Date: 2026-06-06
Time: 16:07
counting-elements/solution.pyThis file solves LeetCode problem "Counting Elements" (problem 1426). It owns exactly one responsibility: given an array of integers, count how many elements x have x + 1 also present in the array. It's one of ~500+ solution files in the leetcode-implementations repo, each solving a single problem behind a uniform interface.
count_elements(arr: list[int]) -> int — The sole public function. Contract: accepts a list of integers, returns the count of elements whose immediate successor exists anywhere in the array. Duplicates are counted independently — if arr = [1, 1, 2], both 1s contribute to the count because 2 exists.
Set-based lookup: The solution converts the array to a set once (O(n) space), then iterates the original array checking membership (O(1) per check). This is the canonical "trade space for time" idiom for existence queries — total O(n) time and space.
Generator expression with sum: sum(1 for x in arr if x + 1 in s) is a Python idiom for conditional counting. It's equivalent to len([x for x in arr if x + 1 in s]) but avoids materializing the intermediate list.
Iterating arr, not s: This is deliberate. The iteration is over the original list, not the set, so duplicate values are counted multiple times. For [1, 1, 2], iterating the set would yield 1 (just the element 1), but iterating the list correctly yields 2.
Imports: None — pure standard library, no external dependencies.
Imported by: The "Imported By" list in the prompt is misleading — those are test files across the entire repo that import their *own* solution.py, not this one. The actual dependent is counting-elements/testsolution.py, which imports countelements to verify it.
1. Build a set s from arr — deduplicates values for O(1) lookups.
2. Iterate every element x in the original arr.
3. For each x, check if x + 1 is in s.
4. Sum the count of elements passing the check.
Single-pass over arr after the set construction. No mutation of input.
arr.arr is evaluated independently — the count reflects multiplicity, not distinct values.x + 1, not x - 1 or any other neighbor.None. The function assumes valid input per the LeetCode contract (list of integers). An empty list returns 0 naturally since the generator produces nothing. No explicit validation, no exceptions raised.
counting-elements/test_solution.py — See the edge cases tested (empty array, all duplicates, no successors)counting-elements/review.md — Review notes on this solution's approach and alternativestwo-sum/solution.py:twoSum — Another set/hash-based lookup pattern for comparisonset-vs-sort-for-existence — When sorting + binary search beats set construction (e.g., memory-constrained scenarios)counting-elements/plan.md — The problem decomposition that led to the set-based approachcounting-elements-counts-with-multiplicity — count_elements counts each array element independently; duplicates contribute separately to the total (iterates arr, not the set).counting-elements-linear-complexity — The solution runs in O(n) time and O(n) space via a single set construction and one pass over the input.counting-elements-no-mutation — The function never modifies the input list; the set is a separate allocation.counting-elements-successor-only — The check is strictly x + 1 in s; predecessor existence (x - 1) does not contribute to the count.