0777MediumArrayDynamic Programming

Tallest Reachable Relay Tower

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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 line of signal relay towers stands left to right, and heights[i] is the height of tower i. A drone carrying a signal repeater can hop between towers, and from tower i it may hop to tower j when either

j > i and heights[j] > heights[i], so the hop goes to the right and lands on a taller tower, or
j < i and heights[j] < heights[i], so the hop goes to the left and lands on a shorter tower.

The two towers of a hop need not be neighbours. The drone may make any number of hops, including none.

For each tower in turn, report the greatest height the drone can be standing on after starting there, counting the starting tower itself. Return those heights in tower order, so the entry at position i belongs to tower i.

Examples

Example 1

Input: heights = [12,40,7,25]
Output: [40, 40, 40, 40]

From tower 2, height 7, the drone hops right to tower 3, height 25, then left to tower 0, height 12, which is shorter, then right to tower 1, height 40, which is taller. Each of the other starts can reach tower 1 as well, so every answer here is 40.

Example 2

Input: heights = [15,60,15]
Output: [60, 60, 15]

Tower 0 hops right to tower 1. Tower 2 has nothing to its right, and a left hop needs a strictly shorter tower while both towers to its left are 15 or more, so it stays put and answers with its own height.

Example 3

Input: heights = [80,60,40,20]
Output: [80, 60, 40, 20]

The heights fall from left to right, so no tower has a taller tower on its right or a shorter tower on its left. Each answer is the tower's own height.

Constraints

1 <= heights.length <= 10^5
1 <= heights[i] <= 10^9

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(heights: list[int]) -> list[int]:

Java

public int[] maxValue(int[] heights)

The intended approach

Time O(n): one backward pass for the suffix maxima and one forward pass whose fill loops together touch each position once.Space O(n) for the suffix maxima, plus the returned array.

Precompute the tallest height at or after each position. Then sweep left to right carrying the shortest and tallest height of the run being built, and close the run at `i` when its shortest height is at least the tallest height after `i`, because then no hop can leave the run. Every position in a closed run is answered with that run's tallest height, and the sweep restarts from the next position.

Pythonentry point: max_value

def max_value(heights):
    # A hop right needs a taller tower and a hop left needs a shorter one, so the
    # same comparison governs both directions of a pair: positions i < j are
    # linked exactly when heights[i] < heights[j], and the linkage is mutual.
    # A run of positions therefore closes at i as soon as the shortest tower in
    # the run so far is at least as tall as every tower after i, and each tower
    # of a closed run answers with that run's tallest height.
    # Time O(n), space O(n).
    n = len(heights)
    tallest_after = [0] * (n + 1)
    for i in range(n - 1, -1, -1):
        tallest_after[i] = heights[i] if heights[i] > tallest_after[i + 1] else tallest_after[i + 1]

    ans = [0] * n
    start = 0
    low = heights[0]
    high = heights[0]
    for i in range(n):
        if heights[i] < low:
            low = heights[i]
        if heights[i] > high:
            high = heights[i]
        if low >= tallest_after[i + 1]:
            for j in range(start, i + 1):
                ans[j] = high
            start = i + 1
            if start < n:
                low = heights[start]
                high = heights[start]
    return ans

The obvious version

Start a separate search at every tower, following the two hop rules literally and marking towers as they are stood on, then report the tallest tower the search touched. Cubic in the number of towers.

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(heights):
    n = len(heights)
    out = []
    for s in range(n):
        seen = [False] * n
        seen[s] = True
        stack = [s]
        best = heights[s]
        while stack:
            i = stack.pop()
            if heights[i] > best:
                best = heights[i]
            for j in range(n):
                if seen[j]:
                    continue
                if j > i and heights[j] > heights[i]:
                    seen[j] = True
                    stack.append(j)
                elif j < i and heights[j] < heights[i]:
                    seen[j] = True
                    stack.append(j)
        out.append(best)
    return out

How this problem was checked

Reference solution executed against 15 test cases — 15 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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