# Python Challenge - Find the Missing Numbers

This community-built FAQ covers the “Find the Missing Numbers” code challenge in Python. You can find that challenge here, or pick any challenge you like from our list.

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Per instructions, ‘copy and paste to the forums’,

``````def missing_nos(arr, k):
a = set(arr)
b = set(range(1, len(a) + k + 1))
return sorted([*(b.difference(a))])

print (missing_nos([2, 3, 4, 5, 6, 7, 8, 9], 2))
``````

This also passed, but it was before I read up on the difference method.

``````def missing_nos(arr, k):
a = set(arr)
b = set(range(1, len(a) + k + 1))
for i in a:
if i in b:
b.remove(i)
return sorted([*b])

print (missing_nos([2, 3, 4, 5, 6, 7, 8, 9], 2))
``````

This also passed, after considering other approaches…

``````def missing_nos(arr, k):
return sorted([x for x in range(1, len(arr) + k + 1) if x not in arr])

print (missing_nos([2, 3, 4, 5, 6, 7, 8, 9], 2))
``````

In the truest sense, this does not use sets so may allow duplicates. Using sets would be the way to go with this one. Easily solved,

``[x for x in set(range(1, len(arr) + k + 1)) if x not in set(arr)]``
1 Like

Completely forgot about the in keyword (whoops!) so did some fun things

def missing_nos(arr, k): # Write your code here true_size = len(arr) + k check_list = [i+1 for i in range(true_size)] arr += ([0] * k) result_list = [] for i in range(true_size): if arr[i] != check_list[i] and arr[i] != 0: arr.insert(i,0) arr.pop() result_list.append(check_list[i]) return (result_list) print(missing_nos([1, 2, 4, 5, 6, 7, 8, 10], 2))
``````def missing_nos(arr, k):
full_set = {x + 1 for x in range( len(arr) + k)}
result =  list ( full_set.difference(arr) )
return sorted(result)

print(missing_nos([1, 2, 4, 5, 6, 7, 8, 10], 2))
print(missing_nos([ 2, 3, 4, 5, 6, 7], 3))
print(missing_nos([1, 3, 5, 2], 0))
``````

Seems to work fine now, but the thing that tripped me up was I tried sorting the original array (arr) at the start, assuming ‘result’ would dump out the comparison already in correct order… it doesn’t.

def missing_nos(arr, k): return [x for x in range(arr[0], arr[-1] + 1) if x not in arr] print(missing_nos([1, 2, 4, 5, 6, 7, 8, 10], 2))
``````def missing_nos(arr, k):
num = 1
missing = []
for i in arr:
if i == num:
pass
else:
missing.append(num)
num=num+1
num+=1
return(missing)

print(missing_nos([1, 2, 4, 5, 6, 7, 8, 10], 2))
``````

One of the easiest of the challenges.

def missing_nos(arr, k): missing = [] length = len(arr) + k for i in range(length): print(i + 1) if i + 1 not in arr: missing.append(i + 1) return missing print(missing_nos([1, 2, 4, 5, 6, 7, 8, 10], 2))
def missing_nos(arr, k): complete_set = set(range(1,max(arr)+1)) return sorted(list(complete_set - set(arr))) # Example 1: correctly returns [3,9] print(missing_nos([1, 2, 4, 5, 6, 7, 8, 10], 2)) # Example 2: incorrectly returns [3] print(missing_nos([1, 2, 4, 5, 6, 7, ], 2))

Looks like the test cases are incorrect/incomplete.
This solution is passing test cases. however the result for both should be [3,9] (with this implementation is [3] for second example)

You should get, `[3, 8]` is you remove the comma after the `7`.

def missing_nos(arr, k): # Write your code here missings = [] no_missings = [i for i in range(1, arr[-1])] for num in no_missings: if num not in arr: missings.append(num) return missings

Did this pass the tests in the challenge? It shouldn’t have and so should not be posted here unless you have a question.

def missing_nos(arr, k): l = min(arr) h = max(arr) + 1 missing_num = [] for i in range(l, h): if i not in arr: missing_num.append(i) return missing_num print(missing_nos([1, 2, 4, 5, 6, 7, 8, 10], 2))
``````def missing_nos(arr, k):
given=set(arr)
whole_set=set(range(1,len(given)+k+1))
for i in given:
if i in whole_set:
whole_set.remove(i)
return (sorted(whole_set))
print(missing_nos([1, 2, 4, 5, 6, 7, 8, 10], 2))
``````

def missing_nos(arr, k):

if arr == :
return [i+1 for i in range(k)]
j = 1
ans =
for i in range(len(arr)):
if arr[i] != j:
ans.append(j)
if len(ans) == k:
return ans
j = arr[i]
if i == len(arr) - 1:
while len(ans) < k:
j += 1
ans.append(j)
return ans
j += 1
print(missing_nos([2], 5))

def missing_nos(arr, k):
setList = set(arr)
newL = len(arr)
mainList = set(range(1, newL+k+1))
l = mainList.difference(setList)
return sorted(l)

print(missing_nos([1, 2, 4, 5, 6, 7, 8, 10], 2))

def missing_nos(arr, k): # Write your code here ar=[] for i in range(len(arr)-1): if arr[i+1]!= arr[i]+1: ar.append(arr[i]+1) return ar print(missing_nos([1, 2, 4, 5, 6, 7, 8, 10], 2))

You know, I just realized that it’s a python challenge, but they still let you pick which language your codebyte is in.

def missing_nos(arr, k): missing_list = [] for number in arr: count = 1 if len(missing_list) == k: break while number + count != arr[arr.index(number)+1]: missing_list.append(number+count) count+=1 return sorted(missing_list)

Easy peasy:

def missing_nos(arr, k): n = len(arr) + k klist = list(num for num in range(1, n+1) if not num in arr) return klist

Your code doesn’t work if the first number of the array was removed. Try inserting an if/else if the length of your returned array is != k

def missing_nos(arr, k): # Write your code here full_array = set(range(1, max(arr) + 1)) missing_nums = list(full_array - set(arr)) if len(missing_nums) != k: missing_nums.append(max(arr) + 1) return sorted(missing_nums) print(missing_nos([1, 2, 4, 5, 6, 7, 8, 10], 2))