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2020/solutions/10/README.md

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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>
+  <ol>
+    <li>2475</li>
+    <li>442136281481216</li>
+  </ol>
+</details>

2020/solutions/10/python/main.py

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+from os.path import join, dirname
+from typing import List, Optional, Set, Tuple
+from itertools import combinations, count
+from math import floor, prod
+
+
+def parse(s: str) -> List[int]:
+    numbers: List[int] = sorted(map(int, s.strip().split('\n')))
+    numbers.insert(0, 0)  # The wall
+    numbers.append(numbers[-1] + 3)  # Phone itself
+    return numbers
+
+
+def diff(l: List[int]) -> List[int]:
+    return [
+        l[x] - l[x-1]
+        for x in range(1, len(l))
+    ]
+
+
+def calc(d: List[int]) -> int:
+    one = d.count(1)
+    three = d.count(3)
+    return one * three
+
+
+def find_valid_permutations(d: List[int]) -> int:
+    i = 0
+    l = len(d)
+    slices: List[int] = []
+    while i < l:
+        if d[i] != 3:
+            try:
+                n = d.index(3, i + 1)  # Find the next three
+                diff = n - i
+                if diff > 1:
+                    slices.append(diff)
+                    i = n
+                    continue
+            except:
+                pass
+        i += 1
+    return prod([
+        2**(s-1) - floor(s/4)
+        for s in slices
+    ])
+
+
+data = join(dirname(__file__), '../data.txt')
+with open(data) as f:
+    numbers: List[int] = parse(f.read())
+    d = diff(numbers)
+    print(calc(d))
+    print(find_valid_permutations(d))