0639EasyMathBinary Search

Complete Tiers Of The Tin Display

Tracked in this browser only

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 shop builds a stepped display of tins. Tier 1 across the top holds one tin, tier 2 below it holds two tins, and in general tier i holds i tins.

With n tins in stock, the staff work downwards and fill each tier completely before starting the next one, stopping when the stock runs short of what the next tier needs. Return how many tiers end up complete.

Examples

Example 1

Input: n = 7
Output: 3

Tier 1 takes 1 tin, tier 2 takes 2 and tier 3 takes 3, which uses 6 of the 7 tins. Tier 4 needs 4 tins and only 1 is left, so 3 tiers are complete.

Example 2

Input: n = 999999999
Output: 44720

The first 44720 tiers take 999961560 tins altogether, leaving 38439 tins. Tier 44721 needs 44721 tins, which is more than what is left, so 44720 tiers are complete.

Example 3

Input: n = 6
Output: 3

Tiers 1, 2 and 3 take 1, 2 and 3 tins, using the stock exactly, so all 3 tiers are complete and nothing is left over.

Constraints

1 <= n <= 2^31 - 1

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 arrange_coins(n: int) -> int:

Java

public int arrangeCoins(int n)

The intended approach

Time O(log n) — the search halves the tier range each step.Space O(1).

Filling the first r tiers consumes r * (r + 1) // 2 tins, and that demand only grows as r grows, so the values of r the stock can cover form an unbroken range starting at one. Binary search that range for its largest member, testing the demand of a candidate against the stock at each step. The upper end of the search never needs to pass 65536, since even the largest allowed stock cannot fill that many tiers.

Pythonentry point: arrange_coins

def arrange_coins(n):
    # O(log n) time, O(1) space: r complete tiers need r * (r + 1) // 2 tins,
    # which grows with r, so the tiers that fit form a range starting at 1 and a
    # binary search finds its last member. At the largest allowed stock the
    # answer is 65535 tiers needing 2147450880 tins, both exact as integers.
    lo = 1
    hi = n
    if hi > 65536:
        hi = 65536
    while lo < hi:
        mid = (lo + hi + 1) // 2
        if mid * (mid + 1) // 2 <= n:
            lo = mid
        else:
            hi = mid - 1
    return lo

The obvious version

Hand out tins tier by tier: while the remaining stock covers the next tier's demand, subtract that demand and count the tier.

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_arrange_coins(n):
    rows = 0
    need = 1
    left = n
    while left >= need:
        left -= need
        rows += 1
        need += 1
    return rows

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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in