现已上线！2025 年 Telegram 研究 — 年度关键洞察 
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 22 394 名订阅者，在 教育 类别中位列第 8 823，并在 印度 地区排名第 18 959 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 22 394 名订阅者。
根据 27 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 -383，过去 24 小时变化为 -21，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 4.45%。内容发布后 24 小时内通常能获得 1.42% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 997 次浏览，首日通常累积 318 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 29 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。
22 397
订阅者
-2124 小时
-927 天
-38330 天
帖子存档
22 394
import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Enter a string: ");
String str = sc.next();
HashMap<String, Integer> maps = new HashMap<String, Integer>();
String[] cha = new String[str.length()];
cha = str.split("");
for(int i=0; i<cha.length; i++)
{
int count = 0;
if(maps.containsKey(" "+cha[i]))
{
maps.put(" "+cha[i], 1+maps.get(" "+cha[i]));
}
else if(maps.containsKey(cha[i]))
{
if(!(cha[i].equals(cha[i-1])) && i > 0)
{
maps.put(" "+cha[i], 1);
}else{
count = 1 + maps.get(cha[i]);
maps.put(cha[i], count);
}
}else{
maps.put(cha[i], 1);
}
}
String updated = "";
for(String key: maps.keySet())
{
System.out.print(key+""+maps.get(key));
updated = updated+""+key+""+maps.get(key);
}
System.out.println();
if(str.length() > updated.length())
{
System.out.println("String Accepted");
System.out.println("Inputted String: "+str);
System.out.println("Compressed String: "+updated);
}else{
System.out.println("String Rejected");
System.out.println("Inputted String: "+str);
System.out.println("Compressed String: "+updated);
}
}
}
String Compression Code in Java
Telegram:- @allcoding1
22 394
import java.io.*;
import java.util.*;
class Main {
static void desc(int[] arr, int num, int total)
{
int temp =0 ;
for(int i=0;i<total; i++)
{
for(int j=0;j<total;j++)
{
if(arr[j] < arr[i])
{
temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
}
topScorer(arr, num);
}
static void topScorer(int[] arr, int num)
{
int sum=0;
for(int i=0; i<num; i++)
{
sum = sum+arr[i];
}
System.out.println(sum);
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Enter total number of participants: ");
int total = sc.nextInt();
int[] arr= new int[total];
System.out.print("Enter number of top scorer: ");
int num = sc.nextInt();
System.out.print("Enter "+total+" elements: ");
for(int i=0;i<total; i++)
{
arr[i] = sc.nextInt();
}
desc(arr, num, total);
}
}
Marathon code
Java
Telegram:-@allcoding1
22 394
Pattern code in java
import java.io.*;
import java.util.*;
class Main {
static void myCode(String str)
{
int num = Integer.parseInt(str);
for(int i=1; i<=num; i++)
{
for(int j=num-1; j>=i; j--)
{
System.out.print(" ");
}
for(int k=1; k<=2*i-1; k++)
{
System.out.print("*");
}
for(int l=num-1; l>=i; l--)
{
System.out.print(" ");
}
System.out.println();
}
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = sc.next();
myCode(str);
}
}
Telegram:-@allcoding1
22 394
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✅Qualification : B.E/B.Tech/✅Any Degree
Batch : 2019, 2020, 2021, or 2022
Telegram:-@allcoding1
22 394
🎯 Hexaware Technologies Off Campus Drive 2022 : Hiring for Freshers as Trainee – IMS With 3.15 LPA
Designation : Trainee – IMS
Job Location : Across India
Experience : Freshers
Salary : Rs. 3.15 LPA
Apply Now:- http://www.allcoding1.com/2022/07/hexaware-off-campus-drive.html
Telegram:-@allcoding1
22 394
Guys ❤️
Easy to discuss your Friends
Hindi students:- @Hindisoftwarejobs
Telugu students:- @TeluguSoftwarejobs
Tamil students:- @Tamilsoftwarejobs
Kannada students:- @kannadasowfarejobs
Telegram:-@allcoding1
22 394
// 1074. Number of Submatrices That Sum to Target
class Solution
{
public:
int numSubmatrixSumTarget(vector<vector<int>> &matrix, int target)
{
int m = matrix.size();
int n = matrix[0].size();
for (int i = 0; i < m; i++)
{
for (int j = 1; j < n; j++)
{
matrix[i][j] += matrix[i][j - 1];
}
}
int count = 0;
for (int index = 0; index < n; index++)
{
for (int j = index; j < n; j++)
{
unordered_map<int, int> mp{{0, 1}};
int sum = 0;
for (int i{}; i < m; i++)
{
if (index > 0)
sum -= matrix[i][index - 1];
sum += matrix[i][j];
if (mp.count(sum - target))
count += mp[sum - target];
mp[sum]++;
}
}
}
return count;
}
};
// C++ unordered_map TC O(MN + MN*MN) || Easy Solution
Telegram:-@allcoding1
22 394
Guys ❤️
Easy to discuss your Friends
Hindi students:- @Hindisoftwarejobs
Telugu students:- @TeluguSoftwarejobs
Tamil students:- @Tamilsoftwarejobs
Kannada students:- @kannadasowfarejobs
Telegram:-@allcoding_1
22 394
VIRTUSA OFF CAMPUS DRIVE
Position Title: Associate Engineer
CTC: 4,00,00 Per Annum
Batch: 2018/19/20/21
Apply Now:- http://www.allcoding1.com/2022/07/virtusa-off-campus-drive.html
Telegram:-@allcoding1
22 394
leecode A
class Solution
{
public:
vector<int> numberOfPairs(vector<int> &nums)
{
int p = 0, u = 0;
vector<int> vec;
unordered_map<int, int> mp;
for (auto x : nums)
mp[x]++;
for (auto x : mp)
{
p += x.second / 2;
u += x.second % 2;
}
vec.push_back(p);
vec.push_back(u);
return vec;
}
};
MAP easy Solution || TC O(N)
C++
Telegram:-@allcoding1
22 394
leetcode B
class Solution
{
public:
int sumofNumber(int n)
{
int sum = 0;
while (n)
{
sum += (n % 10);
n /= 10;
}
return sum;
}
int maximumSum(vector<int> &nums)
{
int ans = -1;
int n = nums.size();
sort(nums.begin(), nums.end());
map<int, vector<int>> m;
for (int i = 0; i < n; i++)
{
m[sumofNumber(nums[i])].push_back(nums[i]);
}
for (auto i : m)
{
if (i.second.size() > 1)
{
int k = i.second.size();
ans = max(ans, i.second[k - 1] + i.second[k - 2]);
}
}
return ans;
}
};
Easy Solution by sum of number ,sorting
C++
Telegram:-@allcoding1
