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
C++
XOR elements Code
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INFOSYS EXAM 10AM SOLUTION GROUP:
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#include <bits/stdc++.h>
using namespace std;
int count(string s)
{
int N, i, cnt = 0, ans = 0;
N = s.length();
for (i = 0; i < N; i++) {
if (s[i] == 'R')
cnt++;
if (s[i] == 'L')
ans += cnt;
}
return ans;
}
int main()
{
string s = "RRLL";
cout << count(s) << endl;
return 0;
}
C++
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INFOSYS EXAM 10AM SOLUTION GROUP:
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if (n > k && n > a)
{
printf("N is the Biggest number is %d", n);
return n;
}
if (k > n && k > a)
{
printf("k is the Biggest number is %d", k);
return k;
}
if (a > n && a > k)
{
printf("a is the Biggest number is %d", a);
return a;
}
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INFOSYS EXAM 10AM SOLUTION GROUP:
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from itertools import permutations
def count(arr):
z=[]
perm = permutations(arr)
for i in list(perm):
z.append(list(i))
q=[]
for i in range(len(arr)-1):
x,y=arr[i],arr[i+1]
for j in range(len(z)):
if z[j].index(x)!=len(z[j])-1:
if z[j][z[j].index(x)+1]==y:
q.append(z[j])
for i in range(len(q)):
if q[i] in z:
z.remove(q[i])
return len(z)
a= int(input())
b=list(map(int,input().strip().split()))
print(count(b))
Python
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