1160MediumArrayDivide and ConquerGreedySortingQuickselect

Laying the Readings Out in a Zigzag

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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 row holds the readings readings. A zigzag is an arrangement of them where the first is below the second, the second is above the third, the third is below the fourth, and so on, strictly alternating all the way along.

Several zigzags may be possible, so the one wanted is pinned down like this. Put the readings in non-decreasing order and cut them into a lower part holding the first half, rounded up when the count is odd, and an upper part holding the rest. Fill positions 0, 2, 4 and onwards from the lower part taken back to front, and positions 1, 3, 5 and onwards from the upper part taken back to front.

Return that arrangement. The readings always allow a zigzag.

Examples

Example 1

Input: readings = [1, 2, 3]
Output: [2, 3, 1]

In order the readings are 1, 2 and 3. The lower part holds 1 and 2 and the upper part holds 3, and taking each from the back gives 2, then 3, then 1.

Example 2

Input: readings = [1, 1, 2]
Output: [1, 2, 1]

The lower part holds the two 1 readings and the upper part the 2, so the 2 lands between them.

Example 3

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

The lower part holds the two zeros and the upper part the two ones, and taking each from the back alternates them.

Constraints

1 <= readings.length <= 5 * 10^4
0 <= readings[i] <= 5000
the readings always allow a zigzag

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 wiggle_sort(readings: list[int]) -> list[int]:

Java

public int[] wiggleSort(int[] readings)

The intended approach

Time O(n log n) for the ordering.Space O(n) for the halves and the answer.

Splitting the ordered readings down the middle and dealing the two halves into alternate positions is what makes a zigzag work: every position filled from the upper part holds a reading at least as large as every reading in the lower part, and its two neighbours both come from the lower part. Taking each half from its back rather than its front is what handles repeats. Read front to back and equal readings straddling the cut would land next to each other, breaking the strict alternation. Read back to front, the largest of the lower part sits at the very start, furthest from the smallest of the upper part, and equal readings end up as far apart as the arrangement allows. Rounding the lower part up when the count is odd gives it the extra reading, which is what lets the arrangement start and end on a low position.

Pythonentry point: wiggle_sort

def wiggle_sort(readings):
    order = sorted(readings)
    n = len(order)
    mid = (n + 1) // 2
    low = order[:mid]
    high = order[mid:]
    out = [0] * n
    at = mid - 1
    for i in range(0, n, 2):
        out[i] = low[at]
        at -= 1
    at = len(high) - 1
    for i in range(1, n, 2):
        out[i] = high[at]
        at -= 1
    return out

The obvious version

Order the readings by tallying how many of each value there are and reading the tallies upwards, then cut them into the two parts and deal them out by repeatedly taking from the back of the lower part and then the back of the upper part until both run dry. Costs the range of the values rather than a sort.

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_wiggle_sort(readings):
    tally = {}
    for value in readings:
        tally[value] = tally.get(value, 0) + 1
    order = []
    for value in sorted(tally):
        for _ in range(tally[value]):
            order.append(value)
    n = len(order)
    mid = (n + 1) // 2
    low = order[:mid]
    high = order[mid:]
    out = []
    while low or high:
        if low:
            out.append(low.pop())
        if high:
            out.append(high.pop())
    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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