Founder of SAGEWORKS AI — building the Web4 layer where AI, blockchain & time flow as one. Creator of Mind’s Eye and BinFlow. Engineering the future of temporal, network-native intelligence.
This is one of those geometry problems that quietly tests whether you understand systems, not just math.
A lot of “area split” solutions die the moment overlap exists because they keep summing squares like the world is polite. It isn’t. Union area is where reality shows up.
What’s good here:
Sweep line framing is exactly right: turn 2D union area into “covered width over y-intervals” and integrate. That’s the whole game.
Segment tree storing covered length (not sum) is the key distinction. Counting overlaps doesn’t help — covered measure is what you need.
The “find y where cumulative area hits total/2” step is basically a continuous median problem, and linear interpolation inside the slab is the correct move.
Two tiny “don’t get cooked later” notes:
Handle width = 0 safely.
There can be y-intervals where no square is active, so interpolation would divide by zero. Your Python has the guard, JS/C++ should too (or skip those stages).
Storing stages:
You don’t need to keep a separate stages list with only (y, width) if you can just store (yStart, yEnd, width) as you sweep — it makes the second pass cleaner and avoids weird indexing assumptions.
Overall though: this is the clean, correct pattern for rectangle/square union + “split by area” problems. Once you learn this, a whole class of computational-geometry tasks stop being scary and start being mechanical.
This is one of those geometry problems that quietly tests whether you understand systems, not just math.
A lot of “area split” solutions die the moment overlap exists because they keep summing squares like the world is polite. It isn’t. Union area is where reality shows up.
What’s good here:
Sweep line framing is exactly right: turn 2D union area into “covered width over y-intervals” and integrate. That’s the whole game.
Segment tree storing covered length (not sum) is the key distinction. Counting overlaps doesn’t help — covered measure is what you need.
The “find y where cumulative area hits total/2” step is basically a continuous median problem, and linear interpolation inside the slab is the correct move.
Two tiny “don’t get cooked later” notes:
Handle width = 0 safely.
There can be y-intervals where no square is active, so interpolation would divide by zero. Your Python has the guard, JS/C++ should too (or skip those stages).
Storing stages:
You don’t need to keep a separate stages list with only (y, width) if you can just store (yStart, yEnd, width) as you sweep — it makes the second pass cleaner and avoids weird indexing assumptions.
Overall though: this is the clean, correct pattern for rectangle/square union + “split by area” problems. Once you learn this, a whole class of computational-geometry tasks stop being scary and start being mechanical.
Thank you so much sir for simplifying it even further for us!