# Weekly Challenge 215

## Odd one Out

You are given a list of words (alphabetic characters only) of same size.

Write a script to remove all words not sorted alphabetically and print the number of words in the list that are not alphabetically sorted.

### My solution

I would have thought that by definition the first world would have being defined as sorted alphabetically. However, the second example shows that this is not the case. So I have a bit of code that will return the length of the list if the first word is not the first alphabetically.

I then set the `current_word` variable to the first word, and the `unsorted_words` variable to 0. I loop through the remaining words, and update the `current_word` variable if it is greater than the current `current_word` or add one to `unsorted_words` if it isn't.

### Examples

``````\$ ./ch-1.py abc xyz tsu
1

3

\$ ./ch-1.py x y z
0
``````

You are given a list of numbers having just 0 and 1. You are also given placement count (>=1).

Write a script to find out if it is possible to replace 0 with 1 in the given list. The only condition is that you can only replace when there is no 1 on either side. Print 1 if it is possible otherwise 0.

### My solution

Sometimes it is easier to brute force the solution, and that is the approach I took with this one. I start by taken the last value from the input, and assign this to the variable `to_place`.

I then loop through the array from the second position to the second last one. If the preceding value, the current value and the next value are all zero, I change the current value to one, and remove from from the `to_place` variable. If this is zero (i.e. we have placed all numbers), I print `1 and exit.

If I have ended the loop and `to_place` is not zero, I cannot place all the numbers, and print `0` to represent this.

### Examples

`bash
\$ ./ch-2.py 1 0 0 0 1 1
1

\$ ./ch-2.py 1 0 0 0 1 2
0

\$ ./ch-2.py 1 0 0 0 0 0 0 0 1 3
1
`