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

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[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:

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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:

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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:

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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

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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([]),[])