0497MediumArrayPrefix Sum

Ink Drops on the Tile Strip

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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 press decorates a strip of n tiles laid out in a line and numbered 1 through n.

The head makes one pass for each entry of passes. The entry passes[i] = [first, last, drops] means that pass left drops drops of ink on tile first, on tile last, and on every tile lying between the two. A tile visited by several passes keeps the ink from all of them.

Return an array total of length n, where total[j] is the number of drops sitting on tile j + 1 once every pass has run.

Examples

Example 1

Input: passes = [[1, 3, 20], [2, 5, 7], [4, 5, 30]], n = 5
Output: [20, 27, 27, 37, 37]

Tile 1 is reached by the first pass alone, so it holds 20 drops. Tile 2 is reached by the first two passes, giving 27. Tile 4 is reached by the second and third passes, giving 37.

Example 2

Input: passes = [[2, 6, 4], [6, 6, 11], [1, 6, 1]], n = 6
Output: [1, 5, 5, 5, 5, 16]

The second pass covers tile 6 only. Tile 6 is the last tile of all three passes, so it collects 4 + 11 + 1 drops, while tile 1 is covered by the third pass alone.

Example 3

Input: passes = [[1, 2, 6], [5, 6, 6]], n = 6
Output: [6, 6, 0, 0, 6, 6]

No pass reaches tile 3 or tile 4, so those tiles stay bare, and the four tiles that are covered each get 6 drops.

Constraints

1 <= n <= 2 * 10^4
1 <= passes.length <= 2 * 10^4
passes[i].length == 3
1 <= first <= last <= n
1 <= drops <= 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 ink_per_tile(passes: list[list[int]], n: int) -> list[int]:

Java

public int[] inkPerTile(int[][] passes, int n)

The intended approach

Time O(m + n) for m passes: each pass is touched once, then one sweep of the strip.Space O(n) for the scratch array and the returned totals.

Record each pass twice instead of tile by tile: add its drops at the tile where it starts and subtract them one tile past where it stops, into a scratch array of length n + 1 so the extra slot absorbs a pass that stops on tile n. Then sweep the strip once, keeping a running total of the scratch entries seen so far; that total is the ink on the current tile, since a pass contributes to it from its first tile onwards and its subtraction cancels the contribution as soon as the tile is out of range.

Pythonentry point: ink_per_tile

def ink_per_tile(passes, n):
    # one slot per tile plus one spare, so a pass that reaches the far tile needs no test
    edge = [0] * (n + 1)
    for entry in passes:
        first, last, drops = entry[0], entry[1], entry[2]
        edge[first - 1] += drops
        edge[last] -= drops
    tiles = []
    running = 0
    for j in range(n):
        running += edge[j]
        tiles.append(running)
    return tiles

The obvious version

Replay every pass literally, walking from its first tile to its last tile and adding its drops to each tile on the way. O(m * n).

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_ink_per_tile(passes, n):
    tiles = [0] * n
    for entry in passes:
        first, last, drops = entry[0], entry[1], entry[2]
        for tile in range(first, last + 1):
            tiles[tile - 1] += drops
    return tiles

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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