Rectangle Fit Queries

For this exercise, use the callable contract below.
Process the rows of operations from left to right. Each row has one of two forms:
[0, a, b]: create and save a rectangle of size a × b.
[1, a, b]: determine whether every rectangle saved by earlier operations can fit inside a box of size a × b.
Test each saved rectangle separately; the rectangles do not need to fit in the box at the same time. You may rotate a rectangle by 90 degrees.
Return one boolean for each query operation, in query order.

Function
solution(operations: int[][]) → boolean[]

Examples
Example 1
operations = [[1,1,1]]
return = [true]
No rectangles have been saved, so every saved rectangle vacuously fits and the answer is true.
Example 2
operations = [[0,1,3],[0,4,2],[1,3,4],[1,3,2]]
return = [true,false]
Both saved rectangles fit the 3 × 4 box after choosing the appropriate orientation. The 4 × 2 rectangle cannot fit the later 3 × 2 box, so the answers are [true,false].

Constraints
operations contains at least one row.
operations[i].length = 3.
operations[i][0] is either 0 or 1.
operations[i][1] and operations[i][2] are positive integers.

Normalize every rectangle so its sides are (min, max). A rectangle fits a box if, after normalizing the box the same way, rect.min <= box.min and rect.max <= box.max. Rotation is already handled by sorting the sides. Now the key: you don't need to store rectangles. Keep maxMin, the largest of all the smaller sides, and maxMax, the largest of all the larger sides. A query passes only if box.min >= maxMin and box.max >= maxMax. Every rectangle fits separately, so the two maxima fully decide it. With no saves, both start at 0 and the answer is true. The common pitfall is looping over every saved rectangle per query, which turns O(n) into O(n^2). Another is forgetting to normalize the box. Each operation is O(1). StealthCoder is the hedge if the live OA makes you second-guess the reduction.

StealthCoder is the hedge for the one pattern you didn't drill. It runs invisibly during the screen share.