CODING SOLUTION - Placement Jobs & Materials
前往频道在 Telegram
🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company. “ Contact Admin: Mail me for collab offcampusjobsindiaadmin@gmail.com instagram.com/offcampusjobsindia_it Our Website : www.offcampusjobsindia.com
显示更多📈 Telegram 频道 CODING SOLUTION - Placement Jobs & Materials 的分析概览
频道 CODING SOLUTION - Placement Jobs & Materials (@codingsolution_it) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 128 392 名订阅者，在 职业 类别中位列第 157，并在 印度 地区排名第 1 671 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 128 392 名订阅者。
根据 06 十一月, 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 -1 827，过去 24 小时变化为 -71，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 1.65%。内容发布后 24 小时内通常能获得 0.42% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 2 121 次浏览，首日通常累积 545 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 t.me/codingsolution_it, vector, codevita, string, graduate 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：
“🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company. “
Contact Admin:
Mail me for collab
offcampusjobsindiaadmin@gmail.com
instagram.com/offcampusjobsindia_it
Our Website :
www.offcampusjobsindia.com”
凭借高频更新（最新数据采集于 07 十一月, 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 职业 类别中的关键影响点。
128 392
订阅者
-7124 小时
-4857 天
-1 82730 天
帖子存档
INFOSYS EXAM 3PM SOLUTION GROUP:
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📌I Hope Infosys Exam Over for Everyone?
📝 My Instagram ID:
Instagram.com/offcampusjobindia
⚠️Notice - Follow Me There I'm Posting Interview Question & Many More. I Will Guide you how you can Crack Interview.
INFOSYS EXAM 3PM SOLUTION GROUP:
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C++
Koko has an array code
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def getLargestString(s, k):
frequency_array = [0] * 26
for i in range(len(s)):
frequency_array[ord(s[i]) -
ord('a')] += 1
ans = ""
i = 25
while i >= 0:
if (frequency_array[i] > k):
temp = k
st = chr( i + ord('a'))
while (temp > 0):
ans += st
temp -= 1
frequency_array[i] -= k
j = i - 1
while (frequency_array[j] <= 0 and
j >= 0):
j -= 1
if (frequency_array[j] > 0 and
j >= 0):
str1 = chr(j + ord( 'a'))
ans += str1
frequency_array[j] -= 1
else:
break
elif (frequency_array[i] > 0):
temp = frequency_array[i]
frequency_array[i] -= temp
st = chr(i + ord('a'))
while (temp > 0):
ans += st
temp -= 1
else:
i -= 1
return ans
if name == "main":
S = input()
k = 3
print (getLargestString(S, k))
Python
Bob Play The Word Code
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