Single Element in a Sorted Array
A medium Binary Search problem included in Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Binary Search
- Sheets
- 1
- Core for
- 7 roles
- Platform
- LeetCode
The problem
Given a sorted array where every element appears exactly twice except for one element that appears once, find and return that single element. The solution must run in O(log n) time and O(1) space.
Example 1
- Input
- nums = [1,1,2,3,3,4,4,8,8]
- Output
- 2
- Why
- Every element except 2 appears twice. 2 is the single element.
Example 2
- Input
- nums = [3,3,7,7,10,11,11]
- Output
- 10
- Why
- Every element except 10 appears twice. 10 is the single element.
Constraints
- 1 <= nums.length <= 10^5
- 0 <= nums[i] <= 10^5
- nums.length is odd
How to think about it
Updated 2026-09-09Pairs before the single element always start on an even index and end on an odd index. The single element shifts the parity: every pair appearing after it starts on an odd index and ends on an even index. Checking index parity of any pair tells you which side contains the disruptor.
Approaches, worst first
Bitwise XOR scan
time O(n) · space O(1)
XOR all elements together; identical paired values cancel to zero, leaving the unpaired element. Operates in linear time without extra space, but fails to use the sorted property to achieve logarithmic speed.
Binary search on index parityWrite this one
time O(log n) · space O(1)
Probe mid and align it to an even index. If nums[mid] equals nums[mid + 1], the disruption has not happened yet, so search to the right (low = mid + 2); otherwise the unique item is at or to the left of mid (high = mid).
Where people lose marks · 3
- Accessing mid + 1 when mid is the final element triggers an index out-of-bounds exception unless the search interval bounds or even-index alignment prevent it.
- Setting high = mid - 1 on a pair mismatch skips the single element when mid itself is the unique item; keep high = mid.
- An input of length 1 should immediately return the only element without executing boundary pair checks.
The theory behind it
Binary Search — the ground this problem stands on. All Binary Search problems
What Binary Search is
Binary search is the guessing strategy used when searching a thick telephone directory or guessing a secret number between one and a hundred. Rather than inspecting names one by one from the first page, the search opens straight to the middle page. If the target precedes that middle entry, the entire back half is discarded; if it follows, the front half is eliminated. Repeating this halved split finds the item with remarkable speed.
When to reach for it
Reach for binary search when queries target sorted collections, rotated sorted arrays, or monotonic answer spaces. Strong hints include logarithmic time constraints like O(log n) or search ranges exceeding one billion where stepping one value at a time times out. It also applies when validating whether a guessed solution is possible via a monotonic boolean check, known as binary search on answer.
How the pattern works
Define the search territory with two inclusive pointers, low and high. Compute the midpoint using low plus half the difference to high, avoiding integer overflow. Formulate an exact boolean condition that divides the range into true and false halves. Decide whether the boundary condition includes the midpoint or shifts strictly past it. The loop invariant states that the sought target, if it exists, remains trapped inside the interval throughout every iteration.
What each operation costs
| Operation | Time |
|---|---|
| find target in sorted array | O(log n) |
| find boundary in monotonic range | O(log n) |
What usually goes wrong with Binary Search
- Triggering integer overflow by computing middle using low plus high divided by two instead of low plus half of high minus low in fixed-width numeric types.
- Creating an infinite loop when the interval shrinks to two elements by setting low equal to mid when mid was rounded down.
- Mismatched loop condition and bounds updates, such as pairing low less than or equal to high with non-advancing pointer assignments that never terminate.
Which roles need this problem
Binary Search is a core topic for these 7 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 17 more roles, including Full-Stack Developer, Data Engineer, Android Developer.
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