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allcoding1 (@allcoding1) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 22 590 obunachidan iborat bo'lib, Taʼlim toifasida 8 822-o'rinni va Hindiston mintaqasida 19 518-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 22 590 obunachiga ega bo‘ldi.

12 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -437 ga, so‘nggi 24 soatda esa -6 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 5.99% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.25% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 1 353 marta ko‘riladi; birinchi sutkada odatda 283 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 2 ta reaksiya keladi.
  • Tematik yo‘nalishlar: Kontent dsa, stack, namaste, javascript, learning kabi asosiy mavzularga jamlangan.

📝 Tavsif va kontent siyosati

Kanal uchun tavsif kiritilmagan.

Yuqori yangilanish chastotasi (oxirgi ma’lumot 13 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

22 590
Obunachilar
-624 soatlar
-967 kunlar
-43730 kunlar
Postlar arxiv
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Athletes code Accenture Hackdiva
Athletes code Accenture Hackdiva

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Latin Alpha Numerals Accenture Hackdiva Code
Latin Alpha Numerals Accenture Hackdiva Code

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LI.GCD code in another way of solving
LI.GCD code in another way of solving

LI Gcd Code in python Infosys
LI Gcd Code in python Infosys

All codes are available To easily find out ur codes Once check it 👇👇 https://www.instagram.com/allcoding1_official?igsh=ZHJpNXdpeWh1d2No

def f(S): n = len(S) lf, rf = {}, {} ls, rs = set(), set() # Initialize the rf map and rs with the entire string S for c in S: rf[c] = rf.get(c, 0) + 1 rs.add(c) mx = 0 # Traverse the string and adjust the ls and rs sets and maps for i in range(n - 1): c = S[i] lf[c] = lf.get(c, 0) + 1 rf[c] -= 1 if rf[c] == 0: rs.remove(c) ls.add(c) cs = len(ls) + len(rs) mx = max(mx, cs) return n - mx // spilit screen @allcoding1_official

public static int solve(int N, int[] A) { int t = 0; for (int num : A) { t += num; } int x = 0; int y = 0; for (int i = 0; i < N; i++) { int res = t - x - A[i]; if (x == res) { y++; } x += A[i]; } return y; } //Equilibrium Point

def GetAnswer(N, A, B, P): dp = [0] * (N + 1) max_d = 0 for i in range(N - 1, -1, -1): max_p = P[i] min_p = P[i] for j in range(1, B + 1): if i + j <= N: max_p = max(max_p, P[i + j - 1]) min_p = min(min_p, P[i + j - 1]) max_d = max(max_d, max_p - min_p) if i + 1 == N: dp[i] = max(dp[i], max_d) else: dp[i] = max(dp[i], max_d, dp[i + 1]) return dp[0] import sys input = sys.stdin.read data = input().split() N = int(data[0]) A = int(data[1]) B = int(data[2]) P = list(map(int, data[3:])) result = GetAnswer(N, A, B, P) print(result) // XOR formaating code