## Day 10, Year 2015: Elves Look, Elves Say

First read the problem description.

Discussions about Conway’s video.

Memory allocation is the slowest part. We estimate the upper bound of the length of the nth sequence and preallocate two lists with the needed space and compute the new sequence populating each time the other list.

```
def look_and_say(current, new):
= current[0]
prev_d = 0
run_len = 0
new_i for d in current:
if d != prev_d:
= run_len
new[new_i] +1] = prev_d
new[new_i+= 2
new_i = 0
run_len = d
prev_d if d == 0:
break
+= 1
run_len
def estimate_length(times):
= 1.30357726903429639125709911215255189073070250465940487575486139062855088785246155712681576686442522555
λ return λ**times
```

```
= 50
times = int(10 * estimate_length(times))
upper_bound
= [0]*upper_bound
sequence 0:10] = [1,1,1,3,1,2,2,1,1,3]
sequence[= [0]*upper_bound
tmp
for _ in range(times):
look_and_say(sequence, tmp)= tmp, sequence
sequence, tmp
# remove trailing 0s
= sequence[:sequence.index(0)]
sequence
len(sequence)
```

`5103798`

### Comments

Source code of the solution(s):