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 Education15 294...

📈 Analytical overview of Telegram channel ACCENTURE EXAM HELP ! CISCO EXAM !

Channel ACCENTURE EXAM HELP ! CISCO EXAM ! (@coding_are) in the English language segment is an active participant. Currently, the community unites 13 208 subscribers, ranking 15 294 in the Education category and 31 490 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 13 208 subscribers.

According to the latest data from 30 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -144 over the last 30 days and by 6 over the last 24 hours, overall reach remains high.

Verification status: Not verified
Engagement rate (ER): The average audience engagement rate is 3.08%. Within the first 24 hours after publication, content typically collects 1.29% reactions from the total number of subscribers.
Post reach: On average, each post receives 407 views. Within the first day, a publication typically gains 170 views.
Reactions and interaction: The audience actively supports content: the average number of reactions per post is 2.
Thematic interests: Content is focused on key topics such as placement, gaurntee, suree, capgemini, infosy.

📝 Description and content policy

The author describes the resource as a platform for expressing subjective opinions:
“🔥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...”

Thanks to the high frequency of updates (latest data received on 01 July, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.

13 208

Subscribers

+624 hours

-127 days

-14430 days

407

Post views

~ 17024 hours

~ 24048 hours

3.08%

Engagement rate

~ 3

Posts per day

Ads index

beta

Posts Archive

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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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