allcoding1_official
前往频道在 Telegram
📈 Telegram 频道 allcoding1_official 的分析概览
频道 allcoding1_official (@allcoding1_official) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 78 703 名订阅者,在 技术与应用 类别中位列第 1 622,并在 印度 地区排名第 3 757 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 78 703 名订阅者。
根据 30 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 -1 600,过去 24 小时变化为 -27,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 1.38%。内容发布后 24 小时内通常能获得 0.62% 的反应,占订阅者总量。
- 帖子覆盖: 每篇帖子平均可获得 1 091 次浏览,首日通常累积 486 次浏览。
- 互动与反馈: 受众积极参与,单帖平均反应数为 0。
- 主题关注点: 内容集中在 dsa, stack, namaste, javascript, dev 等核心主题上。
📝 描述与内容策略
尚未提供频道描述。
凭借高频更新(最新数据采集于 31 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 技术与应用 类别中的关键影响点。
78 703
订阅者
-2724 小时
-3217 天
-1 60030 天
帖子存档
78 690
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🎯L&T Technology Mega off Campus Hiring For Multiple Roles | Rs 4-6 LPA
Job Title Multiple Roles
Qualification B.E/B.Tech
Batch 2023
Salary 4-6 LPA
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Telegram:- @allcoding1_official
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🎯L&T Technology Mega off Campus Hiring For Multiple Roles | Rs 4-6 LPA
Job Title Multiple Roles
Qualification B.E/B.Tech
Batch 2023
Salary 4-6 LPA
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Work Experience : 0 – 6 Years
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Telegram:- @allcoding1_official
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🎯Tech Mahindra off Campus Hiring As Associate Software Engineer | Rs.3.25 LPA & 5.5LPA
Designation : Associate Software Engineer
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Salary CTC : Rs.3.25 LPA & 5.5LPA (for super coders)
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Company Name: Zoho
Role: Software Developer
Batch eligible: 2022, 2023 and 2024 passouts
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🎯Harmonic Off Campus Drive 2024 for Associate Software Engineer | 4 LPA
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💻IT company's
👉Previous exams question papers and answers
👉Not books
👉courses and certificates
💰300
Contact:- @meterials_available
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Meesho extended there exam date
And calling students to virtual interview based on there exam dates
Company name : MEESHO Role: BUSINESS ANALYST APPLY LINK:- https://docs .google.com/forms/d/e /1FAIpQLSc3DvqJDjZeYjFZX2 OgVPwm5shjCX4a0APzYBa TBoPzTf4khg/viewform?pli= 1
While applying make sure in your CV mention python and sql.
Telegram:-
78 690
def PowerLimit(a, b):
current_resource, max_seconds = 1, 0
while a > 0 and b > 0:
if a > b:
current_resource, a, b = 1, a - 2, b + 1
else:
current_resource, b, a = 2, b - 2, a + 1
max_seconds += 1
max_seconds += a // 2 if current_resource == 1 else b // 2
return max_seconds
power limit ✅
