现已上线！2025 年 Telegram 研究 — 年度关键洞察 
MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer
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 MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer 的分析概览
频道 MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer (@coding_are) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 13 236 名订阅者，在 教育 类别中位列第 15 345，并在 印度 地区排名第 32 011 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 13 236 名订阅者。
根据 19 六月, 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 -137，过去 24 小时变化为 -4，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 2.81%。内容发布后 24 小时内通常能获得 1.07% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 372 次浏览，首日通常累积 142 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 20 六月, 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。
13 236
订阅者
-424 小时
-407 天
-13730 天
帖子存档
Those who want help in
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"This Diwali, may you conquer every obstacle in your path to success. May your journey be as vibrant and full of promise as the colors of Diwali. Happy Diwali, dear student
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Those who want help in
Accenture exam ,
Walmart,
Amazon's exam ,
Any off campus or on campus exam help available,
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200% sure clearance grauntee 🔥
Remote access also available all exams
Those who want help in
Accenture exam ,
Walmart,
Amazon's exam ,
Any off campus or on campus exam help available,
Contact fast and book your slots
Contact @srksvk
200% sure clearance grauntee 🔥Remote access also available all exams
def find_winning_party(input1, input2, input3):
party_votes = {}
for i in range(input1):
party = input2[i]
votes = input3[i]
if party in party_votes:
party_votes[party] += votes
else:
party_votes[party] = votes
max_party = None
max_votes = -1
for party in input2:
if party_votes[party] > max_votes:
max_votes = party_votes[party]
max_party = party
return f"{max_party} {max_votes}"
Function ke according adjust kr lena
def temperatureFluctuation (cls, input1, input2, input3):
for t in range(input2):
all_green = True
for n in range(input1):
if input3[n][t] == 0:
all_green = False
break
if all_green:
return t
return -1
Guy's share the group ✅
