0138MediumArrayDynamic ProgrammingGreedySorting

Drop the Clashing Kiln Runs

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Trains the technique from

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 books firings on one shared kiln. Every request comes in as a pair [start, finish] inside runs, both measured in minutes from the studio's reference mark; a request that sits earlier than the mark uses negative minutes, and a firing always finishes after it starts.

Two requests clash when the stretches of kiln time they occupy share at least one minute. A firing that finishes on the exact minute another begins does not clash with it, since the kiln is free again at that mark.

Cancel as few requests as possible so that no two survivors clash. Return how many requests you had to cancel.

Examples

Example 1

Input: runs = [[1, 4], [2, 6], [5, 9]]
Output: 1

Cancelling the request that runs from 2 to 6 leaves the other two, which share no minute. Keeping all three is impossible because the middle firing clashes with both of the others.

Example 2

Input: runs = [[-3, 0], [0, 5], [5, 8]]
Output: 0

Each firing finishes exactly when the next one begins, so the kiln is never double booked and nothing has to go.

Example 3

Input: runs = [[0, 10], [9, 12], [11, 20]]
Output: 1

Dropping the short middle request keeps the two long firings, which never overlap. Cancelling one of the long firings instead would leave the remaining pair still clashing.

Constraints

1 <= runs.length <= 10^5
runs[i].length == 2
-5 * 10^4 <= start < finish <= 5 * 10^4
Requests that only touch at a single minute mark do not clash

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 min_runs_to_cancel(runs: list[list[int]]) -> int:

Java

public int minRunsToCancel(int[][] runs)

The intended approach

Time O(n log n)Space O(n) for the sorted copy

Cancelling the fewest requests is the same as keeping the most, so build the largest clash-free set instead. Order the requests by finishing minute and keep every request that begins at or after the last kept finish: freeing the kiln as early as possible never costs a later chance, so this greedy pass is optimal. The answer is the total minus the number kept.

Pythonentry point: min_runs_to_cancel

def min_runs_to_cancel(runs):
    order = sorted(runs, key=lambda run: run[1])
    kept = 0
    limit = None
    for start, finish in order:
        if limit is None or start >= limit:
            kept += 1
            limit = finish
    return len(runs) - kept

The obvious version

Order by starting minute and grow the longest clash-free chain with a quadratic scan: for each request, extend the best chain that already finishes by the time it begins. Subtract that length from the total.

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_min_cancels(runs):
    order = sorted(runs, key=lambda run: (run[0], run[1]))
    n = len(order)
    chain = [1] * n
    for i in range(n):
        for j in range(i):
            if order[j][1] <= order[i][0] and chain[j] + 1 > chain[i]:
                chain[i] = chain[j] + 1
    longest = 0
    for value in chain:
        if value > longest:
            longest = value
    return n - longest

How this problem was checked

Reference solution executed against 10 test cases — 10 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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