0536HardArrayStackMonotonic Stack

Sightlines Down the Mast Row

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Trains the technique from

LeetCode 1944Number of Visible People in a Queue

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 survey line carries radio masts, listed left to right in masts by height. No two masts are the same height.

An engineer standing on top of mast i faces the right-hand end of the line. For a mast j further along, that is with j > i, the engineer can see mast j when every mast standing between them is shorter than mast i and shorter than mast j. Nothing else blocks the view.

Return an array whose entry i is the number of masts to the right of mast i that the engineer on mast i can see.

Examples

Example 1

Input: masts = [7, 3, 5, 2, 4]
Output: [2, 1, 2, 1, 0]

From mast 0, of height 7, the masts of height 3 and 5 are visible, while the ones of height 2 and 4 are hidden behind the mast of height 5. From mast 2, of height 5, the masts of height 2 and 4 are both visible.

Example 2

Input: masts = [4, 3, 2, 1]
Output: [1, 1, 1, 0]

The heights fall away to the right, so from any mast every mast beyond it is visible: 3 from the first, then 2, then 1, then none.

Example 3

Input: masts = [1, 2, 3, 4]
Output: [1, 1, 1, 0]

From mast 0 the mast of height 2 is visible; the mast of height 3 is not, because the mast of height 2 between them is not shorter than mast 0. The same holds along the line.

Constraints

n == masts.length
1 <= n <= 10^5
1 <= masts[i] <= 10^5
The heights are all different.

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 visible_masts(masts: list[int]) -> list[int]:

Java

public int[] visibleMasts(int[] masts)

The intended approach

Time O(n): every height is pushed once and popped at most once.Space O(n) for the stack, beyond the returned array.

Work from the far end of the line backwards, keeping a stack of the heights still worth remembering, arranged so it increases as you go down. For the mast being handled, every stacked height below its own is a mast it can see, and popping them off counts them; whatever height is left on top after that is one further mast it can see, the first one taller than itself. Pushing the current height keeps the stack in shape, and each height is pushed and popped at most once.

Pythonentry point: visible_masts

def visible_masts(masts):
    # O(n): walk from the far end holding a stack of heights that increases from
    # the top down, so each mast pops exactly the masts it can see past.
    n = len(masts)
    out = [0] * n
    stack = []
    for i in range(n - 1, -1, -1):
        here = masts[i]
        count = 0
        while stack and stack[-1] < here:
            stack.pop()
            count += 1
        if stack:
            count += 1
        out[i] = count
        stack.append(here)
    return out

The obvious version

For each mast, walk to the right keeping the tallest mast met so far, count a mast whenever it beats that running tallest, and stop as soon as one of them beats the mast being looked from.

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_visible_masts(masts):
    n = len(masts)
    out = []
    for i in range(n):
        count = 0
        tallest = 0
        for j in range(i + 1, n):
            if masts[j] > tallest:
                count += 1
                tallest = masts[j]
                if masts[j] > masts[i]:
                    break
        out.append(count)
    return out

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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