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easyDSA

How do you find the single number where every other element appears twice?

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01

Understand the problem

Explain the XOR trick.

arraysbit-manipulation
02

Attempt it yourself

Sketch your approach before reading the solution — that's what interviews test.

Nudge consolestandby

Stuck? Beam a request up — the console returns a conceptual nudge that guides your logic without spoiling the implementation.

03

Study the solution

XOR all elements together. Pairs cancel (x ^ x = 0) and x ^ 0 = x, so the lone element remains. O(n) time, O(1) space — no hash set needed. Variants with triples need bit-count tricks.

Solution ready — 2 min read

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04

Read the code

XOR fold
Run Playground
def single_number(nums):
    result = 0
    for x in nums:
        result ^= x
    return result


# --- demo ---  (one-liner: reduce(xor, nums))
print(single_number([4, 1, 2, 1, 2]))   # 4
05

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