ch
Feedback
ACCENTURE EXAM SOLUTIONS

ACCENTURE EXAM SOLUTIONS

前往频道在 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

显示更多

📈 Telegram 频道 ACCENTURE EXAM SOLUTIONS 的分析概览

频道 ACCENTURE EXAM SOLUTIONS (@coding_are) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 14 086 名订阅者,在 教育 类别中位列第 14 131,并在 印度 地区排名第 28 155 位。

📊 受众指标与增长动态

自 невідомо 创建以来,项目保持高速增长,吸引了 14 086 名订阅者。

根据 05 十月, 2026 的最新数据,频道保持稳定运转。过去 30 天订阅人数变化为 -120,过去 24 小时变化为 -12,整体触达仍然可观。

  • 认证状态: 未认证
  • 互动率 (ER): 平均受众互动率为 3.51%。内容发布后 24 小时内通常能获得 1.81% 的反应,占订阅者总量。
  • 帖子覆盖: 每篇帖子平均可获得 495 次浏览,首日通常累积 255 次浏览。
  • 互动与反馈: 受众积极参与,单帖平均反应数为 2。
  • 主题关注点: 内容集中在 placement, gaurntee, suree, capgemini, infosy 等核心主题上。

📝 描述与内容策略

作者将该频道定位为表达主观观点的平台:
“🔥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...”

凭借高频更新(最新数据采集于 06 十月, 2026),频道始终保持新鲜度与高覆盖。分析显示受众积极互动,使其成为 教育 类别中的关键影响点。

14 086
订阅者
-1224 小时
-327 天
-12030 天
帖子存档
I will upload code 💯% running

I will upload all code Afte 11k ...so do fast everyone✅✅ @codeing_area

Second code ✅💯💯💯 Infosys 🔥🥇🥇🥇
Second code ✅💯💯💯 Infosys 🔥🥇🥇🥇

I will upload all code Afte 11k ...so do fast everyone

If you want answer then share group Make it 11k 🥺👇🥺 Join @codeing_area

Frist short 🔥🔥🔥🔥🔥🔥🔥 Infosys done ✅✅✅
Frist short 🔥🔥🔥🔥🔥🔥🔥 Infosys done ✅✅✅

Share this screenshot of big group everyone ✅✅✅✅✅
Share this screenshot of big group everyone ✅✅✅✅✅

Share the group everyone for 2nd answer 🔥🔥 Make it 11k ...then I will upload all answers

int solve(const string& S) { vector freq(26, 0); for (char c : S) { freq[c - 'a']++; } int total = 0; for (int i = 0; i &
int solve(const string& S) { vector freq(26, 0); for (char c : S) { freq[c - 'a']++; } int total = 0; for (int i = 0; i < 26; ++i) { for (int j = i + 1; j < 26; ++j) { total += freq[i] * freq[j] * abs(i - j); } } return total; } cost of string s

....

Just share group everyone ✅🥺🥺 After 11k I will uploaded first answer

First' answer will uploaded... after 11k So share fast group ✅✅✅ if you want answer

I will start now exam .....if am done then I will share....

Share this screenshot in big group everyone ✅✅✅ And join the group members ✅ Fast guys ✅✅
Share this screenshot in big group everyone ✅✅✅ And join the group members ✅ Fast guys ✅✅

Share FAstt .....✅✅ In your friends and big group se

Infosys exam answer will uploaded free Join and share group 👇👇👇 @codeing_area

Final..🔥 11am Infosys 1 slot available Contact @srksvk Note : paid help available 200% suree clearance guarantee🔥🔥 Priviously helped proof 👇👇👇 https://t.me/codeing_area/7369?single

Final..🔥 11am Infosys 1 slot available Contact @srksvk 200% suree clearance guarantee

Encora exam successfully done by remote access ✅✅ All MCQ done with 100% correct answer ✅🔥 Contact for placement exam @srksv
Encora exam successfully done by remote access ✅✅ All MCQ done with 100% correct answer ✅🔥 Contact for placement exam @srksvk

Power Bi exam successfully done by remote access ✅✅ Mark 48/50 ..🔥🔥🔥🔥🔥96%/100✅🔥🔥 Contact for placement exam help avail
Power Bi exam successfully done by remote access ✅✅ Mark 48/50 ..🔥🔥🔥🔥🔥96%/100✅🔥🔥 Contact for placement exam help available @srksvk