0124MediumArrayBinary Search

Kiln Load Ceiling

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Trains the technique from

LeetCode 1011Capacity To Ship Packages Within D Days

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A ceramics studio has one kiln and a queue of unfired pieces. The queue is given as masses, where masses[i] is the clay mass in grams of the piece standing at position i. Pieces leave the queue from the front only, so they are fired in exactly the listed order and are never rearranged.

The studio has booked sessions kiln runs. A single run takes some number of pieces off the front of the queue, which means each run fires one contiguous block of the queue, and it may take none at all. The kiln is rated for a load ceiling: the combined mass loaded into any one run may not exceed it.

Return the smallest load ceiling for which sessions runs are enough to fire the entire queue. A run is never split across sessions, and a piece is never fired twice.

Examples

Example 1

Input: masses = [9, 4, 6, 3, 8], sessions = 3
Output: 11

With a ceiling of 11 the runs can be 9 | 4, 6 | 3, 8. Dropping to 10 forces the trailing 3 and 8 apart, which needs a fourth run.

Example 2

Input: masses = [5, 5, 5, 5], sessions = 4
Output: 5

Four runs are booked for four pieces, so each run can hold one piece and the ceiling only has to cover the heaviest piece.

Example 3

Input: masses = [2, 4, 6, 1], sessions = 1
Output: 13

A single run must swallow the whole queue, so the ceiling equals the total mass of 13 grams.

Constraints

1 <= sessions <= masses.length <= 5 * 10^4
1 <= masses[i] <= 500

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def kiln_load_ceiling(masses: list[int], sessions: int) -> int:

Java

public int kilnLoadCeiling(int[] masses, int sessions)

The intended approach

Time O(n log S) where S is the total massSpace O(1)

Feasibility is monotone: if a ceiling clears the queue in `sessions` runs, every larger ceiling does too. So binary search the ceiling over the range from the heaviest single piece (nothing smaller can ever fit) up to the total mass (one run for everything). Test a candidate greedily by walking the queue and starting a new run only when the next piece would overflow the current one, which uses the fewest runs possible for that ceiling.

Pythonentry point: kiln_load_ceiling

def kiln_load_ceiling(masses, sessions):
    low = max(masses)
    high = sum(masses)
    while low < high:
        mid = (low + high) // 2
        runs = 1
        load = 0
        for m in masses:
            if load + m > mid:
                runs += 1
                load = m
            else:
                load += m
        if runs <= sessions:
            high = mid
        else:
            low = mid + 1
    return low

The obvious version

Enumerate every way to cut the queue into contiguous blocks by iterating over all subsets of cut positions, discard the ones needing more than `sessions` blocks, score each layout by its heaviest block and keep the minimum. Exponential, and it never reasons about monotonicity or greedy packing.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_kiln_load_ceiling(masses, sessions):
    n = len(masses)
    best = None
    for mask in range(1 << (n - 1)):
        cuts = [i + 1 for i in range(n - 1) if (mask >> i) & 1]
        if len(cuts) + 1 > sessions:
            continue
        edges = [0] + cuts + [n]
        worst = 0
        for a, b in zip(edges, edges[1:]):
            block = sum(masses[a:b])
            if block > worst:
                worst = block
        if best is None or worst < best:
            best = worst
    return best

How this problem was checked

Reference solution executed against 12 test cases — 12 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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