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 22 558 名订阅者,在 教育 类别中位列第 8 854,并在 印度 地区排名第 19 507 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 22 558 名订阅者。
根据 14 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 -445,过去 24 小时变化为 -14,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 6.31%。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 1 423 次浏览,首日通常累积 282 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 15 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。
22 558
订阅者
-1424 小时
-947 天
-44530 天
帖子存档
22 548
🎯SAS Off Campus Drive 2024 – Associate Software Engineer – Freshers | Rs 4-8 LPA
Job Role : Associate Software Developer
Qualification : B.E/B/Tech/M.E. / M.Tech
Experience : Freshers
Job Location : Pune
Package : Rs 4-8 LPA
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Telegram:- @allcoding1
22 548
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Batch:- 2019, 2020, 2021, 2022, 2023 & 2024
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Telegram:- @allcoding1
22 548
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22 548
🎯Atos Is Hiring
Degree:- Any Graduate
Batch:- 2019, 2020, 2021, 2022, 2023 & 2024
Apply Now:- https://www.allcoding1.com/2024/02/atos-off-campus-drive-for-any-graduate.html
Telegram:- @allcoding1
22 548
def helper(reward):
nums = []
for i in reward:
heappush(nums, -i)
ans = -heappop(nums)
i = 1
while nums:
temp = -heappop(nums)
if temp - i <= 0: break
ans += temp - i
i += 1
return ans
@allcoding1
22 548
int
getMinlength(vector<int>a,int k)
{
int ans=1;
long long mul=1;
for(int i=0;i<a.size();i++){
if(mul*a[i]>k){
ans++;
mul=a[i];
}
else{
mul*=a[i];
}
}
return ans;
}
L&T
@allcoding1
Q) compressing array
22 548
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Telegram:- @allcoding1
22 548
public class Solution {
public static int longestArraySegment(int array_length, List<Integer> arr) {
int maxLength = 1;
int currentLength = 1;
for (int i = 1; i < array_length; i++) {
if (arr.get(i) > arr.get(i - 1)) {
currentLength++;
} else {
currentLength = 1;
}
maxLength = Math.max(maxLength, currentLength);
}
currentLength = 1;
for (int i = 1; i < array_length; i++) {
if (arr.get(i) < arr.get(i - 1)) {
currentLength++;
} else {
currentLength = 1;
}
maxLength = Math.max(maxLength, currentLength);
}
return maxLength;
}
Ever increasing and ever decreasing - L&T
Java
Telegram:- @allcoding1
22 548
int solve(vector<int>& arr) { int n = arr.size(); if (n == 0) return 0;
int maxLen = 1; int currLen = 1; bool f = true;
for (int i = 1; i < n; ++i) { if ((arr[i] > arr[i-1] && f) II (arr[i] < arr[i-1] && !f)) { ++currLen; } else {
maxLen = max(maxLen, currLen); currLen = 2; f = arr[i] > arr[i-1]; }
maxLen = max(maxLen, currLen); return maxLen; }
C++
Telegram:- @allcoding1
L&T
22 548
int solve(vector<int>& arr) { int narr.size(); if (n == 0) return 0;
int maxLen = 1;
int currLen = 1;
bool f = true;
for (int i = 1; i < n; ++i) { if ((arr[i] > arr[i-1] && f) || (arr[i] < arr[i-1] && !f)) { ++currLen; } else { maxLen = max(maxLen, currLen); currLen = 2;
f = arr[i] > arr[i-1]
}
}
maxLen = max(maxLen, currLen); return maxLen;
}
C++
Telegram:- @allcoding1
L&t
Q) Ever - increasing & ever decreasing array segments
22 548
int solve(vector<int>& A) { int s = 0; for (const auto&n: A) { string sn = to_string(n); char md = *max_element(sn.begin().
sn.end());
s += md-'0';
}
return s;
}
C++
Telegram:- @allcoding1
L&t
Q) Encryption by digits
22 548
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Experience: Fresher
Qualification: B.Com. 2023 passed out
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22 548
22 548
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Job Position:- IT Software Developer (UX/UI/Fullstack)
Job ID: 3058496
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Job Location:- Hyderabad, India
Department: Software Engineering
Salary Package:- As per Company Standards
Job Type:- Full-time
Last Date:- ASAP
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Telegram:- @allcoding1_official
22 548
def check_first_and_last_char(Str):
if Str[0].isdigit() and not Str[-1].isdigit():
print("Yes")
elif Str[-1].isdigit() and not Str[0].isdigit():
print("No")
else:
print("Invalid input")
Str = input()
check_first_and_last_char(Str)
@allcoding1
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