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allcoding1_official (@allcoding1_official) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 84 838 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 497-o'rinni va Hindiston mintaqasida 3 505-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 84 838 obunachiga ega bo‘ldi.
06 Iyul, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -1 554 ga, so‘nggi 24 soatda esa -61 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 2.83% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.90% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 2 405 marta ko‘riladi; birinchi sutkada odatda 762 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 07 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.
84 838
Obunachilar
-6124 soatlar
-3767 kunlar
-1 55430 kunlar
Postlar arxiv
84 838
![n=int(input()) 4 a,c=[],o for i in range(n): I b=int(input()) a.append(b) for i in range(n-1): for j in range (i + 1, n) : if](https://cdn.tlmtr.io/_next/static/media/empty-channel.35ed232f.svg)
n=int(input())
4 a,c=[],o
for i in range(n): I
b=int(input())
a.append(b)
for i in range(n-1):
for j in range (i + 1, n) :
if a[i] > a[j] :
C+=1
a[i] , a[j] = a[j] , a[i]
print(c)
Telegram:-
84 838
#include<bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
string str;
cin >> str;
int n = str.length(), ans = 0;
for(int i = 0; i < n; i++){
if(str[i] == '5' || str[i] == '6')
continue;
else
ans += 1;
}
cout << ans << endl;
}
return 0;
}
Beautiful Number
Telegram:- @allcoding1_official
84 838
array AR of size N
perfect_sum(arr, s, result) :
x = [0]*len(arr)
j = len(arr) - 1
while (s > 0) :
x[j] = s % 2
s = s // 2
j -= 1
sum = 0
for i in range(len(arr)) :
if (x[i] == 1) :
sum += arr[i]
if (sum == result) :
print("{",end="");
for i in range(len(arr)) :
if (x[i] == 1) :
print(arr[i],end= ", ");
print("}, ",end="")
def print_subset(arr, K) :
x = pow(2, len(arr))
for i in range(1, x):
perfect_sum(arr, i, K)
# Driver code
arr = [ ]
n = int(input("Enter length of array : "))
s=int(input("Enter sum : "))
for i in range(n):
ele=int(input("Enter element : "))
arr.append(ele)
print_subset(arr, s)
Python
Telegram:- @allcoding1_
84 838
def minOperations(arr1, arr2, i, j):
# Base Case
if arr1 == arr2:
return 0
if i >= len(arr1) or j >= len(arr2):
return 0
# If arr[i] < arr[j]
if arr1[i] < arr2[j]:
# Include the current element
return 1 \
+ minOperations(arr1, arr2, i + 1, j + 1)
# Otherwise, excluding the current element
return max(minOperations(arr1, arr2, i, j + 1),
minOperations(arr1, arr2, i + 1, j))
# Function that counts the minimum
# moves required to sort the array
def minOperationsUtil(arr):
brr = sorted(arr);
# If both the arrays are equal
if(arr == brr):
# No moves required
print("0")
# Otherwise
else:
# Print minimum operations required
print(minOperations(arr, brr,)
Python
Q) minimum operations
Telegram - @allcoding1_official
84 838
# A Naive recursive python program to find minimum of coins
# to make a given change V
import sys
# m is size of coins array (number of different coins)
def minCoins(coins, m, V):
# base case
if (V == 0):
return 0
# Initialize result
res = sys.maxsize
# Try every coin that has smaller value than V
for i in range(0, m):
if (coins[i] <= V):
sub_res = minCoins(coins, m, V-coins[i])
# Check for INT_MAX to avoid overflow and see if
# result can minimized
if (sub_res != sys.maxsize and sub_res + 1 < res):
res = sub_res + 1
return res
# Driver program to test above function
coins = [9, 6, 5, 1]
m = len(coins)
V = 11
print("Minimum coins required is",minCoins(coins, m, V))
Python 3
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