0120EasyArrayHash TableDivide and ConquerSortingCountingBoyer–Moore Majority Vote Algorithm

Quorum Drift Value

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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 monitoring service asks every node in a cluster for the same figure: how far that node's clock has drifted from the reference, in signed milliseconds. Nodes with a broken sensor answer with whatever number their hardware happens to produce, so the array reports mixes trustworthy answers with junk, in no particular order.

The cluster is provisioned so that the healthy nodes are always a strict majority: one value occupies more than half of the entries of reports. Return that value.

The array is streamed in from the wire, so solve it with a single left-to-right pass over reports and only a constant amount of bookkeeping. Do not sort and do not tally every distinct value.

Examples

Example 1

Input: reports = [4, 7, 4]
Output: 4

Two of the three nodes answered 4, which is more than half, so 7 is the faulty reading. Note that the answer is not the entry sitting in the middle.

Example 2

Input: reports = [-6, -6, 2, -6, 9]
Output: -6

Three entries out of five hold -6. Drift can run either side of the reference, so a negative figure is a perfectly ordinary answer.

Constraints

n == reports.length
1 <= n <= 5 * 10^4
-10^9 <= reports[i] <= 10^9
The input is provisioned so that some value appears more than n / 2 times

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 quorum_drift(reports: list[int]) -> int:

Java

public int quorumDrift(int[] reports)

The intended approach

Time O(n)Space O(1)

Hold one tentative value and a lead counter. Each entry equal to the tentative value strengthens its lead by one; each different entry cancels one unit of that lead. When the lead is spent, adopt the current entry as the new tentative value. Because the true answer outnumbers everything else combined, it can never be cancelled out overall, so whatever survives the pass is it.

Pythonentry point: quorum_drift

def quorum_drift(reports):
    tentative = reports[0]
    lead = 0
    for value in reports:
        if lead == 0:
            tentative = value
            lead = 1
        elif value == tentative:
            lead += 1
        else:
            lead -= 1
    return tentative

The obvious version

For each entry, rescan the whole array counting how often that exact value occurs, and return the first entry whose tally clears half the length. O(n^2) and it needs no insight at all.

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_quorum_drift(reports):
    n = len(reports)
    for candidate in reports:
        tally = 0
        for value in reports:
            if value == candidate:
                tally += 1
        if tally > n // 2:
            return candidate
    return reports[0]

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
2 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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