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Firibgarlarga uchmang! Telemetrio bunday kanallarni topadi va belgilaydi 👉 Belgini ko‘rmoqchi bo‘lsangiz, obuna bo‘ling

allcoding1_official

allcoding1_official

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📈 Telegram kanali allcoding1_official analitikasi

allcoding1_official (@allcoding1_official) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 78 690 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 624-o'rinni va Hindiston mintaqasida 3 768-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 78 690 obunachiga ega bo‘ldi.

31 Oktabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -1 635 ga, so‘nggi 24 soatda esa -91 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 1.60% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.61% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 1 263 marta ko‘riladi; birinchi sutkada odatda 481 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 0 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 01 Noyabr, 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.

78 690
Obunachilar
-9124 soatlar
-3887 kun
-1 63530 kun
Postlar arxiv
return min(e-b+1, numChapters) Python 3✅ IBM Telegram:-
+1
return min(e-b+1, numChapters) Python 3✅ IBM Telegram:-

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#scammer Please report 👇 @offical1allcoding , Vds , @Allcodingoffical, @coding1officalyyh The changing names Guys ♥️ be care
+1
#scammer Please report 👇 @offical1allcoding , Vds , @Allcodingoffical, @coding1officalyyh The changing names Guys ♥️ be careful don't pay money

Guys ♥️ @allcoding, @Allcodingoffical1, @offical1allcoding @Allcodingoffical, @codingoffical and @coding1officalyyh @Allcodin
Guys ♥️ @allcoding, @Allcodingoffical1, @offical1allcoding @Allcodingoffical, @codingoffical and @coding1officalyyh @Allcodingofficalmain it's not me don't lose your money #scammer Please report Please share with your friends and telegram Group's

🎯Phone pe hiring Job role:- Advisor,ONDC Qualification:- Any Experience:- 0-2years Location:- Bangalore Apply Now:- www.allcoding1.com Telegram:- @allcoding1_official

#scammer Please report 👇 @offical1allcoding , Vds , @Allcodingoffical The changing names Guys ♥️ be careful don't pay money
+1
#scammer Please report 👇 @offical1allcoding , Vds , @Allcodingoffical The changing names Guys ♥️ be careful don't pay money

Guys ♥️ @allcoding, @Allcodingoffical1, @offical1allcoding @Allcodingoffical, @codingoffical and @Allcodingofficalmain it's n
Guys ♥️ @allcoding, @Allcodingoffical1, @offical1allcoding @Allcodingoffical, @codingoffical and @Allcodingofficalmain it's not me don't lose your money #scammer Please report Please share with your friends and telegram Group's

#scammer Please report 👇 @offical1allcoding , Vds , @Allcodingoffical The changing names
+1
#scammer Please report 👇 @offical1allcoding , Vds , @Allcodingoffical The changing names

Guys ♥️ @allcoding, @Allcodingoffical1, @offical1allcoding @Allcodingoffical, @codingoffical and @Allcodingofficalmain it's n
Guys ♥️ @allcoding, @Allcodingoffical1, @offical1allcoding @Allcodingoffical, @codingoffical and @Allcodingofficalmain it's not me don't lose your money #scammer Please report Please share with your friends and telegram Group's

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sticker.webp0.13 KB

sticker.webp0.13 KB

sticker.webp0.13 KB

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int solve(vector> arr) {     int n = arr.size();     vector> dp(n, vector(n));  &nbs
int solve(vector> arr) {     int n = arr.size();     vector> dp(n, vector(n));     int maxi = 0;     for(int i = 0; i < n; i++) {         for(int j = 0; j < n; j++) {             if(i == 0 || j == 0) {                 dp[i][j] = arr[i][j];             } else if(arr[i][j] == 1) {                 dp[i][j] = min({dp[i-1][j], dp[i][j-1], dp[i-1][j-1]}) + 1;             }             maxi = max(maxi, dp[i][j]);         }     }     return maxi; } IBM