0679HardArrayMathDynamic ProgrammingSorting

Water Taps Along the Lane

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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 lane runs dead straight, and houses[i] is the position of one house along it, measured in metres from the lane's start. No two houses stand at the same position, and the positions arrive in no particular order.

The council will install exactly taps water taps, each at a whole-metre position on the lane. A tap may stand at the same position as a house, and two taps may stand at the same position. Every household then draws water from whichever tap is closest to its own house, walking the distance between the two positions.

Return the smallest total walking distance over all households that any placement of the taps can achieve.

Examples

Example 1

Input: houses = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10000], taps = 2
Output: 20

Put one tap at position 5 and the other at position 10000. The nine close households walk 4, 3, 2, 1, 0, 1, 2, 3 and 4 metres, and the far household walks nothing.

Example 2

Input: houses = [7, 2, 9, 1, 4], taps = 2
Output: 5

Put one tap at position 2 and the other at position 9. The households at 1, 2 and 4 walk 1, 0 and 2 metres to the first tap, and the households at 7 and 9 walk 2 and 0 metres to the second.

Example 3

Input: houses = [5, 11, 30, 31, 32, 33], taps = 6
Output: 0

There are as many taps as houses, so a tap can stand at every house and nobody has to walk.

Constraints

1 <= taps <= houses.length <= 100
1 <= houses[i] <= 10^4
The positions in houses 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 least_total_walk(houses: list[int], taps: int) -> int:

Java

public int leastTotalWalk(int[] houses, int taps)

The intended approach

Time O(taps * n^2): the stretch costs take O(n^2) to build and each of the taps * n states scans up to n split points.Space O(n^2) for the stretch costs plus O(taps * n) for the sweep.

Sort the positions first, because the houses served by one tap are always a stretch of neighbouring positions. For a stretch, the cheapest tap position is the middle house of the stretch, and the resulting total is the sum of the distances between the outermost pair, the next pair in, and so on, which lets every stretch cost be built from the stretch two shorter inside it. Then sweep over how many taps have been used and how many houses have been covered, trying every position where the last stretch could begin, and keep the cheapest.

Pythonentry point: least_total_walk

def least_total_walk(houses, taps):
    # O(n^2) block costs, then an O(taps * n^2) sweep over where each block ends
    spots = sorted(houses)
    n = len(spots)
    block = [[0] * n for _ in range(n)]
    for span in range(2, n + 1):
        for i in range(0, n - span + 1):
            j = i + span - 1
            block[i][j] = block[i + 1][j - 1] + spots[j] - spots[i]
    huge = float("inf")
    best = [[huge] * (n + 1) for _ in range(taps + 1)]
    best[0][0] = 0
    for t in range(1, taps + 1):
        for j in range(t, n + 1):
            room = huge
            for i in range(t - 1, j):
                if best[t - 1][i] == huge:
                    continue
                total = best[t - 1][i] + block[i][j - 1]
                if total < room:
                    room = total
            best[t][j] = room
    return best[taps][n]

The obvious version

Some cheapest placement puts every tap at a house position, so try every choice of that many house positions as the tap sites. For each choice, walk the houses, measure the distance to the nearest chosen site and add it up, then keep the smallest total found.

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_least_total_walk(houses, taps):
    from itertools import combinations
    spots = sorted(houses)
    best = None
    for sites in combinations(spots, taps):
        total = 0
        for h in houses:
            near = None
            for s in sites:
                gap = h - s if h > s else s - h
                if near is None or gap < near:
                    near = gap
            total += near
        if best is None or total < best:
            best = total
    return best

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

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