🧭 Conceptual Foundation & Pattern Intuition

The answer here is a height, and a height can be any real number: 1, 7 / 6, 999999999.5. Raise a horizontal line and the area below it only grows, so the question "is at least half of the area below y = mid?" answers No, No, ..., then Yes from the lowest balancing line upward. That is a binary search, but over a continuum: there is no next height after mid, so neither end can step past it with + 1 or - 1, and a loop that waits for lo and hi to meet never stops.

📐 The Analogy: Setting a Shelf by Halving the Gap

You are fixing a shelf at the height where the boxes stacked on a wall are split into two equal halves by weight. You know one height that is too low and one that is high enough. Try the middle: if it is high enough, it becomes your new "high enough" mark; if not, it becomes your new "too low" mark. The gap between the two marks halves every time. There is no "next notch" on a smooth wall, so you do not wait for the marks to touch: after a fixed number of tries the gap is thinner than a hair, and you fix the shelf at the "high enough" mark.

🪄 The Mathematical Harmony / Magic Trick

lo = min(y for _, y, _ in squares)      # the check fails here
hi = max(y + l for _, y, l in squares)  # the check passes here
for _ in range(100):                    # count rounds, never while lo < hi
    mid = (lo + hi) / 2                 # a real midpoint
    full = sum(l * l for _, y, l in squares if y + l <= mid)             # exact integer
    part = sum(l * (mid - y) for _, y, l in squares if y < mid < y + l)  # the only float
    if 2 * part >= total - 2 * full:    # >=: the lowest balancing line
        hi = mid                        # move onto mid, never past it
    else:
        lo = mid
return hi
 

lo always fails and hi always passes, so the answer stays in (lo, hi] while the window halves every round: after t rounds it is (hi - lo) / 2^t wide. The heights here are at most about 2 * 109 apart, so 48 rounds already reach 10^-5, and by round 100 lo and hi are neighbouring floats. From then on mid equals one of them and the rounds change nothing, which is exactly why while lo < hi would never end. The check is split so rounding cannot undo that precision: full, the squares wholly below the line, is an exact integer, and only part, the squares the line cuts, is a float. A single float sum of every area would be off by up to about 6e-5 area units once the total nears 1012, and with a thin square at the line that is a height error above 10^-5.

💡 Summary

When the answer is a real number, bisect the value itself: move an end onto mid (never mid ± 1), run a fixed number of rounds chosen from the bounds and the tolerance, and return hi. One pass over the squares per round: $O(N \cdot R)$ time, $O(1)$ space.