Intersection of Two Rectangles

Each axis-aligned rectangle is represented as [left, bottom, right, top], where left < right and bottom < top.
Return the positive-area intersection of first and second in the same format. If they do not overlap with positive area, return an empty array. Rectangles that only touch along an edge or at one point have no positive-area intersection.

Function
intersectRectangles(first: int[], second: int[]) → int[]

Examples
Example 1
first = [0,0,4,4]
second = [2,1,6,3]
return = [2,1,4,3]
The common horizontal interval is [2, 4], and the common vertical interval is [1, 3].
Example 2
first = [-3,-2,5,7]
second = [0,0,2,2]
return = [0,0,2,2]
The second rectangle lies completely inside the first.
Example 3
first = [0,0,2,2]
second = [2,0,5,3]
return = []
The rectangles touch at the vertical edge x = 2 but share no positive area.

Constraints
first.length == second.length == 4.
-10^9 <= left < right <= 10^9.
-10^9 <= bottom < top <= 10^9.

The trick is to treat each axis independently. The intersection's left is max(first left, second left). Its right is min(first right, second right). Do the same for bottom (max) and top (min). Then check that the new left is strictly less than the new right, and the new bottom is strictly less than the new top. If either check fails, return an empty array. The pitfall is using <= instead of <, which would return a zero-width rectangle for the touching case in Example 3. Another pitfall is returning null or [0,0,0,0] when the problem wants an empty array. Values go up to 10^9 in magnitude, and since you only compare and never add or multiply, overflow isn't a concern in most languages. If you freeze on the strictness check during the live OA, StealthCoder is the hedge that gives you the working code. Complexity is O(1) time and space.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.