0098MediumArrayBinary Search

Unmatched Component Serial

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Trains the technique from

LeetCode 540Single Element in a Sorted Array

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 repair depot stocks components in matched pairs, one for the machine and one held as a spare, and both members of a pair carry the same serial. The stock sheet serials lists the serial of every component on the shelf in non-decreasing order.

One component arrived without its twin, so exactly one serial appears once on the sheet while every other serial appears twice. Return the serial of that unmatched component.

The sheet can be very long, so your routine must finish in O(log n) time and use O(1) extra space.

Examples

Example 1

Input: serials = [4, 4, 12, 12, 19, 19, 23]
Output: 23

Serials 4, 12 and 19 each sit on the shelf twice, and 23 sits at the far end on its own.

Example 2

Input: serials = [6, 11, 11, 30, 30]
Output: 6

The unmatched component opens the sheet, so every pair after it starts at an odd position.

Example 3

Input: serials = [2, 2, 7, 15, 15]
Output: 7

The pair 2, 2 sits before the loose component and the pair 15, 15 sits after it.

Constraints

1 <= serials.length <= 10^5
serials.length is odd
0 <= serials[i] <= 10^5
serials is sorted in non-decreasing order
Every serial appears exactly twice except one, which appears once

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 unmatched_serial(serials: list[int]) -> int:

Java

public int unmatchedSerial(int[] serials)

The intended approach

Time O(log n)Space O(1)

Before the loose component, every pair starts at an even position; after it, every pair starts at an odd position. So halve the sheet, snap the probe to an even position, and read the entry beside it: if the two match, the loose component lies further along, otherwise it lies at the probe or before it. The window shrinks by half each round and never loses the answer.

Pythonentry point: unmatched_serial

def unmatched_serial(serials):
    low = 0
    high = len(serials) - 1
    while low < high:
        probe = (low + high) // 2
        if probe % 2 == 1:
            probe -= 1
        if serials[probe] == serials[probe + 1]:
            low = probe + 2
        else:
            high = probe
    return serials[low]

The obvious version

Tally how often each serial appears with a dictionary and return the one tallied once, O(n) time and O(n) space, ignoring the required bounds.

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_unmatched_serial(serials):
    tally = {}
    for s in serials:
        tally[s] = tally.get(s, 0) + 1
    for s in serials:
        if tally[s] == 1:
            return s
    return -1

How this problem was checked

Reference solution executed against 9 test cases — 9 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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