1165HardArrayHash TableMathDynamic ProgrammingBit ManipulationMeet in the MiddleBitmask

Splitting the Tray Into Two Equal Averages

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Trains the technique from

LeetCode 805Split Array With Same Average

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 tray holds the weights weights. Put every weight into one of two bins, and neither bin may end up empty.

Return true when some way of doing that leaves the two bins with the same average weight.

Examples

Example 1

Input: weights = [1, 2, 3]
Output: true

Putting the 2 in one bin and the 1 and the 3 in the other gives both bins an average of 2.

Example 2

Input: weights = [1, 2, 4]
Output: false

The tray totals 7 across three weights. A bin of one weight would have to hold seven thirds and a bin of two would have to total fourteen thirds, and neither is whole.

Example 3

Input: weights = [5, 5, 5, 5, 5]
Output: true

Every weight is the same, so any split at all leaves both bins on that average.

Constraints

1 <= weights.length <= 30
0 <= weights[i] <= 10^4

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 split_array_same_average(weights: list[int]) -> bool:

Java

public boolean splitArraySameAverage(int[] weights)

The intended approach

Time O(n^2 * total) bit operations, done a machine word at a time.Space O(n * total) bits for the reachable totals.

If the two bins share an average then each bin shares it with the whole tray, because the tray's average is a blend of the two and a blend of two equal things is that thing. So a bin of `k` weights totalling `s` works exactly when `s / k` equals the tray's total over its count, which rearranges to `s * n = total * k` with no division at all. Each bin is the other's leftovers, so only bins holding at most half the weights need checking. For each size up to half, the question is whether some selection of that many weights hits the required total. That is a reachability question, answered by carrying, for each size, the set of totals selections of that size can reach. Bringing in one more weight extends every recorded total by it, and walking the sizes downwards while doing so keeps a weight from being used twice in one bin. Holding those sets as bit patterns, one bit per total, makes each extension a single shift, which is what keeps thirty weights of ten thousand within reach. A required total that does not come out whole rules its size out at once.

Pythonentry point: split_array_same_average

def split_array_same_average(weights):
    n = len(weights)
    if n < 2:
        return False
    total = 0
    for value in weights:
        total += value
    half = n // 2
    reach = [0] * (half + 1)
    reach[0] = 1
    for value in weights:
        for k in range(half - 1, -1, -1):
            reach[k + 1] |= reach[k] << value
    for k in range(1, half + 1):
        if (total * k) % n == 0:
            want = total * k // n
            if (reach[k] >> want) & 1:
                return True
    return False

The obvious version

Try every way of filling the first bin as a bit pattern, leaving out the empty pattern and the full one, and check whether the two bins' averages match by cross-multiplying their totals and counts so no division is needed. Costs two to the power of the weight count.

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_split_array_same_average(weights):
    n = len(weights)
    if n < 2:
        return False
    total = 0
    for value in weights:
        total += value
    for mask in range(1, (1 << n) - 1):
        count = 0
        part = 0
        for i in range(n):
            if mask >> i & 1:
                count += 1
                part += weights[i]
        if part * (n - count) == (total - part) * count:
            return True
    return False

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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