The problem
Write a perform that takes a shuffled listing of distinctive numbers from 1 to n with one ingredient lacking (which might be any quantity together with n). Return this lacking quantity.
Observe: big lists can be examined.
Examples:
[1, 3, 4] => 2
[1, 2, 3] => 4
[4, 2, 3] => 1
The answer in Python code
Choice 1:
def find_missing_number(a):
n = len(a) + 1
return n * (n + 1) // 2 - sum(a)
Choice 2:
def find_missing_number(nums):
return sum(vary(1,len(nums)+2))-sum(nums)
Choice 3:
def find_missing_number(numbers):
if numbers == []:return 1
diff = listing(set(vary(1, len(numbers)+1))- set(numbers))
if diff == []:return max(numbers)+1
return diff[0]
Check circumstances to validate our resolution
take a look at.assert_equals(find_missing_number([2, 3, 4]), 1)
take a look at.assert_equals(find_missing_number([1, 3, 4]), 2)
take a look at.assert_equals(find_missing_number([1, 2, 4]), 3)
take a look at.assert_equals(find_missing_number([1, 2, 3]), 4)
