0953HardArrayDivide and ConquerBinary Indexed TreeSegment TreeSweep LineSortingHeap (Priority Queue)Ordered Set

Outline of a Row of Blocks

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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.

Blocks stand in a row along a line. Each entry of buildings is [left, right, height] for one block, standing on the ground between those two positions with that height. The list is given in non-decreasing order of left.

The outline is the shape of the row seen from a distance. Return it as a list of [position, height] turning points, in increasing order of position: each turning point marks a position where the outline's height changes, giving the new height. The height after the last block is 0, so the final turning point always has height 0. No two turning points share a position, and no two neighbouring turning points have the same height.

Examples

Example 1

Input: buildings = [[3, 9, 14], [5, 12, 27], [11, 18, 6], [20, 24, 31]]
Output: [[3, 14], [5, 27], [12, 6], [18, 0], [20, 31], [24, 0]]

The outline rises to 14 at position 3, to 27 at 5 where the taller block starts, drops to 6 at 12 where that block ends, falls to nothing at 18, and then rises to 31 at 20 before ending at 24.

Example 2

Input: buildings = [[0, 5, 3], [5, 10, 3]]
Output: [[0, 3], [10, 0]]

The two blocks are the same height and meet exactly, so the outline shows one step up and one step down with no turning point in the middle.

Example 3

Input: buildings = [[0, 1, 1]]
Output: [[0, 1], [1, 0]]

A single block gives one step up and one step back down to nothing.

Constraints

1 <= buildings.length <= 10^4
buildings[i].length == 3
0 <= buildings[i][0] < buildings[i][1] <= 2^31 - 1
1 <= buildings[i][2] <= 2^31 - 1
buildings is given in non-decreasing order of buildings[i][0]

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 get_skyline(buildings: list[list[int]]) -> list[list[int]]:

Java

public List<List<Integer>> getSkyline(int[][] buildings)

The intended approach

Time O(n log n) for n blocks: sorting the events and a heap step each.Space O(n) for the events and the heap.

The outline can only change where a block starts or ends, so gather those positions as events: a block starting adds its height to the set of heights in play, and a block ending takes it away. Sweeping the events from left to right, the outline's height at any moment is the tallest height still in play, or nothing when none are. So keep the heights in play in a heap ordered tallest first, together with the position each one stops mattering. Before reading off the tallest, drop any entries whose block has already ended. Process every event sharing a position together, then compare the tallest height against the last one recorded and add a turning point only when it has changed, which is what keeps neighbouring points from repeating a height. Sorting the events with starts before ends at the same position, and taller starts first, makes the sweep read cleanly.

Pythonentry point: get_skyline

def get_skyline(buildings):
    import heapq
    events = []
    for left, right, height in buildings:
        events.append((left, -height, right))
        events.append((right, 0, 0))
    events.sort()
    out = []
    live = [(0, float("inf"))]     # tallest first, as negated height with its end
    at = 0
    n = len(events)
    while at < n:
        here = events[at][0]
        while at < n and events[at][0] == here:
            pos, neg, right = events[at]
            if neg != 0:
                heapq.heappush(live, (neg, right))
            at += 1
        while live[0][1] <= here:
            heapq.heappop(live)
        tallest = -live[0][0]
        if not out or out[-1][1] != tallest:
            out.append([here, tallest])
    return out

The obvious version

Take every position where a block starts or ends, and for each one work out the outline's height by walking the whole list of blocks and keeping the tallest that covers it. Then keep only the positions where that height differs from the one before. Costs the block count per position and uses no heap.

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_get_skyline(buildings):
    spots = set()
    for left, right, height in buildings:
        spots.add(left)
        spots.add(right)
    order = sorted(spots)
    out = []
    last = 0
    for pos in order:
        tallest = 0
        for left, right, height in buildings:
            if left <= pos < right and height > tallest:
                tallest = height
        if tallest != last:
            out.append([pos, tallest])
            last = tallest
    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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