Date: 2026-06-06
Time: 17:55
minimum-cost-to-move-chips-to-the-same-position/solution.pyThis file solves LeetCode 1217: Minimum Cost to Move Chips to the Same Position. It's the canonical solution module for this problem within the repo's one-directory-per-problem structure.
The problem: you have chips at various positions on a number line. Moving a chip by 2 costs 0; moving by 1 costs 1. Find the minimum total cost to stack all chips at one position.
sort_array(position: list[int]) -> int — Despite the name (likely a leftover from the repo's code generation pipeline), this function computes the minimum cost. It takes a list of chip positions and returns an integer cost.
The function body is two lines:
1. odd = sum(p % 2 for p in position) — counts how many chips sit at odd positions.
2. return min(odd, len(position) - odd) — returns the smaller of the odd-count and even-count.
Parity reduction. The key insight is that moving a chip by 2 (any even distance) is free, so all chips on even positions can be consolidated to any single even position at zero cost, and likewise for odd positions. The only real cost is moving chips across the parity boundary — each such move costs 1. So the minimum cost is min(countodd, counteven): move whichever group is smaller into the other.
Generator expression inside sum() — idiomatic Python for counting elements matching a predicate without materializing a list.
Imports: None. Pure function, no standard library or third-party dependencies.
Imported by: minimum-cost-to-move-chips-to-the-same-position/testsolution.py directly. The massive "Imported By" list in the prompt is an artifact of the repo's test infrastructure — every testsolution.py likely imports a shared test runner or uses a common pattern that shows up in static analysis, not a real dependency on this specific solution.
1. Iterate over every position in the input list.
2. Compute p % 2 for each — yields 1 for odd, 0 for even.
3. Sum those values to get the count of odd-positioned chips.
4. The even count is implicit: len(position) - odd.
5. Return the minimum of the two counts.
Time complexity: O(n) — single pass. Space complexity: O(1) — no auxiliary data structures.
position is non-empty (LeetCode guarantees 1 <= len(position)).position are non-negative integers (per problem constraints).None. An empty list would return min(0, 0) = 0, which happens to be reasonable but isn't explicitly guarded. Invalid inputs (non-integers, negative values) would propagate exceptions from % or sum() naturally.
minimum-cost-to-move-chips-to-the-same-position/test_solution.py — See what edge cases the tests cover (single chip, all same parity, etc.)minimum-cost-to-move-chips-to-the-same-position/plan.md — Understand the problem decomposition that led to this parity-based approachparity-reduction-pattern — Other LeetCode solutions in this repo that reduce a problem to odd/even counting (e.g., minimum-changes-to-make-alternating-binary-string)minimum-cost-to-move-chips-to-the-same-position/solution.py:sort_array — The function name doesn't match its purpose; worth checking if this naming is systematic across the repo or a one-off artifactchips-parity-cost-zero — Moving a chip by any even distance costs 0, so all chips sharing parity can be consolidated for free; the minimum cost equals the size of the smaller parity group.sort-array-misnaming — The function is named sort_array but performs no sorting; it computes a minimum cost via parity counting.single-pass-o1-space — The solution runs in O(n) time and O(1) space with no auxiliary data structures.no-input-validation — The function performs no validation and relies on the caller to provide a non-empty list of non-negative integers per LeetCode constraints.