1034MediumMathBinary SearchGreedy

The Tallest Post at a Chosen Place

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Trains the technique from

LeetCode 1802Maximum Value at a Given Index in a Bounded Array

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 fence has n posts in a row, numbered from 0. Every post's height is a whole number of at least 1, posts standing next to each other differ in height by at most 1, and all the heights added together come to at most budget.

Return the greatest height the post at spot can be given.

Examples

Example 1

Input: n = 4, spot = 2, budget = 6
Output: 2

Four posts and a budget of six. Making the third post two tall needs heights of one, one, two and one, which adds to five and fits. Making it three tall would need one, two, three and two, adding to eight.

Example 2

Input: n = 6, spot = 1, budget = 10
Output: 3

Six posts and a budget of ten. Heights of two, three, two, one, one and one add to exactly ten. Pushing the second post to four would need three either side and cost fourteen.

Example 3

Input: n = 2, spot = 0, budget = 2
Output: 1

Two posts and a budget of two leaves nothing over: both posts stand at the minimum of one.

Constraints

1 <= n <= 10^9
1 <= budget <= 10^9
n <= budget
0 <= spot <= n - 1

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 max_value(n: int, spot: int, budget: int) -> int:

Java

public int maxValue(int n, int spot, int budget)

The intended approach

Time O(log(budget)).Space O(1).

For a chosen height at the spot, the cheapest fence allowing it steps down by exactly one on each side until it reaches `1`, then holds at `1` for whatever posts remain. Nothing cheaper works, because a neighbour may not drop by more than one, and nothing is gained by going higher anywhere else. That cost has a closed form. On one side, the heights step down over as many posts as the height allows, which is a run of consecutive whole numbers, and any posts past that each cost `1`. Add the two sides and the post at the spot. A taller post never costs less, so the tallest affordable height can be closed in on by halving the range from `1` up to the budget. Each test is a few pieces of arithmetic, so the whole thing costs the logarithm of the budget and nothing depends on the number of posts.

Pythonentry point: max_value

def max_value(n, spot, budget):
    def cost(top, reach):
        if top - 1 >= reach:
            return (top - 1 + top - reach) * reach // 2
        run = top - 1
        return run * (run + 1) // 2 + (reach - run)
    left = spot
    right = n - 1 - spot
    lo = 1
    hi = budget
    while lo < hi:
        mid = (lo + hi + 1) // 2
        spend = mid + cost(mid, left) + cost(mid, right)
        if spend <= budget:
            lo = mid
        else:
            hi = mid - 1
    return lo

The obvious version

Build the fence outright for each candidate height, starting at one and climbing. For a height, work out every post as that height less its distance from the spot, floored at `1`, add them all up, and stop as soon as the total passes the budget. The last height that fitted is the answer. Costs the posts times the answer rather than a closed form and a halving search.

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_max_value(n, spot, budget):
    best = 0
    top = 1
    while True:
        total = 0
        for at in range(n):
            gap = at - spot if at > spot else spot - at
            height = top - gap
            if height < 1:
                height = 1
            total += height
        if total > budget:
            break
        best = top
        top += 1
    return best

How this problem was checked

Reference solution executed against 14 test cases — 14 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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