Date: 2026-06-06
Time: 19:06
shuffle-the-array/solution.pyThis file solves LeetCode 1470 — Shuffle the Array. It takes an array of 2n elements structured as [x1, x2, ..., xn, y1, y2, ..., yn] and interleaves the two halves to produce [x1, y1, x2, y2, ..., xn, yn].
Solution.shuffle(nums, n) — The sole method. Takes the full array and the half-length n, returns a new list with elements interleaved.
nums has exactly 2n elements. n splits the array into two equal halves. The caller (LeetCode harness) guarantees 1 <= n <= 500.nums[:n] and nums[n:] and zipping, it uses a single loop with offset indexing (nums[i] and nums[n + i]). This avoids allocating intermediate lists.result from scratch rather than mutating nums in-place. This is the simplest correct approach — in-place interleaving for this problem requires bit-packing tricks that aren't worth the complexity.typing.List — used only for the type annotation on the method signature.shuffle-the-array/test_solution.py and (per the provided list) hundreds of other test files — likely an artifact of a shared test harness or import pattern across the repo, not actual usage of this solution class.1. Initialize empty result list.
2. Loop i from 0 to n - 1:
nums[i] (the i-th element from the first half).nums[n + i] (the i-th element from the second half).3. Return result.
For input [2, 5, 1, 3, 4, 7] with n = 3: the loop pairs (2,3), (5,4), (1,7) → output [2, 3, 5, 4, 1, 7].
i iterations of the loop, result contains exactly 2i elements: the first i pairs interleaved.nums.None. The method trusts its inputs — no bounds checking on n vs len(nums). This is standard for LeetCode solutions where input constraints are guaranteed by the problem.
shuffle-the-array/test_solution.py — See what edge cases the tests cover (n=1, large arrays, duplicate values)shuffle-the-array/plan.md — Read the planning notes to understand if alternative approaches (zip, in-place) were consideredshuffle-the-array/review.md — Check if the review flagged any complexity or style issuesin-place-interleave — The O(1) space version of this problem uses cycle-leader or encoding tricks; worth comparingshuffle-string/solution.py:Solution.restoreString — A related permutation-based shuffle problem in this reposhuffle-returns-new-list — shuffle always returns a newly allocated list and never mutates the input numsshuffle-output-length-equals-input — The returned list has exactly 2n elements, same as the inputshuffle-linear-time — The algorithm runs in O(n) time with a single pass over the indicesshuffle-no-validation — No runtime checks exist for len(nums) == 2*n; the method relies on caller-guaranteed preconditions