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How do you find the missing number in 0..n?

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01

Understand the problem

Explain missing-number approaches.

arraysmath
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

Sum 0..n with n(n+1)/2 and subtract the actual array sum — the difference is the missing number, O(n) time / O(1) space. XOR of indices and values also works and avoids overflow risk.

Solution ready — 2 min read

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04

Read the code

Gauss sum + XOR
Run Playground
def missing_number(nums):
    n = len(nums)
    return n * (n + 1) // 2 - sum(nums)

def missing_number_xor(nums):
    result = len(nums)
    for i, x in enumerate(nums):
        result ^= i ^ x
    return result


# --- demo ---
print(missing_number([3, 0, 1]))       # 2
print(missing_number_xor([3, 0, 1]))   # 2
05

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