Date: 2026-06-06
Time: 16:55
get-maximum-in-generated-array/solution.pySolves LeetCode 1646: Get Maximum in Generated Array. The file owns the generation of an array defined by a specific recurrence relation and finding its maximum value. It follows the repo's convention of one Solution class per problem directory.
Solution.getMaximumGenerated(self, n: int) -> int — the sole method. Given n, it constructs an array nums of length n + 1 using these rules:
| Index i | Value |
|-----------|-------|
| 0 | 0 |
| 1 | 1 |
| even i (≥ 2) | nums[i // 2] |
| odd i (≥ 3) | nums[i // 2] + nums[i // 2 + 1] |
Then returns max(nums).
Bottom-up DP. The array is filled iteratively from index 2 to n. Each element depends only on previously computed elements, so a single forward pass suffices. This is the standard approach for recurrence-based problems — no memoization overhead, O(n) time and space.
Early return for edge case. The n == 0 guard at line 13 avoids an out-of-bounds write to nums[1] when the array has length 1.
Imports: None — pure Python, no standard library or third-party dependencies.
Imported by: get-maximum-in-generated-array/test_solution.py directly. The "Imported By" list in the prompt is misleadingly large — those are other problems' test files that share a common test harness pattern importing from their own solution.py, not this file.
1. If n == 0, return 0 immediately.
2. Allocate nums as a zero-filled list of size n + 1.
3. Set nums[1] = 1 (base case).
4. Iterate i from 2 through n:
i: copy from nums[i // 2] (the "parent" in the recurrence).i: sum nums[i // 2] and nums[i // 2 + 1] (adjacent pair in the "parent" region).5. Return max(nums).
The key insight for the odd case: integer division of an odd number i gives (i-1)/2, so the formula combines two adjacent elements from the first half of the array.
n >= 0 is assumed by the problem constraints. No negative-input guard exists.i, all indices < i are already computed. This holds because both i // 2 and i // 2 + 1 are strictly less than i for i >= 2.nums never shrinks or grows after allocation — fixed-size array.None. The method trusts LeetCode's constraint that 0 <= n <= 100. Passing n < 0 would return 0 (the early return checks n == 0, but the range(2, n+1) would be empty for n = 1 too, which is fine since nums[1] = 1 is set and max([0, 1]) returns 1). Negative n would create a zero-length list and then fail at nums[1] = 1 with an IndexError.
nums array. Could be reduced with a sliding window if only the max were needed, but at the cost of clarity for no practical gain given n <= 100.generated-array-recurrence-correctness — For odd index i, nums[i // 2 + 1] is always in bounds because i // 2 + 1 <= i - 1 for all i >= 3, and the array has length n + 1generated-array-n0-guard-required — The n == 0 early return is necessary; without it, nums[1] = 1 would raise IndexError on a length-1 listgenerated-array-single-pass-dp — The solution computes all values in one forward pass because every dependency index (i // 2, i // 2 + 1) is strictly less than the current indexgenerated-array-no-external-deps — The solution uses no imports and depends only on Python built-in max() and integer arithmetic