Python already has garbage collectors. For example, CPython counts the number of references to a specific memory field, and once the memory is not referenced anymore, the memory will be freed.
If my understanding is correct, I have no idea when and how to use
gc module. I can see that my colleagues use it to reduce
the memory usage, but I need to check if it's right.
gc - Garbage Collector Interface - Python 3.12.4 Documentation
This module provides an interface to the optional garbage collector. It provides the ability to disable the collector, tune the collection frequency, and set debugging options. It also provides access to unreachable objects that the collector found but cannot free. Since the collector supplements the reference counting already used in Python, you can disable the collector if you are sure your program does not create reference cycles.
gc.collect(generation=2)With no arguments, run a full collection. The effect of calling gc.collect() while the interpreter is already performing a collection is undefined.
In general, calling gc.collect() is dangerous because
CPython's garbage collection is already running, and they would
conflict.
However, I found this thread.
Celery doesn't release memory after finishing a task
It's an old thread, but our application seems to face the same issue,
running gc.collect() at the end of a task frees huge amount
of memory.
So, in our case, calling gc.collect() to free memory
used in Celery makes sense.
greedy approach
I should have noticed that I could check if the person could jump for
each position. m is at most 10, so for-looping
10 times for each position does not affect the time complexity. 10 is a
constant.
def possible(n: int, m: int, k: int, a: list[str]) -> bool:
posi: int = -1
while posi < n:
if posi >= 0 and a[posi] == "C":
return False
elif posi >= 0 and a[posi] == "W":
posi += 1
k -= 1
else: # posi == -1 or a[posi] == "L"
# if there is a log in the range, jump to it
# if there is no log, jump to to the farthest water
logi: int = -1
wateri: int = -1
# at most m = 10, and this loop does not affect the time complexity
for jumplen in range(1, m + 1):
if posi + jumplen >= n:
logi = n # goal
break
if a[posi + jumplen] == "L":
logi = posi + jumplen
elif a[posi + jumplen] != "C":
wateri = posi + jumplen
if logi != -1:
posi = logi
elif wateri != -1:
posi = wateri
else:
return False
return k >= 0
for _ in range(int(input())):
n, m, k = map(int, input().split())
a: list[str] = list(input())
if possible(n, m, k, a):
print("YES")
else:
print("NO")Ketone 60 mg/dl
Cheese 10g Snacks 10g Protein shake 10g Bacon Egg 0g
Total carbohydrate 30g
walking
MUST:
huyfififi/dashboard-backendTODO: