ACCENTURE EXAM HELP ! CISCO EXAM !

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🔥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

India31 490 Educación15 294...

📈 Análisis del canal de Telegram ACCENTURE EXAM HELP ! CISCO EXAM !

El canal ACCENTURE EXAM HELP ! CISCO EXAM ! (@coding_are) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 13 208 suscriptores, ocupando la posición 15 294 en la categoría Educación y el puesto 31 490 en la región India.

📊 Métricas de audiencia y dinámica

Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 13 208 suscriptores.

Según los últimos datos del 30 junio, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -144, y en las últimas 24 horas de 6, conservando un alto alcance.

Estado de verificación: No verificado
Tasa de interacción (ER): El promedio de interacción de la audiencia es 3.08%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 1.29% de reacciones respecto al total de suscriptores.
Alcance de las publicaciones: Cada publicación recibe en promedio 407 visualizaciones. En el primer día suele acumular 170 visualizaciones.
Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 2.
Intereses temáticos: El contenido se centra en temas clave como placement, gaurntee, suree, capgemini, infosy.

📝 Descripción y política de contenido

El autor describe el recurso como un espacio para expresar opiniones subjetivas:
“🔥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...”

Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 01 julio, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Educación.

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Suscriptores

+624 horas

-127 días

-14430 días

407

Visitas de la publicación

~ 17024 horas

~ 24048 horas

3.08%

Tasa de compromiso

~ 3

Mensajes por día

Ads index

beta

Archivo de publicaciones

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Fully paased ✅🎉🎉🎉🎉🎉🎉✅💯💯

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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 ✅✅✅✅

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Next code will upload soon....✅✅

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Share group everyone fastttt✅✅✅✅ @codeing_area

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Warehouse Fully paased ✅💯👍

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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

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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

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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

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I will post all code after 10am 💯💯 So don't worry everyonee ✅✅✅ Share @codeing_are Fastt

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Sahre fasttt everyone group I am not asking money....you need to sahare group only

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#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; }

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If you more code then share group @codeing_areaa Make it 5.5 k then I will share codes

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Full passed

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Full workings ✅✅✅✅✅

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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="")

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