The challenge#

You will be given an array of numbers. You have to sort the odd numbers in ascending order while leaving the even numbers at their original positions.

Examples#

``````[7, 1]  =>  [1, 7]
[5, 8, 6, 3, 4]  =>  [3, 8, 6, 5, 4]
[9, 8, 7, 6, 5, 4, 3, 2, 1, 0]  =>  [1, 8, 3, 6, 5, 4, 7, 2, 9, 0]
``````

The solution in Python code#

Option 1:

``````def sort_array(arr):
odds = sorted((x for x in arr if x%2 != 0), reverse=True)
return [x if x%2==0 else odds.pop() for x in arr]
``````

Option 2:

``````def sort_array(source_array):
result = sorted([l for l in source_array if l % 2 == 1])
for index, item in enumerate(source_array):
if item % 2 == 0:
result.insert(index, item)
return result
``````

Option 3:

``````def sort_array(source_array):
odd = sorted(list(filter(lambda x: x % 2, source_array)))
l, c = [], 0
for i in source_array:
if i in odd:
l.append(odd[c])
c += 1
else:
l.append(i)
return l
``````

Test cases to validate our solution#

``````test.assert_equals(sort_array([5, 3, 2, 8, 1, 4]), [1, 3, 2, 8, 5, 4])
test.assert_equals(sort_array([5, 3, 1, 8, 0]), [1, 3, 5, 8, 0])
test.assert_equals(sort_array([]),[])
``````