File: min-max-game/solution.py

Date: 2026-06-06

Time: 17:50

min-max-game/solution.py

Purpose

This file solves LeetCode 2293 — Min Max Game. It simulates a tournament-style reduction of an array: pair up adjacent elements, alternately take the min or max of each pair, and repeat until one element remains.

Key Components

min_steps(nums: List[int]) -> int — Despite the name suggesting "steps," this function returns the *last remaining value*, not a count. It takes an array whose length is a power of 2 and reduces it by half each round until a single element is left.

The pairing rule per round:

• Even-indexed pairs (i % 2 == 0): take min(nums[2*i], nums[2*i+1])
• Odd-indexed pairs (i % 2 == 1): take max(nums[2*i], nums[2*i+1])

Patterns

Iterative simulation — Rather than using recursion or a mathematical shortcut, it directly simulates the game by building a new half-sized array each round. This is the most straightforward approach: O(n) total work across all rounds (n + n/2 + n/4 + ... = 2n), O(n) space for the working array.

In-place reassignment — nums = new_nums at the end of each iteration replaces the old array, letting the GC reclaim previous rounds. No mutation of the input.

Dependencies

• Imports: List from typing (type annotation only).
• Imported by: min-max-game/testsolution.py — the test harness for this problem. The massive "Imported By" list in the prompt is an artifact of the repo's test infrastructure importing a shared test runner, not actual callers of minsteps.

Flow

1. Enter the while loop — runs log2(len(nums)) times.

2. Each iteration pairs elements (nums[0], nums[1]), (nums[2], nums[3]), etc.

3. For pair index i: apply min if i is even, max if i is odd.

4. Collect results into new_nums, reassign nums.

5. When len(nums) == 1, return nums[0].

For [1, 3, 5, 2, 4, 8, 2, 2]:

• Round 1: min(1,3)=1, max(5,2)=5, min(4,8)=4, max(2,2)=2 → [1, 5, 4, 2]
• Round 2: min(1,5)=1, max(4,2)=4 → [1, 4]
• Round 3: min(1,4)=1 → [1] → return 1

Invariants

• Power-of-2 length: The algorithm assumes len(nums) is a power of 2. If not, the final unpaired element would be silently dropped — no validation guards this.
• Non-empty input: If nums is empty, nums[0] on the return would raise IndexError.
• Alternating min/max is index-based: The min/max alternation resets each round (pair 0 is always min).

Error Handling

None. The function trusts its caller to provide a valid power-of-2-length array. Invalid inputs produce silent wrong answers or index errors rather than descriptive exceptions. This is typical for LeetCode solutions where the problem guarantees valid input.

Naming Note

The function is named min_steps, which is misleading — it returns the surviving *value*, not a step count. This likely came from a naming template applied across the repo and wasn't corrected for this problem.

Topics to Explore

• [file] min-max-game/test_solution.py — Test cases revealing edge behavior (single-element input, all-equal arrays)
• [file] min-max-game/plan.md — The planning doc may explain why iterative simulation was chosen over recursion
• [function] delete-greatest-value-in-each-row/solution.py:min_steps — Another iterative-reduction problem to compare structural similarity
• [general] tournament-reduction-patterns — How other problems (e.g., segment trees, elimination games) use the same halving structure
• [file] min-max-game/review.md — May flag the naming issue or discuss alternative approaches

Beliefs

• min-max-game-returns-value-not-count — min_steps returns the last surviving element, not the number of reduction rounds
• min-max-alternation-resets-per-round — Pair index i resets to 0 each round, so the first pair always uses min
• min-max-game-linear-total-work — Total comparisons across all rounds is O(n) due to geometric halving (n/2 + n/4 + ...)
• min-max-game-no-input-validation — The function does not verify that len(nums) is a power of 2 or that the list is non-empty