Endi mavjud! Telegram Tadqiqoti 2025 — yilning asosiy insaytlari 
ACCENTURE EXAM HELP ! CISCO EXAM !
Kanalga Telegram’da o‘tish
🔥Guys plz Stop fearing for daily exams 📝 👨💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srksvk
Ko'proq ko'rsatish📈 Telegram kanali ACCENTURE EXAM HELP ! CISCO EXAM ! analitikasi
ACCENTURE EXAM HELP ! CISCO EXAM ! (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 13 208 obunachidan iborat bo'lib, Taʼlim toifasida 15 294-o'rinni va Hindiston mintaqasida 31 490-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 13 208 obunachiga ega bo‘ldi.
30 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -144 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 3.08% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.29% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 407 marta ko‘riladi; birinchi sutkada odatda 170 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 placement, gaurntee, suree, capgemini, infosy kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“🔥Guys plz Stop fearing for daily exams 📝
👨💻 @srksvk is here to help you all at lowest cost possible.💪
🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company
🔥Effort from our side = 💯
📱Main Channel: @coding_are
📱Tel I'd : @srks...”
Yuqori yangilanish chastotasi (oxirgi ma’lumot 01 Iyul, 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.
13 208
Obunachilar
+624 soatlar
-127 kunlar
-14430 kunlar
Postlar arxiv
import math
def calculate_distance(point1, point2):
return abs(point1[0] - point2[0]) + abs(point1[1] - point2[1])
def assign_vehicles(passengers, vehicles):
allocated_vehicles = {}
total_distance = 0
for passenger in sorted(passengers):
min_distance = math.inf
closest_vehicle = ""
passenger_coordinates = passengers[passenger]
for vehicle in vehicles:
if vehicles[vehicle] == "":
vehicle_coordinates = vehicles[vehicle + "_coordinates"]
distance = calculate_distance(passenger_coordinates, vehicle_coordinates)
if distance < min_distance or (distance == min_distance and vehicle < closest_vehicle):
min_distance = distance
closest_vehicle = vehicle
allocated_vehicles[passenger] = closest_vehicle
vehicles[closest_vehicle] = passenger
total_distance += min_distance
return total_distance
N, M = map(int, input().split())
passengers = {}
vehicles = {}
for _ in range(N):
name, x, y = input().split()
passengers[name] = (int(x), int(y))
for _ in range(M):
vehicle, x, y = input().split()
vehicles[vehicle] = ""
vehicles[vehicle + "_coordinates"] = (int(x), int(y))
minimum_distance = assign_vehicles(passengers, vehicles)
print(minimum_distance,end="")
Solo rider✅✅
Python 3
@codeing_area ✅✅✅
Share share ✅✅✅✅
def generate_password(number, name):
# Check if name consists of only lowercase letters
if name.isalpha() and name.islower():
# Convert number to absolute value to handle negative numbers
abs_number = abs(number)
# Ensure length of name is within the specified range
name_length = min(max(len(name), 1), 100)
# Generate password using a combination of number and name
password = str(abs_number) + name[:name_length]
return password
else:
return "Invalid"
try:
# Input: Number of test cases
T = int(input())
# Process each test case
for _ in range(T):
# Input: Number and Name separated by space
test_input = input().split()
if len(test_input) == 2:
number = int(test_input[0])
name = test_input[1]
# Output: Print the password for each test case in a new line
result = generate_password(number, name)
print(result)
else:
print("Invalid input format")
except Exception as e:
print("An error occurred:", e)
password generator updated all passed
def minVehicles(weights, limit):
weights.sort() # Sort the weights in ascending order
left, right = 0, len(weights) - 1
vehicles = 0
while left <= right:
if weights[left] + weights[right] <= limit:
left += 1
right -= 1
vehicles += 1
return vehicles
# Python 3 compatibility
try:
# Python 3
raw_input = input
except NameError:
# Python 2
pass
weights = list(map(int, raw_input().split()))
limit = int(raw_input())
result = minVehicles(weights, limit)
# Output the result without extra spaces or newline characters
print(result, end="")
Warehouse public passed ✅✅✅... python 3
Shared @codeing_area fastt everyone
def bubble_sort(a1, a2):
n = len(a1)
for i in range(n):
swapped = False
for j in range(0, n - i - 1):
if a1[j] > a1[j + 1]:
a1[j], a1[j + 1] = a1[j + 1], a1[j]
a2[j], a2[j + 1] = a2[j + 1], a2[j]
swapped = True
if not swapped:
break
return a2
a1 = list(map(int, input().split()))
a2 = list(map(int, input().split()))
result = bubble_sort(a1, a2)
print(*result)
Bubble sort in python
Full working ✅💯💯💯💯
Share @codeing_area
I will post all code after 10am 💯💯
So don't worry everyonee ✅✅✅
Share @codeing_are
Fastt
Sahre fasttt everyone group
I am not asking money....you need to sahare group only
#include <bits/stdc++.h>
using namespace std;
int solve(vector<int>items,int ind,int size,vector<vector<int>>&dp){
if(ind>=items.size()){
return 0;
}
if(dp[ind][size]!=-1){
return dp[ind][size];
}
int take = items[ind] * size + solve(items, ind + 1, size + 1,dp);
int nottake = solve(items, ind + 1, size,dp);
return dp[ind][size]=max(take, nottake);
}
int main()
{
string itemsInput;
getline(cin, itemsInput);
vector<int> items;
stringstream ss(itemsInput);
int item;
while (ss >> item)
{
items.push_back(item);
}
sort(items.begin(),items.end());
vector<vector<int>> dp(items.size()+1, vector<int>(items.size()+1, -1));
int ans = solve(items, 0, 1,dp);
cout << ans << endl;
}
If you more code then share group
@codeing_areaa
Make it 5.5 k then I will share codes
def shyam_and_strings(s1, s2, k):
n, m = len(s1), len(s2)
dp = [[0] * (m + 1) for _ in range(n + 1)]
for i in range(n + 1):
for j in range(m + 1):
if i == 0 or j == 0:
dp[i][j] = 0
else:
cnt = 0
if s1[i - 1] != s2[j - 1]:
cnt = min(abs(ord(s1[i - 1]) - ord(s2[j - 1])), 26 - abs(ord(s1[i - 1]) - ord(s2[j - 1])))
if cnt <= k:
dp[i][j] = 1 + dp[i - 1][j - 1]
dp[i][j] = max(dp[i][j], max(dp[i - 1][j], dp[i][j - 1]))
return dp[n][m]
s1 = input().strip()
s2 = input().strip()
k = int(input().strip())
if not s1 or not s2:
print(0)
else:
print(shyam_and_strings(s1, s2, k), end="")
