From 1431bf863cef5bc31ae00e20731de644a566d78d Mon Sep 17 00:00:00 2001
From: Nicco <hi@nicco.io>
Date: Thu, 10 Dec 2020 12:33:50 +0100
Subject: [PATCH] Update README.md

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 solutions/10/README.md | 32 ++++++++++++++++++++++++++++++++
 1 file changed, 32 insertions(+)

diff --git a/solutions/10/README.md b/solutions/10/README.md
index c37adf0..e2cefe7 100644
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 # 10
 
+# First
+
 The first one is really easy. Just sort, make the diff and count.
+First I take the list and sort it:
+
+```python
+[16, 10, 15, 5, 1, 11, 7, 19, 6, 12, 4]
+[0, 1, 4, 5, 6, 7, 10, 11, 12, 15, 16, 19, 22] # Sorted and added the wall plug (0) and the phone (biggest + 3)
+[1, 3, 1, 1, 1, 3, 1, 1, 3, 1, 3, 3] # The size of each step
+```
+
+Now we can simply count how many `1` and `3` there are with `l.count(1)`.
+
+## Second
+
+This is where it gets tricky.
+
+First lets find all the consecutive `1`s ad only they can be removed. If we have more than 1 consecutive `1` we can remove one of it. However we need to be careful not to remove to many or the step will be higher than `3` and the chain breaks.
+
+```python
+[1, 1, 1, 1] # We can transform this example by adding 2 numbers together and "joining" them.
+
+[1, 2, 1] # Valid
+[1, 1, 2] # Valid
+[1, 3] # Valid
+[4] # Invalid because we can jump a max of 3 steps at a time.
+```
+
+Now we could iterate but I wanted to find a formula. Not sure this is correct but here we go.
+
+Basically we take the length of the consecutive `1` and compute `2**(l-1)` to get all possible combinations.
+Now we need to subtract the possible `4` which can only be achieved if we have at least 4 numbers -> `floor(l/4)`
+For a grand total of `2**(l-1) - floor(l/4)`
 
 <details>
   <summary>Solutions</summary>