پایتون | Data Science | Machine Learning
◀️اینجا با تمرین و چالش با هم پایتون رو یاد می گیریم ⏮بانک اطلاعاتی پایتون پروژه / code/ cheat sheet +ویدیوهای آموزشی +کتابهای پایتون تبلیغات: @alloadv 🔁ادمین : @maryam3771
Ko'proq ko'rsatish📈 Telegram kanali پایتون | Data Science | Machine Learning analitikasi
پایتون | Data Science | Machine Learning (@python4all_pro) Forsiy til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 24 694 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 5 515-o'rinni va Eron mintaqasida 13 715-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 24 694 obunachiga ega bo‘ldi.
18 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni 1 596 ga, so‘nggi 24 soatda esa -10 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 3.81% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 2.09% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 941 marta ko‘riladi; birinchi sutkada odatda 515 ta ko‘rish yig‘iladi.
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- Tematik yo‘nalishlar: Kontent مصنوعی, دنیا, آموزش, پایتون, وبینار kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“◀️اینجا با تمرین و چالش با هم پایتون رو یاد می گیریم
⏮بانک اطلاعاتی پایتون
پروژه / code/ cheat sheet
+ویدیوهای آموزشی
+کتابهای پایتون
تبلیغات:
@alloadv
🔁ادمین :
@maryam3771”
Yuqori yangilanish chastotasi (oxirgi ma’lumot 19 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.
git clone https://github.com/basicW/badjpg.git
#library
#python
🆔 @Python4all_proدروازه ورود به دنیای کوئـریهـا🗓تاریخ شروع: ۱۴ مرداد %۳۰ تخفیف پیش ثبتنام + هـدیه 🎁 💵امکان پرداخت اقساطـی ⭐️ ارائه مدرک دو زبانـه شرکت در دوره 🔗مشاهده سرفصلها و پیش ثبتنام رایگان 🔗مشاهده سرفصلها و پیش ثبتنام رایگان 🔗مشاهده سرفصلها و پیش ثبتنام رایگان https://nkmz.ir/mddnzqgnfq ❤️❤️❤️ 📞 02191070017 💬@nikamooz
# class ListNode:
# def init(self, x):
# self.val = x
# self.next = None
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
if head.next == None:
return None
tmp = head
size = 0
# find the size of the linked list
while tmp:
size += 1
tmp = tmp.next
tmp = head
#if we have to remove the first node:
if n == size:
return head.next
for i in range(size-n-1):
tmp = tmp.next
tmp.next = tmp.next.next
return head
Explanation:
Determining the list size:
First we go through the entire list to find its size. This is necessary to determine which node needs to be removed.
The size variable keeps track of the number of nodes in the list.
Removing the first node:
If n is equal to the size of the list, it means the first node should be removed. In this case, we return the next node from the head of the list.
Search for a node before the one you want to delete:
If n is not equal to the size of the list, we need to find the node that comes before the nth node from the end. We do this by moving size-n-1 nodes forward in the list.
Removing a node:
We set the next node for the found node so that it points to the next node after the next one, effectively bypassing the node to be deleted.
Returning a new header:
Returning the head of the list. If the first node was deleted, the new header will be the starting node of the list after the delete.
Time and space complexity:
Time complexity: O(L), where L is the length of the list. We go through the list twice: once to count nodes and again to remove a node.
Space complexity: O(1) because we use a constant amount of extra space independent of the size of the input data.
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