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allcoding1_official

allcoding1_official

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allcoding1_official (@allcoding1_official) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 84 502 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 495-o'rinni va Hindiston mintaqasida 3 530-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 84 502 obunachiga ega bo‘ldi.

12 Iyul, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -1 602 ga, so‘nggi 24 soatda esa -68 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 2.00% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.84% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 1 688 marta ko‘riladi; birinchi sutkada odatda 712 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 1 ta reaksiya keladi.
  • Tematik yo‘nalishlar: Kontent dsa, stack, namaste, javascript, dev kabi asosiy mavzularga jamlangan.

📝 Tavsif va kontent siyosati

Kanal uchun tavsif kiritilmagan.

Yuqori yangilanish chastotasi (oxirgi ma’lumot 13 Iyul, 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.

84 502
Obunachilar
-6824 soatlar
-3987 kunlar
-1 60230 kunlar
Postlar arxiv
16 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates
16 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

Black box testing 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates
Black box testing 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

155 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates
155 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

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🎯Company name: Apple Inc Job Position: Security Engineer Location: Hyderabad Job Type: Full time Experience: Freshers Qualification: BE/ B.Tech/ ME/ M.Tech Batch: 2017/ 2018/ 2019/ 2020/ 2021 Salary: Min 10 LPA (Expected) Apply:- https://www.joboffersadda.com/2022/04/apple-recruitment-be-btech-me-mtech.html 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

Apply:- https://www.joboffersadda.com/2022/04/wipro-recruitment-any-graduate.html 🔗 Telegram: @allcoding1 🔔Unmute this chan
Apply:- https://www.joboffersadda.com/2022/04/wipro-recruitment-any-graduate.html 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

🎯 Cognizant Recruitment For Process Executive - Data Location: Hyderabad Qualification: Any Graduate Work Experience: 0 - 2 Year Apply - https://www.joboffersadda.com/2022/04/cognizant-recruitment-2022-for-process.html 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

🎯Capgemini Engineering Off Campus Drive For Software Analyst Location: Bangalore Work Experience: Fresher Apply - https://www.joboffersadda.com/2022/04/capgemini-engineering-off-campus-drive.html 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

🎯IBM OFF CAMPUS DRIVE Degree: B.Tech/BE, MCA, M.Tech/ME 📌 link:- https://www.joboffersadda.com/2022/04/ibm-off-campus-drive.html 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

2nd code in Python
2nd code in Python

2nd code in C++
2nd code in C++

Executed 1st code python
Executed 1st code python

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def smallest(n): arr = list() x = 0 for i in range(n): x = int(input()) arr.append(x) arr = sorted(arr) count = 1 temp = list() for j in range(1, n): if arr[j] == arr[j-1] + 1: count += 1 else: temp.append(count) count = 1 temp.append(count) return min(temp) n = int(input()) res = smallest(n) print(res) def isP(n): if n<=3: return True if ((n%2==0)|(n%3==0)): return False for d in range(5,int(n*5)+1,6): if ((n%d==0)|(n%(d+2)==0)): return False return True def getP(m): P=[1,2] for n in range(3,m+1,2): if isP(n): P.append(n) return P P=getP(int(105+1000)) g = int(input().strip()) for a0 in range(g): n = int(input().strip()) for i in range(len(P)): if P[i]>n: c=i-1 break if c%2==1: print("Alice") else: print("Bob") Python Alice & Bob Code TCS answers 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

Pseudo: 1) True 2)12 38 3) compilation error 4)the valve of each 5)Construction of prototype 6)Prototyping 7)bytes 8) weighted 9)any greedy .... 10) option A

Test of Knowledge 1.True

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