{"results":[{"id":"abstraction-cost-predicts-convergence-strength","text":"Strategy convergence strength in an uncoordinated repo appears predictable from abstraction overhead: streaming (zero abstractions) converges strongest, hash-then-stream (one preprocessing step with Counter/set) converges next, sort-then-scan (ordering prerequisite plus pointer management) converges weakest — the adoption barrier gradient closely tracks the abstraction cost gradient.","truth_value":"IN","justification_count":1,"dependent_count":2,"challenges":[],"last_reviewed":"2026-06-07T22:02:22","review_result":"pass","source_type":""},{"id":"alien-dict-assumes-valid-order","text":"The alien dictionary solution assumes `order` covers all characters in `words`; a missing character produces an unhandled `KeyError` — no input validation.","truth_value":"IN","justification_count":0,"dependent_count":0,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"alien-dict-rank-map-idiom","text":"The alien dictionary solution builds a `{char: index}` dict from the ordering string, converting custom-alphabet comparison to integer comparison with O(1) lookups.","truth_value":"IN","justification_count":0,"dependent_count":0,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"anagram-mappings-lifo-index-order","text":"When duplicates exist, `anagramMappings` assigns indices in LIFO order (last-appended index consumed first) because `deque.pop()` removes from the right.","truth_value":"IN","justification_count":0,"dependent_count":0,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"apples-mutates-input","text":"`maxNumberOfApples` mutates the caller's list via `weight.sort()` rather than using `sorted()`, so callers cannot rely on original order being preserved.","truth_value":"IN","justification_count":0,"dependent_count":0,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"arithmetic-progression-mutates-input","text":"`can_construct` calls `arr.sort()`, mutating the input list in place; callers needing the original order must pass a copy.","truth_value":"IN","justification_count":0,"dependent_count":0,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"assign-cookies-mutates-inputs","text":"`find_content_children` mutates both input lists via in-place `.sort()`; callers cannot assume list order is preserved.","truth_value":"IN","justification_count":0,"dependent_count":1,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"ba-substring-equivalence","text":"`\"ba\" not in s` is equivalent to \"every 'a' precedes every 'b'\" when the input contains only 'a' and 'b' — the only way to violate the ordering is a b-to-a transition, which is exactly the substring \"ba\".","truth_value":"IN","justification_count":0,"dependent_count":0,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"bfs-guarantees-manhattan-order","text":"BFS from a center cell on a grid with 4-directional edges produces cells in non-decreasing Manhattan distance order, because every edge has weight 1 and BFS explores layer by layer.","truth_value":"IN","justification_count":0,"dependent_count":0,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"binary-watch-deterministic-order","text":"`readBinaryWatch` output is ordered hours ascending, then minutes ascending within each hour, as a direct consequence of nested `range()` iteration order.","truth_value":"IN","justification_count":0,"dependent_count":0,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"bst-inorder-no-materialized-list","text":"The BST minimum-difference solution computes the answer in O(h) stack space during traversal without collecting all node values into a list first.","truth_value":"IN","justification_count":0,"dependent_count":0,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"bst-min-diff-between-inorder-neighbors","text":"The minimum absolute difference in a BST always occurs between two values adjacent in the inorder (sorted) traversal; the solution exploits this by comparing only consecutive visits rather than all pairs.","truth_value":"IN","justification_count":0,"dependent_count":0,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"build-array-encode-order-safety","text":"The `% n` in `nums[nums[i]] % n` during the encode pass ensures correctness regardless of whether `nums[nums[i]]` has already been encoded earlier in the same loop iteration.","truth_value":"IN","justification_count":0,"dependent_count":0,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"build-tree-bfs-from-level-order","text":"Tree solutions use a `build_tree` utility that constructs a `TreeNode` tree from a level-order list (LeetCode's serialization format) via BFS queue traversal, with `None` representing absent nodes.","truth_value":"IN","justification_count":0,"dependent_count":0,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"build-tree-level-order","text":"`_build_tree(values)` constructs a tree from a level-order (BFS) list where `None` represents absent children, matching LeetCode's standard serialization format; it is used across tree-problem test suites","truth_value":"IN","justification_count":0,"dependent_count":1,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"build-tree-level-order-convention","text":"`build_tree` constructs trees from level-order lists (LeetCode's serialization format) using BFS, where `None` entries represent missing nodes — this is the standard tree construction interface across the repo.","truth_value":"IN","justification_count":0,"dependent_count":1,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"build-tree-levelorder-serialization","text":"`build_tree` constructs trees level-by-level using a queue, matching LeetCode's standard level-order serialization format where `None` marks absent nodes.","truth_value":"IN","justification_count":0,"dependent_count":0,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"build-tree-uses-leetcode-level-order","text":"`_build_tree` constructs trees from LeetCode's standard level-order serialization format (BFS order with `None` for missing nodes), making test cases directly copy-pasteable from problem examples.","truth_value":"IN","justification_count":0,"dependent_count":0,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"build-tree-uses-level-order","text":"`build_tree` deserializes LeetCode's bracket/level-order format using a FIFO queue (`list.pop(0)`), assigning left then right children per node; `None` entries represent absent children.","truth_value":"IN","justification_count":0,"dependent_count":0,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""},{"id":"can-place-flowers-greedy-is-optimal","text":"Greedy left-to-right placement is provably optimal: planting at the earliest valid slot never reduces the number of remaining valid slots compared to any alternative placement order.","truth_value":"IN","justification_count":0,"dependent_count":1,"challenges":[],"last_reviewed":null,"review_result":null,"source_type":""}],"count":168,"limit":20,"offset":0}