loop body never executes for zerosingle-pass-complexity — Time complexity is O(N * D) where N is array length and D is max digit count per element; space is O(1)no-string-conversion — Digit extraction uses modular arithmetic exclusively, avoiding str() conversion and intermediate allocations