allcoding1
前往频道在 Telegram
📈 Telegram 频道 allcoding1 的分析概览
频道 allcoding1 (@allcoding1) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 22 575 名订阅者,在 教育 类别中位列第 8 822,并在 印度 地区排名第 19 518 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 22 575 名订阅者。
根据 12 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 -437,过去 24 小时变化为 -6,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 5.99%。内容发布后 24 小时内通常能获得 1.25% 的反应,占订阅者总量。
- 帖子覆盖: 每篇帖子平均可获得 1 353 次浏览,首日通常累积 283 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 dsa, stack, namaste, javascript, learning 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 13 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。
22 575
订阅者
-624 小时
-967 天
-43730 天
帖子存档
22 575
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22 575
22 575
Telegram:- @allcoding1
Juspay
=int(input())
dic={}
def addnode(u,v):
dic[u].append(v)
for i in range(N):
n1=int(input())
dic[n1]=[]
edge=int(input())
for i in range(edge):
u,v=map(int,input().split())
addnode(u,v)
start=int(input())
end=int(input())
tr=[]
tr.append(start)
def traverse(dic,start):
for i in dic[start]:
if(i==end):
return
if(len(dic[start])==0):
return
else:
tr.append(i)
traverse(dic,i)
traverse(dic,start)
tr=sorted(tr)
for i in tr:
print(i,end=" ")
22 575
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