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allcoding1_official

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📈 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

📊 受众指标与增长动态

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
-1 55630
帖子存档
Another divisor problem Java 8 Telegram:-@allcoding1
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Another divisor problem Java 8 Telegram:-@allcoding1

Gcd code in java 8 Telegram:-@allcoding1
Gcd code in java 8 Telegram:-@allcoding1

Perfect Subsequence Code Python Telegram:-@allcoding1
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Perfect Subsequence Code Python Telegram:-@allcoding1

Emil's Function Java 8 All Test Cases Passed Telegram:-@allcoding1
Emil's Function Java 8 All Test Cases Passed Telegram:-@allcoding1

def isPalindrome(Str): Len = len(Str) if (Len == 1): return True ptr1 = 0 ptr2 = Len - 1 while (ptr2 > ptr1): if (Str[ptr1] != Str[ptr2]): return False ptr1 += 1 ptr2 -= 1 return True def noOfAppends(s): if (isPalindrome(s)): return 0 del s[0] return 1 + noOfAppends(s) //allcoding1 se = "abede" s = [i for i in se] print(noOfAppends(s)) Make palindrome code in python Telegram:-@allcoding1

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import java.util.Arrays; public class GFG { public static void main(String[] args) { int max = 1000000; int[] facs = new int[max]; for (int i = 2; i < max; ++i) { for (int j = i; j < max; j += i) { facs[j]++; } } // System.out.println(Arrays.toString(facs)); int count = 0; for (int i = 0; i < max; ++i) if (facs[i] == 7)// 1 is a factor of all number so check for count 7 { if (primeFactors(i) == 2 + 1) { System.out.println("YESSSSSSS"); break; } } System.out.println(count); } //allcoding1 static int primeFactors(int n) { // Print the number of 2s that divide n while (n % 2 == 0) { n /= 2; } // n must be odd at this point. So we can // skip one element (Note i = i +2) for (int i = 3; i <= Math.sqrt(n); i += 2) { // While i divides n, print i and divide n while (n % i == 0) { n /= i; } } return n; } }

INFOSYS, TATA STEEL EXAM ANSWERS (24/4/22) All Slot INFOSYS:- http://www.joboffersadda.com/2022/01/infosys-exam-answer.html TATA STEEL:- http://www.joboffersadda.com/2022/04/tata-steel-exam-ans.html Telegram - @allcoding1 🔔Unmute this channel to never miss any updates

Python A lot of sum Code Telegram - @allcoding1
Python A lot of sum Code Telegram - @allcoding1

Maximise multiplication code Python 3 Telegram:-@allcoding1
Maximise multiplication code Python 3 Telegram:-@allcoding1

Positive sum segment code in python3 Telegram:-@allcoding1
Positive sum segment code in python3 Telegram:-@allcoding1

Maximise multiplication code Python 3 telegram:-@allcoding1
Maximise multiplication code Python 3 telegram:-@allcoding1

INFOSYS, TATA STEEL EXAM ANSWERS (24/4/22) All Slot INFOSYS:- http://www.joboffersadda.com/2022/01/infosys-exam-answer.html TATA STEEL:- http://www.joboffersadda.com/2022/04/tata-steel-exam-ans.html Telegram - @allcoding1 🔔Unmute this channel to never miss any updates

// C++ program to minimize subtree sum // difference by one edge deletion #include <bits/stdc++.h> using namespace std; /* DFS method to traverse through edges, calculating subtree sum at each node and updating the difference between subtrees */ void dfs(int u, int parent, int totalSum, vector<int> edge[], int subtree[], int& res) { int sum = subtree[u]; /* loop for all neighbors except parent and aggregate sum over all subtrees */ for (int i = 0; i < edge[u].size(); i++) { int v = edge[u][i]; if (v != parent) { dfs(v, u, totalSum, edge, subtree, res); sum += subtree[v]; } } // store sum in current node's subtree index subtree[u] = sum; /* at one side subtree sum is 'sum' and other side subtree sum is 'totalSum - sum' so their difference will be totalSum - 2*sum, by which we'll update res */ if (u != 0 && abs(totalSum - 2*sum) < res) res = abs(totalSum - 2*sum); } // Method returns minimum subtree sum difference int getMinSubtreeSumDifference(int vertex[], int edges[][2], int N) { int totalSum = 0; int subtree[N]; // Calculating total sum of tree and initializing // subtree sum's by vertex values for (int i = 0; i < N; i++) { subtree[i] = vertex[i]; totalSum += vertex[i]; } // filling edge data structure vector<int> edge[N]; for (int i = 0; i < N - 1; i++) { edge[edges[i][0]].push_back(edges[i][1]); edge[edges[i][1]].push_back(edges[i][0]); } int res = INT_MAX; // calling DFS method at node 0, with parent as -1 dfs(0, -1, totalSum, edge, subtree, res); return res; } // Driver code to test above methods int main() { int vertex[] = {4, 2, 1, 6, 3, 5, 2}; int edges[][2] = {{0, 1}, {0, 2}, {0, 3}, {2, 4}, {2, 5}, Telegram:@allcofing1

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#include<bits/stdc++.h> using namespace std; // Function to check whether given sequence is // Jolly Jumper or not bool isJolly(int a[], int n) { // Boolean vector to diffSet set of differences. // The vector is initialized as false. vector<bool> diffSet(n, false); // Traverse all array elements for (int i=0; i < n-1 ; i++) { // Find absolute difference between current two int d = abs(a[i]-a[i+1]); // If difference is out of range or repeated, // return false. if (d == 0 d > n-1 diffSet[d] == true) return false; // Set presence of d in set. diffSet[d] = true; } return true; } // Driver Code int main() { int a[] = {11, 7, 4, 2, 1, 6}; int n = sizeof(a)/ sizeof(a[0]); isJolly(a, n)? cout << "Yes" : cout << "No"; return 0; } C++ Jolly Jumper Sequence Code Telegram - @allcoding1

#include<bits/stdc++.h> using namespace std; // Function to check whether given sequence is // Jolly Jumper or not bool isJolly(int a[], int n) { // Boolean vector to diffSet set of differences. // The vector is initialized as false. vector<bool> diffSet(n, false); // Traverse all array elements for (int i=0; i < n-1 ; i++) { // Find absolute difference between current two int d = abs(a[i]-a[i+1]); // If difference is out of range or repeated, // return false. if (d == 0 d > n-1 diffSet[d] == true) return false; // Set presence of d in set. diffSet[d] = true; } return true; } // Driver Code int main() { int a[] = {11, 7, 4, 2, 1, 6}; int n = sizeof(a)/ sizeof(a[0]); isJolly(a, n)? cout << "Yes" : cout << "No"; return 0; } C++ Jolly num code

Python Special number after aurgument Code Telegram - @allcoding1
Python Special number after aurgument Code Telegram - @allcoding1