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allcoding1_official (@allcoding1_official) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 85 687 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 509-o'rinni va Hindiston mintaqasida 3 512-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 85 687 obunachiga ega bo‘ldi.
20 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -1 460 ga, so‘nggi 24 soatda esa -39 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 3.36% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.73% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 2 882 marta ko‘riladi; birinchi sutkada odatda 625 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 21 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.
85 687
Obunachilar
-3924 soatlar
-3267 kunlar
-1 46030 kunlar
Postlar arxiv
85 687
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🔹Learning language resources English , 🇫🇷
𝐂𝐘𝐁𝐄𝐑 𝐒𝐄𝐂𝐔𝐑𝐈𝐓𝐘 𝐀𝐋𝐋 𝐂𝐎𝐔𝐑𝐒𝐄
⚡️ Basics
⚡️ Reconnaissance and Footprinting
⚡️ Network Scanning
⚡️ Enumeration
⚡️ Firewalls HIDs Honeypot
⚡️ Malware and Threats
⚡️ Mobile Platform
⚡️ Pentesting
⚡️ Sql Injection
⚡️ System Hacking
⚡️ Web Application
⚡️ Wireless Network
⚡️ Cloud Computing
⚡️ Web Server
⚡️ Social Engineering
⚡️ Session Hijacking
⚡️ Sniffing
⚡️ BufferOverflow
⚡️ Cryptography
⚡️ Denial Of Service
All courses (100 rupees)
Contact:- @meterials_available
85 687
📌IT learning courses
📌All programing courses
📌Abdul bari courses
📌Ashok IT
📌Linux
📌Networking
📌Design patterns
📌Donet
📌Docker
📌Entity framework
📌Node.js
📌ASP. Net
📌Aps. Net cro
📌java
📌JavaScript
📌full stack developer
Tutorials + Books + Courses + Trainings + Workshops + Educational Resources
🔹Data science
🔹Python
🔹Artificial Intelligence
🔹AWS Certified
🔹Cloud
🔹BIG DATA
🔹Data Analytics
🔹BI
🔹Google Cloud Platform
🔹IT Training
🔹MBA
🔹Machine Learning
🔹Deep Learning
🔹Ethical Hacking
🔹SPSS
🔹Statistics
🔹Data Base
🔹Learning language resources English , 🇫🇷
𝐂𝐘𝐁𝐄𝐑 𝐒𝐄𝐂𝐔𝐑𝐈𝐓𝐘 𝐀𝐋𝐋 𝐂𝐎𝐔𝐑𝐒𝐄
⚡️ Basics
⚡️ Reconnaissance and Footprinting
⚡️ Network Scanning
⚡️ Enumeration
⚡️ Firewalls HIDs Honeypot
⚡️ Malware and Threats
⚡️ Mobile Platform
⚡️ Pentesting
⚡️ Sql Injection
⚡️ System Hacking
⚡️ Web Application
⚡️ Wireless Network
⚡️ Cloud Computing
⚡️ Web Server
⚡️ Social Engineering
⚡️ Session Hijacking
⚡️ Sniffing
⚡️ BufferOverflow
⚡️ Cryptography
⚡️ Denial Of Service
All courses (100 rupees)
Contact:- @meterials_available
85 687
All codes are available
To easily find out ur codes
Once check it 👇👇
https://www.instagram.com/allcoding1_official?igsh=ZHJpNXdpeWh1d2No
85 687
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
85 687
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
85 687
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
85 687
def main():
import sys
input = sys.stdin.read
data = input().split('\n')
S = data[0].strip()
N = int(data[1].strip())
W = []
for i in range(N):
W.append(data[2 + i].strip())
freqS = get_frequency(S)
count = 0
for w in W:
if is_valid_anagram_subsequence(freqS, w):
count += 1
print(count)
def get_frequency(S):
freq = [0] * 26
for c in S:
freq[ord(c) - ord('a')] += 1
return freq
def is_valid_anagram_subsequence(freqS, w):
freqW = [0] * 26
for c in w:
freqW[ord(c) - ord('a')] += 1
for i in range(26):
if freqW[i] > freqS[i]:
return False
return True
if name == "main":
main()
// MInimal subarray length
85 687
MOD = 10**9 + 7
def f(X, P, L, R):
s = [(x, x + p) for x, p in zip(X, P)]
s.sort()
c = 0
i = 0
e = L
f = L
while e <= R:
while i < len(s) and s[i][0] <= e:
f = max(f, s[i][1])
i += 1
if f == e:
return -1
c += 1
e = f + 1
return c
def g(N, M, X, P, Q, Qs):
t = 0
for L, R in Qs:
r = f(X, P, L, R)
t = (t + r) % MOD
return t
// Lighting lamp
@allcoding1_official
85 687
def f(N, L, R, A):
M = 10**9 + 7
dp = [0] * (N + 1)
dp[0] = 1 # There's one way to partition an empty array
for i in range(1, N + 1):
xv = 0
for j in range(i, 0, -1):
xv ^= A[j - 1]
if L <= xv <= R:
dp[i] = (dp[i] + dp[j - 1]) % M
return dp[N]
// divide the array
@allcoding1_official
