(I need help codding this problem; an explanation would be appreciated. In Python, please, Thank you.)

### Part A

Write a function `findAllOddIndices` that takes in a list of integers `lst` of type `List[Int]`

Return a list of indices $[i_1, \ldots, i_k]$ from the original list `lst` such that the list element $e_j$ corresponding to each returned index $i_j$ is odd.

Remember indices start at $0$ and range until the length of list -1.

#### Example

~~~ lst: List( 1, 5, 0, 2, 4, 8, 7) ~~~

should return

~~~ List(0, 1, 6) ~~~

since the elements at indices 0, 1 and 6 are odd.

**Restrictions** are same as problem 1. Do not use list API functions other than `length` and `reverse`. You can use operators `:+`, `++`, and `::` to append elements to a list. - Be careful not to use the list indexing operator `list(i)`.

**Challenge:** Make your code as efficient as possible. Ideally, it should have complexity O(n) though we will not deduct points for inefficient code.

// YOUR CODE HERE ???

//BEGIN TESTS val lst1 = List(1, 3, 0, 2, 4, 6, 5) testWithMessage(findAllOddIndices(lst1), List(0, 1, 6), "Test1")

val lst2 = List(4, 6, -2, 8, 10) testWithMessage(findAllOddIndices(lst2), Nil, "Test2")

val lst3 = List(5, 3, 1, 5, 15, 9 ) testWithMessage(findAllOddIndices(lst3), List(0, 1, 2, 3, 4, 5), "Test3")

val lst4 = List() testWithMessage(findAllOddIndices(lst4), Nil, "Test4")

passed(5) //END TESTS

Is your code above efficient? Try using the following code test. It should run in under a second ideally. We have disabled running this cell just in case you have an inefficent solution that crashes your computer or our autograder. ~~~ val lst = (0 to 100000).toList val ret =findAllOddIndices(lst) assert(ret.length == 50000) ~~~