Define a function f(y) = area above y - area below y, which is continuous and monotonically decreasing from total area to -total area, guaranteeing a root. Use binary search on y within the grid's vertical bounds, iterating until the interval width is less than epsilon, and compute f(y) efficiently by summing contributions from each square. Then extend to overlapping squares by handling intersections and to rectangles by adjusting area formulas.
Pro tip: Emphasize that the problem reduces to finding a root of a monotonic function, and that binary search is optimal because f(y) can be evaluated in O(n) time. Mention that for overlapping squares, you can still compute f(y) in O(n) by summing individual contributions, but for rectangles, the same holds; however, if you need to handle arbitrary overlaps, you might need a sweep line or inclusion-exclusion, but since f(y) is linear in y, the sum of individual contributions works.
Define f(y) = total area above y - total area below y. Show that f is continuous and strictly decreasing (or non-increasing) because as y increases, area above decreases and area below increases. Thus, f(y) goes from total area (at y below all cakes) to -total area (at y above all cakes), so by IVT a root exists.
Set initial interval [y_min, y_max] covering all cakes. While (y_max - y_min) > epsilon, compute mid = (y_min + y_max)/2, evaluate f(mid) in O(n) by summing each cake's area above and below mid. If f(mid) > 0, set y_min = mid; else set y_max = mid. Return y* = (y_min + y_max)/2.
Time complexity: O(n log((y_max - y_min)/epsilon)). Space: O(1) extra. Discuss floating-point precision: use epsilon as tolerance, avoid exact equality, and consider using relative error. Stopping criteria: when interval width < epsilon or when |f(mid)| < delta.
If multiple solutions exist (e.g., f(y)=0 over an interval), binary search will find one; return any. For overlapping squares, f(y) is still monotonic and can be computed by summing individual contributions (since area above is additive even with overlaps). For rectangles, same approach works; just adjust area formulas for width and height.
Mention edge cases: cakes entirely above/below line, zero-area cakes. For large n, consider sorting events or using a sweep line to evaluate f(y) faster if needed, but O(n) per evaluation is acceptable. For rectangles, if they overlap, the sum of individual areas above still gives correct total area above because area is additive.
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