0021MediumArrayTwo PointersGreedy

Banner Between Pillars

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 stone walkway is lined with pillars, one pillar every metre. pillars[i] is the height in metres of the pillar at metre mark i. A stump left over from an old pillar has height 0.

A crew hangs one rectangular banner across the walkway. They pick two pillars, run a level rope between them and let the fabric hang from the rope all the way down to the paving. Because the rope has to sit at the same height on both pillars, it can be tied no higher than the lower of the two pillar tops, and the banner's height equals the rope's height. The banner's width is the gap in metres between the two chosen pillars.

Return the largest area, in square metres, that a single banner can cover. The rope is never tied at a tilt, and pillars standing between the chosen two are ignored: the fabric hangs freely past them.

Examples

Example 1

Input: pillars = [2, 9, 8, 1]
Output: 8

Tying to marks 1 and 2 gives a rope at height 8 across a gap of 1 metre. Reaching for the widest gap, marks 0 and 3, only allows height 1 over 3 metres, which is 3 square metres.

Example 2

Input: pillars = [3, 1, 1, 1, 1, 3]
Output: 15

The two end pillars are both 3 metres tall and stand 5 metres apart, so the banner covers 15 square metres. The short pillars in between do not get in the way.

Example 3

Input: pillars = [4, 7]
Output: 4

With only one pair available, the rope sits at 4 metres across a 1 metre gap.

Constraints

n == pillars.length
2 <= n <= 10^5
0 <= pillars[i] <= 10^4

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 largest_banner(pillars: list[int]) -> int:

Java

public int largestBanner(int[] pillars)

The intended approach

Time O(n)Space O(1)

Start with the widest pair, one marker at each end of the walkway, and record its area. That pair already uses the largest possible width, so any pair that keeps the shorter of the two pillars can only lose width and cannot gain height; the shorter pillar is therefore finished and its marker steps inward. Repeating this discards one candidate pillar per step while never discarding the best pair, so a single sweep of the markers toward each other reaches the answer.

Pythonentry point: largest_banner

def largest_banner(pillars):
    left, right = 0, len(pillars) - 1
    best = 0
    while left < right:
        width = right - left
        if pillars[left] <= pillars[right]:
            area = width * pillars[left]
            left += 1
        else:
            area = width * pillars[right]
            right -= 1
        if area > best:
            best = area
    return best

The obvious version

Price every pair of pillars with two nested loops and keep the biggest area, 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_largest_banner(pillars):
    best = 0
    n = len(pillars)
    for i in range(n):
        for j in range(i + 1, n):
            area = (j - i) * min(pillars[i], pillars[j])
            if area > best:
                best = area
    return best

How this problem was checked

Reference solution executed against 13 test cases — 13 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