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Missing number

by Tom
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def missing_number(numbers):
Test Code
import random def test(): for n in range(1, 101): numbers = list(range(n)) real_missing_number = random.randrange(n) numbers = [number for number in numbers if number != real_missing_number] random.shuffle(numbers) result = missing_number(numbers) assert result == real_missing_number, f"Test failed for {n=}. Expected {real_missing_number}, got {result} for {numbers=}" if __name__ == "__main__": test()
Description

Given an array of length n - 1 containing distinct numbers from 0 to n - 1 with one number missing, find the missing number.

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def missing_number(numbers):
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Fastest Solutions

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 Solution by Anonymous User #16 (0.1 seconds)
def missing_number(numbers): # numbers: list of [0..n-1] except one, length n-1 n = len(numbers) + 1 # how many items there should be return n * (n - 1) // 2 - sum(numbers) # missing = expected − got
0
 Solution by Anonymous User #32 (0.2 seconds)
def missing_number(numbers): # numbers: list of [0..n-1] except one, length n-1 n = len(numbers) + 1 # how many items there should be return n * (n - 1) // 2 - sum(numbers) # missing = expected − got

New Solutions

0
 Solution by Anonymous User #32 (0.2 seconds)
def missing_number(numbers): # numbers: list of [0..n-1] except one, length n-1 n = len(numbers) + 1 # how many items there should be return n * (n - 1) // 2 - sum(numbers) # missing = expected − got
0
 Solution by Anonymous User #16 (0.1 seconds)
def missing_number(numbers): # numbers: list of [0..n-1] except one, length n-1 n = len(numbers) + 1 # how many items there should be return n * (n - 1) // 2 - sum(numbers) # missing = expected − got