0908HardArrayBinary Indexed TreeSegment TreeSimulation

Splitting a Feed Across Two Trays

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Trains the technique from

LeetCode 3072Distribute Elements Into Two Arrays II

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.

Readings arrive in the order given by nums and are laid out on two trays. The first reading goes on tray one and the second on tray two.

Every later reading is placed by comparing the trays. Count how many readings already on tray one are strictly greater than the arriving reading, and likewise for tray two. The reading joins whichever tray has the larger count. When the counts are equal it joins the tray holding fewer readings, and when that is equal too it joins tray one.

Return tray one's readings in the order they were placed, followed by tray two's.

Examples

Example 1

Input: nums = [14, 9, 21, 6, 17, 3]
Output: [14, 21, 6, 17, 3, 9]

Tray one opens with 14 and tray two with 9. The 21 beats nothing on either tray, so the counts tie at zero and the trays are the same size, sending it to tray one. The 6 is beaten by both readings on tray one and by the one on tray two, so it joins tray one, and the rest follow the same rule.

Example 2

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

Neither tray holds anything greater than 3, so the counts tie at zero and both trays hold one reading, which sends it to tray one.

Example 3

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

Tray one's 3 beats the arriving 1 while tray two's 2 also beats it, so the counts tie at one apiece and the equal sizes send it to tray one.

Constraints

3 <= nums.length <= 10^5
1 <= nums[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 result_array(nums: list[int]) -> list[int]:

Java

public int[] resultArray(int[] nums)

The intended approach

Time O(n log n) for n readings: the compression sorts once and each reading costs a few tree steps.Space O(n) for the two trees and the two trays.

The only question asked of a tray is how many of its readings beat the arriving one, which is a count over a range of values, so keep each tray as a tally indexed by value and read that count off the tally. A Fenwick tree over the values does exactly this, answering how many readings are at most a given value in logarithmic time, and the count that beats the arrival is the tray's size less that. The values themselves reach into the billions, so compress them first: sort the distinct values and index each reading by its rank, which shrinks the trees to the number of distinct readings without changing any comparison. Keep the actual readings in two lists alongside, since those are what gets returned.

Pythonentry point: result_array

def result_array(nums):
    ranks = sorted(set(nums))
    where = {}
    for i in range(len(ranks)):
        where[ranks[i]] = i + 1
    size = len(ranks)

    def make():
        return [0] * (size + 1)

    def add(tree, at):
        while at <= size:
            tree[at] += 1
            at += at & -at

    def upto(tree, at):
        total = 0
        while at > 0:
            total += tree[at]
            at -= at & -at
        return total

    one_tree = make()
    two_tree = make()
    tray_one = [nums[0]]
    tray_two = [nums[1]]
    add(one_tree, where[nums[0]])
    add(two_tree, where[nums[1]])
    for i in range(2, len(nums)):
        v = nums[i]
        at = where[v]
        beats_one = len(tray_one) - upto(one_tree, at)
        beats_two = len(tray_two) - upto(two_tree, at)
        if beats_one > beats_two:
            pick_one = True
        elif beats_two > beats_one:
            pick_one = False
        elif len(tray_one) <= len(tray_two):
            pick_one = True
        else:
            pick_one = False
        if pick_one:
            tray_one.append(v)
            add(one_tree, at)
        else:
            tray_two.append(v)
            add(two_tree, at)
    return tray_one + tray_two

The obvious version

Hold each tray as a plain list and, for every arriving reading, walk both lists counting the readings that beat it. Quadratic in the reading count and it keeps no structure over the values.

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_result_array(nums):
    tray_one = [nums[0]]
    tray_two = [nums[1]]
    for i in range(2, len(nums)):
        v = nums[i]
        beats_one = 0
        for x in tray_one:
            if x > v:
                beats_one += 1
        beats_two = 0
        for x in tray_two:
            if x > v:
                beats_two += 1
        if beats_one > beats_two:
            tray_one.append(v)
        elif beats_two > beats_one:
            tray_two.append(v)
        elif len(tray_one) <= len(tray_two):
            tray_one.append(v)
        else:
            tray_two.append(v)
    return tray_one + tray_two

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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