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 84 584 名订阅者,在 技术与应用 类别中位列第 1 497,并在 印度 地区排名第 3 527 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 84 584 名订阅者。
根据 10 七月, 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 -1 556,过去 24 小时变化为 -30,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 2.01%。内容发布后 24 小时内通常能获得 0.85% 的反应,占订阅者总量。
- 帖子覆盖: 每篇帖子平均可获得 1 701 次浏览,首日通常累积 723 次浏览。
- 互动与反馈: 受众积极参与,单帖平均反应数为 1。
- 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 11 七月, 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。
84 584
订阅者
-3024 小时
-4257 天
-1 55630 天
帖子存档
84 569
Accenture Velotio & Johnson Off Campus Drive | 4LPA+
Batch - 2019 2020 2021 2022
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Telegram - @allcoding1
84 569
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🔔Jobs notifications
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TELEGRAM :-@allcoding1
84 569
#include <bits/stdc++.h>
using namespace std;
int vowel(string s){
int n = s.length();
vector<int> arr;
for (int i = 0; i < n; i++) {
if (i == 0){
arr.push_back(n);
}
else{
arr.push_back((n - i) + arr[i - 1] - i);
}
}
int sum = 0;
for (int i = 0; i < n; i++)
if (s[i] == 'a' s[i] == 'e' s[i] == 'i' s[i] == 'o' s[i] == 'u'){
sum += arr[i];
}
return sum;
}
int main(){
string s = "abc";
cout << vowel(s) << endl;
}
EPAM
VOWEL Code
C++
TELEGRAM :-@allcoding1
84 569
Repost from allcoding1_official
#include <bits/stdc++.h>
using namespace std;
int vowel(string s){
int n = s.length();
vector<int> arr;
for (int i = 0; i < n; i++) {
if (i == 0){
arr.push_back(n);
}
else{
arr.push_back((n - i) + arr[i - 1] - i);
}
}
int sum = 0;
for (int i = 0; i < n; i++)
if (s[i] == 'a' s[i] == 'e' s[i] == 'i' s[i] == 'o' s[i] == 'u'){
sum += arr[i];
}
return sum;
}
int main(){
string s = "abc";
cout << vowel(s) << endl;
}
EPAM CODE
Telegram👇
@allcoding1 @allcoding1
@allcoding1 @allcoding1
@allcoding1 @allcoding1
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84 569
Repost from allcoding1_official
from bisect import bisect_left, insort_left
n, d = map(int, input().split())
t = list(map(int, input().split()))
noti = 0
lastd = sorted(t[:d])
def med():
return lastd[d//2] if d % 2 == 1 else ((lastd[d//2] + lastd[d//2-1])/2)
for i in range(d, n):
if t[i] >= 2*med():
noti += 1
del lastd[bisect_left(lastd,t[i-d])]
insort_left(lastd, t[i])
print(noti)
EPAM Bank Notification Code
Telegram👇
@allcoding1 @allcoding1
@allcoding1 @allcoding1
@allcoding1 @allcoding1
⚠️ Share post in ur college and telegram Group's
84 569
Repost from allcoding1_official
#include <bits/stdc++.h>
using namespace std;
int vowel(string s){
int n = s.length();
vector<int> arr;
for (int i = 0; i < n; i++) {
if (i == 0){
arr.push_back(n);
}
else{
arr.push_back((n - i) + arr[i - 1] - i);
}
}
int sum = 0;
for (int i = 0; i < n; i++)
if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i'|| s[i] == 'o' || s[i] == 'u'){
sum += arr[i];
}
return sum;
}
int main(){
string s = "abc";
cout << vowel(s) << endl;
}
EPAM
Telegram - https://t.me/allcoding1
