Google Interview Question for SDE1s

Country: United States

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of 0 vote

the question will be, random with what distribution?
I assume with a uniform distribution: each point in the final region has the same probability to be chosen. For a single rectangle this is straight forward:

x = rectangle.left + random(rectangle.width);
y = + random(rectangle.height);

For multiple rectangles that don't overlap, we can say, we choose the rectangle proportionally to it's area and once the rectangle is chosen, pick a point in there.

long long are_sum = 0;
for(Rectangle& r : rectangles) {
	area_sum += r.width * r.height;
	if(random(area_sum) < r.width * r.height) {
		x = rectangle.left + random(rectangle.width);
		y = + random(rectangle.height);

But if they overlap it's tricky. The easiest way I can think of is to union the new rectangle with the existing area and only consider the rectangles that are not already covered and apply method 2.

- Chris December 03, 2017 | Flag Reply

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