easyDSA

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

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

**Intuition.** XOR is the perfect tool: x ^ x = 0 and x ^ 0 = x . XOR every element together — each pair cancels to 0, and the lone element is left standing. O(n) time, O(1) space, no hash set. [4,1,2,1,2] → pairs cancel, 4 remains 4 ^ 1 ^ 2 <text x="260" y="60" fill="currentColo

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XOR fold

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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

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