- was able to solve the first one, completely with code and test-run.
Wrote the code in intellij , interviewer was okay with me using any ide
as long as it did not have any AI features like copilot. 

- discussed the approach and pseudo code for the 2nd one 
 - It started as a geometry problem and evolved to a intervals problem through clever observations
 - Interviewer was very good at giving hints and guiding my direction of thought process. 

Count Non-Overshadowed Triangles

Problem Statement

You are given a list of n points in a 2D plane, 
where each point represents the top vertex of a right-angled triangle.
 For each point (x, y) (with y > 0), two lines are drawn through it: one with slope +1 and another with slope -1. 
These lines intersect the X-axis (where y = 0) at two points,
forming a triangle with vertices at (x, y) and the two intersection points.

A triangle is overshadowed if its top vertex (x, y) lies 
strictly inside or on the boundary of another triangle 
formed by a different point. 
Your task is to count the number of triangles that are not overshadowed by any other triangle.

Input

points: A list of n integer arrays, where points[i] = [x_i, y_i] represents the coordinates of the i-th point.


1 <= n <= 10^5
-10^9 <= x_i <= 10^9
1 <= y_i <= 10^9

Output

Return an integer representing the number of triangles that are not overshadowed.

Examples
Example 1
Input: points = [[0,1], [2,1]]
Output: 2

Explanation:

For point [0,1]:
Lines with slopes +1 and -1 intersect the X-axis at [1,0] and [-1,0].
Forms triangle [[0,1], [1,0], [-1,0]].


For point [2,1]:
Lines intersect at [3,0] and [1,0].
Forms triangle [[2,1], [3,0], [1,0]].


Neither top vertex ([0,1] or [2,1]) lies inside or on the boundary of the other triangle.
Result: Both triangles are not overshadowed, so output is 2.


Example 2
Input: points = [[0,2], [1,1]]
Output: 1

Explanation:

For point [0,2]:
Lines intersect at [2,0] and [-2,0].
Forms triangle [[0,2], [2,0], [-2,0]].


For point [1,1]:
Lines intersect at [2,0] and [0,0].
Forms triangle [[1,1], [2,0], [0,0]].


The top vertex [1,1] lies on the boundary of the triangle formed by [0,2], so it is overshadowed.
Result: Only the triangle from [0,2] is not overshadowed, so output is 1.


Example 3
Input: points = [[0,2], [1,1], [3,3]]
Output: 1

Explanation:

Triangle from [0,2]:
Base interval: [-2,2].
Vertices: [[0,2], [-2,0], [2,0]].


Triangle from [1,1]:
Base interval: [0,2].
Vertices: [[1,1], [0,0], [2,0]].


Triangle from [3,3]:
Base interval: [0,6].
Vertices: [[3,3], [0,0], [6,0]].


Point [1,1] is on the boundary of [0,2]’s triangle, so its triangle is overshadowed.
Point [0,2] is inside [3,3]’s triangle, so its triangle is overshadowed.
Result: Only [3,3]’s triangle is not overshadowed, so output is 1.

Constraints

1 <= points.length <= 10^5
points[i].length == 2
-10^9 <= points[i][0] <= 10^9
1 <= points[i][1] <= 10^9
All points are unique.

Notes

Lines with slopes +1 and -1 through a point (x, y) intersect the X-axis at (x + y, 0) and (x - y, 0), respectively.
A point is considered inside or on the boundary of a triangle if it lies within its interior, on its edges, or at its vertices.
Assume floating-point precision does not affect containment checks.

Solution Hints

Step 1: Remove duplicate points to avoid counting identical triangles.
Step 2: For each point (x, y), compute base points (x - y, 0) and (x + y, 0), forming interval [x - y, x + y].
Step 3: Use a heap to sort points by x - y. Check if a point’s top vertex is inside another triangle by:
Ensuring its X-coordinate is within the other’s base interval.
Checking if its Y-coordinate is below the lines’ Y-values at that X.


Step 4: Count triangles not overshadowed, using a list to track active triangles.

There is a Parking lot. We need to implement  allocateSlot()  and vacateSlot() funcitons .
The parking area has z floors, each floor having x rows and y columns. Entry point is at x = 0, y = 0, z = 0.

For simplification, distance from entry point was taken to be x + y + z. 
If two slots had same distance from the entry point, then the slot with smallest floor number would be considered, followed by column number and row number.

Could be solved using priority queue and matrix of size x * y * z ! 

Given a multi story parking lot with z floors, and each floor with x rows and y columns. A combination of x, y, z defines a unique spot in parking lot.
There is a single entrance to this parking lot at ground floor z=0, x=0, y=0. Write below two functionsallocateNearestSpot()
vacateSpot(z, x, y)