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Two Calibration Cards to a Target

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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 cold store files its calibration cards in a branching index. Every card carries one signed offset in tenths of a degree, and no two cards carry the same offset. Each card holds at most two cards below it, one on its low side and one on its high side, and the filing rule holds everywhere: every offset filed anywhere below a card on its low side is smaller than that card's offset, and every offset filed anywhere below it on its high side is larger.

The index reaches you as index, a band listing, one band of the index at a time from the top down and left to right within a band. Slot 0 holds the topmost card. Each card that appears in the listing takes the next two unclaimed slots for what hangs below it, the low side first and the high side second. A slot holding null means nothing hangs there, and such a slot claims no slots of its own. Trailing null slots are left off the end of the listing.

Return true when the index holds two different cards whose offsets add up to k, and false otherwise. One card cannot be counted twice, so an offset that is exactly half of k is no use on its own.

Examples

Example 1

Input: index = [8, 3, 10, 1, 6, null, 14, null, null, 4, 7, 13], k = 20
Output: true

The card carrying 6 and the card carrying 14 are two different cards and their offsets come to 20.

Example 2

Input: index = [8, 3, 10, 1, 6, null, 14, null, null, 4, 7, 13], k = 100
Output: false

The two largest offsets in the index are 13 and 14, which come to 27, so no pair of cards reaches 100.

Example 3

Input: index = [4], k = 8
Output: false

There is only one card. Its offset is half of 8, but a single card cannot be counted twice.

Example 4

Input: index = [0, -3, 3], k = 0
Output: true

The card carrying -3 and the card carrying 3 are different cards and their offsets cancel out.

Example 5

Input: index = [2, null, 5], k = 7
Output: true

The index has two cards, and 2 with 5 comes to 7.

Constraints

The number of cards is in the range [1, 10^4].
-10^4 <= card offset <= 10^4
All the offsets are different, and index obeys the filing rule.
-10^5 <= k <= 10^5

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 find_target(index: list[int | None], k: int) -> bool:

Java

public boolean findTarget(Integer[] index, int k)

The intended approach

Time O(m + n) for a listing of m slots holding n cards: one sweep to wire the sides, one walk that pushes and pops each card once, and one pass of the two fingers.Space O(m + n) for the two side tables and the offsets read out in order, plus a stack bounded by the index depth.

One sweep over the listing gives every card the slot numbers of its two sides, honouring the rule that an empty slot claims nothing. Reading the index low side, card, high side with an explicit stack then delivers the offsets in increasing order without any sorting. From there a pair is hunted by holding one finger at the smallest offset and one at the largest: a total below `k` can only be helped by raising the low finger, a total above `k` only by lowering the high one, and the fingers meeting means no pair exists. They never land on the same card, which is what keeps a half-of-`k` offset from answering on its own.

Pythonentry point: find_target

def find_target(index, k):
    # Resolve the band listing into low/high slot numbers, read the offsets out
    # in filing order with an explicit stack, which hands them over smallest
    # first, then close in from both ends: too small a total means raising the
    # low end, too large means lowering the high end. Iterative throughout, so a
    # 10000-deep index is safe.
    size = len(index)
    if size == 0:
        return False
    low = [-1] * size
    high = [-1] * size
    free = 1
    for slot in range(size):
        if index[slot] is None:
            continue
        if free < size:
            if index[free] is not None:
                low[slot] = free
            free += 1
        if free < size:
            if index[free] is not None:
                high[slot] = free
            free += 1

    ordered = []
    pending = []
    node = 0
    while pending or node != -1:
        while node != -1:
            pending.append(node)
            node = low[node]
        node = pending.pop()
        ordered.append(index[node])
        node = high[node]

    left = 0
    right = len(ordered) - 1
    while left < right:
        total = ordered[left] + ordered[right]
        if total == k:
            return True
        if total < k:
            left += 1
        else:
            right -= 1
    return False

The obvious version

Collect every offset in the listing, then try each pair of distinct positions and report whether any of them adds up to k. O(n^2).

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_find_target(index, k):
    offsets = []
    for slot in index:
        if slot is not None:
            offsets.append(slot)
    for a in range(len(offsets)):
        for b in range(a + 1, len(offsets)):
            if offsets[a] + offsets[b] == k:
                return True
    return False

How this problem was checked

Reference solution executed against 18 test cases — 18 passed
5 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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