0653MediumArrayBacktracking

Notch Sets for a Key Blank

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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 locksmith cuts notches into a key blank. A notch is cut to one of the nine standard depths 1 through 9, and a blank may not carry two notches of the same depth. A cutting plan is a set of count notch depths whose depths add up to total.

Return every cutting plan. Each plan must list its depths in increasing order, and the plans themselves must be ordered so that, comparing two plans depth by depth from the shallowest, the one with the smaller depth at the first place they differ comes first. Return an empty list when no plan exists.

Examples

Example 1

Input: count = 2, total = 7
Output: [[1, 6], [2, 5], [3, 4]]

Each listed plan cuts two notches of different depths that add up to 7, and the plans are ordered by their shallowest notch.

Example 2

Input: count = 4, total = 30
Output: [[6, 7, 8, 9]]

Every plan shown holds four different depths adding up to 30, for example 6 plus 7 plus 8 plus 9.

Example 3

Input: count = 3, total = 60
Output: []

Three different depths reach at most 9 plus 8 plus 7, which is 24, so no plan reaches 60 and the answer is the empty list.

Constraints

2 <= count <= 9
1 <= total <= 60

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 notch_sets(count: int, total: int) -> list[list[int]]:

Java

public List<List<Integer>> notchSets(int count, int total)

The intended approach

Time O(C(9, count) * count) in the worst case: the search visits each increasing selection of depths at most once and copies the ones that qualify.Space O(count) for the search state, beyond the returned plans.

Build a plan one notch at a time with a depth-first search that is only ever allowed to add a depth strictly greater than the last one it took, which enforces both the no-repeat rule and the increasing order for free. Carry how many notches are still owed and how much depth is still owed; record the plan when both hit zero at the same moment, and abandon a branch as soon as the next depth on offer already exceeds what is still owed. Scanning the depths in increasing order at every level makes the plans come out in the required order.

Pythonentry point: notch_sets

def notch_sets(count, total):
    # Depth-first walk over the depths in increasing order: the search only ever
    # extends with a depth larger than the last one taken, so each set comes out
    # already increasing and the sets come out in lexicographic order.
    out = []
    chosen = []

    def walk(next_depth, left, remaining):
        if left == 0:
            if remaining == 0:
                out.append(list(chosen))
            return
        for d in range(next_depth, 10):
            if d > remaining:
                break
            chosen.append(d)
            walk(d + 1, left - 1, remaining - d)
            chosen.pop()

    walk(1, count, total)
    return out

The obvious version

Enumerate all 511 non-empty subsets of the nine depths as bit patterns, expand each pattern into its list of depths, keep the ones with the right size and the right sum, and sort the survivors.

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_notch_sets(count, total):
    found = []
    for mask in range(1, 1 << 9):
        picks = []
        acc = 0
        for bit in range(9):
            if mask & (1 << bit):
                picks.append(bit + 1)
                acc += bit + 1
        if len(picks) == count and acc == total:
            found.append(picks)
    found.sort()
    return found

How this problem was checked

Reference solution executed against 11 test cases — 11 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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