Don't get caught by a cheater! Telemetrio finds and tags such channels 👉 If you want to see the tag, subscribe

ACCENTURE EXAM SOLUTIONS

Open in Telegram

🔥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

India28 153Education14 642...

📈 Analytical overview of Telegram channel ACCENTURE EXAM SOLUTIONS

Channel ACCENTURE EXAM SOLUTIONS (@coding_are) in the English language segment is an active participant. Currently, the community unites 13 954 subscribers, ranking 14 642 in the Education category and 28 153 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 13 954 subscribers.

According to the latest data from 03 November, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -151 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 3.70%. Within the first 24 hours after publication, content typically collects 1.38% reactions from the total number of subscribers.
Post reach: On average, each post receives 516 views. Within the first day, a publication typically gains 193 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 04 November, 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.

13 954

Subscribers

-624 hours

-377 days

-15130 days

516

Post views

~ 19324 hours

~ 26148 hours

3.70%

Engagement rate

~ 3

Posts per day

Ads index

Posts Archive

13 954

Telescope code All code passed ✅😀😌😌😌 Python3

13 954

Best telescope code Python3 👆👆👆 19 line for i in range(n)

13 954

Python3 Pass the baton code✅✅

13 954

Best telescope code Python3 👆👆👆

13 954

Paint the ceiling 100% executed ✅✅✅✅✅ Python3

13 954

def variantsCount(n, s0, k, b, m, a): if n <=0: return 0 sum_array = [s0] for i in range(1,n): s_i = ((k * sum_array[i - 1] + b) % m) + 1 + sum_array[i - 1] sum_array.append(s_i) # two pointer to calculate the count left = 0 right = len(sum_array) -1 count = 0 while left <= right: area = sum_array[left] * sum_array[right] if area <= a: count += 2*(right - left) + 1 left +=1 elif area > a: right -=1 return count Paint the ceiling code Python3

13 954

Python3 Pass the baton code✅✅

13 954

#include <bits/stdc++.h> using namespace std; vector<int> batonPass(int n, long t){ vector<int> ans; int oddEven = 1,count=0; while(true){ if(oddEven & 1){ if(count==0){ for(int i=0 ; i<n ; i++){ count++; if(t == count){ if(i!=n-1){ // cout << i+1 << " " << i+2; ans.push_back(i+1); ans.push_back(i+2); return ans; }else{ ans.push_back(i+1); ans.push_back(i); return ans; } } } }else{ for(int i=1 ; i<n ; i++){ count++; if(t == count){ if(i!=n-1){ ans.push_back(i+1); ans.push_back(i+2); return ans; }else{ ans.push_back(i+1); ans.push_back(i); return ans; } } } } oddEven++; }else{ for(int i = n-2 ; i>=0 ; i--){ count++; if(t == count){ if(i!=0){ ans.push_back(i+1); ans.push_back(i); return ans; }else{ ans.push_back(i); ans.push_back(i+1); return ans; } } } oddEven++; } } } int main() { int friends,timee; cin >> friends >> timee; vector<int> ans = batonPass(friends,timee); for(auto it : ans) cout << it << " "; return 0; } Pass the baton code 👆👆 C++14

13 954

vector<int> batonPass(int friends, long time) { vector<int> f; int k = 1; bool r = false; long i = 1; while (i <= time) { if (!r) { if (k <= friends) k++; else { r = true; } } else { if (k >= 2) { k--; } else { r = false; } } i++; } if (r) { f.push_back(k); f.push_back(k - 1); } else { f.push_back(k - 1); f.push_back(k); } return f; } Pass the baton code- c++14

13 954

Baton pass

13 954

Atlassian free code help 🔥🔥 Join here guys Join this group guys for free code Atlassian code free 🤘 free h:https://t.me/rituyr https://t.me/rituyr https://t.me/rituyr share ✅share✅ share ✅

13 954

https://t.me/rituyr

13 954

java 4th code ✅✅

13 954

def FindRemainingBalls(direction, strength): asteroids = [] for i in range(0, len(direction)): if direction[i] < 0: asteroids.append(strength[i]*-1) else: asteroids.append(strength[i]) stack = [] collided = False for asteroid in asteroids: while stack and asteroid < 0 < stack[-1]: defending_asteroid = stack.pop() if defending_asteroid != abs(asteroid): asteroid = asteroid if abs(asteroid) > defending_asteroid else defending_asteroid else: asteroid = 0 if asteroid != 0: stack.append(asteroid) return stack Ball collision-python

13 954

mex problem-c++

13 954

int mex(vectorarr,int K) { int n=arr.size(); unordered_map mp; for (int i = 0; i < n; i++) { mp[arr[i] % K]++; } for (int i = 0; i < n; i++) { if (mp.find(i % K) == mp.end()) { return i; } mp[i % K]--; if (mp[i % K] == 0) mp.erase(i % K); } return n; } Mex problem guys ✅✅ Note -type as it is it will work C++

13 954