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allcoding1

allcoding1

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šŸ“ˆ Analytical overview of Telegram channel allcoding1

Channel allcoding1 (@allcoding1) in the English language segment is an active participant. Currently, the community unites 21 564 subscribers, ranking 9 073 in the Education category and 19 010 in the India region.

šŸ“Š Audience metrics and dynamics

Since its creation on невіГомо, the project has demonstrated rapid growth, gathering an audience of 21 564 subscribers.

According to the latest data from 30 August, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -372 over the last 30 days and by -25 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 6.68%. Within the first 24 hours after publication, content typically collects N/A% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 1 441 views. Within the first day, a publication typically gains 0 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 0.
  • Thematic interests: Content is focused on key topics such as dsa, stack, namaste, javascript, learning.

šŸ“ Description and content policy

Channel description not provided.

Thanks to the high frequency of updates (latest data received on 31 August, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.

21 564
Subscribers
-2524 hours
-867 days
-37230 days
Posts Archive
#include <iostream> #include <string> #include <vector> using namespace std; int numDecodings(string msg) { Ā Ā Ā  int MOD = 1000000007; Ā Ā Ā  int n = msg.size(); Ā Ā Ā  Ā Ā Ā  vector<long long> dp(n + 1, 0); Ā Ā Ā  dp[0] = 1; Ā Ā Ā  Ā Ā Ā  if (msg[0] == '0') Ā Ā Ā Ā Ā Ā Ā  dp[1] = 0; Ā Ā Ā  else if (msg[0] == '*') Ā Ā Ā Ā Ā Ā Ā  dp[1] = 9; Ā Ā Ā  else Ā Ā Ā Ā Ā Ā Ā  dp[1] = 1; Ā Ā Ā  Ā Ā Ā  for (int i = 2; i <= n; ++i) { Ā Ā Ā Ā Ā Ā Ā  if (msg[i - 1] == '0') { Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  if (msg[i - 2] == '1' || msg[i - 2] == '2') Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  dp[i] += dp[i - 2]; Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  else if (msg[i - 2] == '*') Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  dp[i] += 2 * dp[i - 2]; Ā Ā Ā Ā Ā Ā Ā  } else if (msg[i - 1] >= '1' && msg[i - 1] <= '9') { Ā Ā Ā Ā Ā  Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  dp[i] += dp[i - 1]; Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  if (msg[i - 2] == '1' || (msg[i - 2] == '2' && msg[i - 1] <= '6')) Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  dp[i] += dp[i - 2]; Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  else if (msg[i - 2] == '*') { Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  if (msg[i - 1] <= '6') Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  dp[i] += 2 * dp[i - 2]; Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  else Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  dp[i] += dp[i - 2]; Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  } Ā Ā Ā Ā Ā Ā Ā  } else if (msg[i - 1] == '*') { Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  dp[i] += 9 * dp[i - 1]; Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  if (msg[i - 2] == '1') Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  dp[i] += 9 * dp[i - 2]; Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  else if (msg[i - 2] == '2') Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  dp[i] += 6 * dp[i - 2]; Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  else if (msg[i - 2] == '*') Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā  dp[i] += 15 * dp[i - 2]; Ā Ā Ā Ā Ā Ā Ā  } Ā Ā Ā Ā Ā Ā Ā  Ā Ā Ā Ā Ā Ā Ā  dp[i] %= MOD; Ā Ā Ā  } Ā Ā Ā  Ā Ā Ā  return dp[n]; } Number of ways decode

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import re a = int(input().strip()) b = set() c = r'\b[A-Za-z0-9._%+-]+@[A-Za-z0-9.-]+\.[A-Z|a-z]{2,}\b' for _ in range(a): li
import re a = int(input().strip()) b = set() c = r'\b[A-Za-z0-9._%+-]+@[A-Za-z0-9.-]+\.[A-Z|a-z]{2,}\b' for _ in range(a): Ā Ā Ā  line = input().strip() Ā Ā Ā  b.update(re.findall(c, line)) print(';'.join(sorted(b)))

int solve(int k, vector&amp; s) { &nbsp;&nbsp;&nbsp; sort(s.rbegin(), s.rend()); &nbsp;&nbsp;&nbsp; int c = 0; &nbsp;&nbsp;&n
int solve(int k, vector<int>& s) {     sort(s.rbegin(), s.rend());     int c = 0;     for (int i = 0; i < s.size(); i++) {         if (i < k && s[i] > 0) {             c++;         } else if (s[i] == s[i - 1] && s[i] > 0) {             c++;         } else {             break;         }     }     return c; } Competitive Gaming Accenture exam Telegram:- @allcoding1

#include <iostream> #include <vector> #include <algorithm> using namespace std; int findLIS(vector<int>& s) {     vector<int> tails;     for (int x : s) {         auto it = lower_bound(tails.begin(), tails.end(), x);         if (it == tails.end()) {             tails.push_back(x);         } else {             *it = x;         }     }     return tails.size(); }   Swiggy LIS

#include <bits/stdc++.h> using namespace std; int jumps(int flagHeight, int bigJump) { Ā Ā Ā  return flagHeight / bigJump+flagheight% bigJump; }.Ā  Swiggy Jump to The Flag

long getMaxPrisonHole(int n, int m, vector x, vector y) {     vector xb(n+1, true);     vector yb(m+1, true);         for(int i : x) {         xb[i] = false;     }         for(int i : y) {         yb[i] = false;     }         long cx = 0, xm = LONG_MIN, cy = 0, ym = LONG_MIN;         for(int i = 0; i < xb.size(); i++) {         if(xb[i]) {             cx = 0;         } else {             cx++;             xm = max(cx, xm);         }     }         for(int i = 0; i < yb.size(); i++) {         if(yb[i]) {             cy = 0;         } else {             cy++;             ym = max(cy, ym);         }     }         return (xm+1) * (ym+1); }  Swiggy Prison Break

public static int selectStock(int saving, int[] currentValue, int[] futureValue) {         int n = currentValue.length;         int[][] dp = new int[n + 1][saving + 1];         for (int i = 1; i <= n; i++) {             for (int j = 0; j <= saving; j++) {                 dp[i][j] = dp[i - 1][j];                 if (j >= currentValue[i - 1]) {                     dp[i][j] = Math.max(dp[i][j], dp[i - 1][j - currentValue[i - 1]] + futureValue[i - 1] - currentValue[i - 1]);                 }             }         }         return dp[n][saving];     }.  Swiggy Selecting  Stocks

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šŸ“ŒIT learning courses šŸ“ŒAll programing courses šŸ“ŒAbdul bari courses šŸ“ŒAshok IT Tutorials + Books + Courses + Trainings + Workshops + Educational Resources šŸ”¹Data science šŸ”¹Python šŸ”¹Artificial Intelligence šŸ”¹AWS Certified šŸ”¹Cloud šŸ”¹BIG DATA šŸ”¹Data Analytics šŸ”¹BI šŸ”¹Google Cloud Platform šŸ”¹IT Training šŸ”¹MBA šŸ”¹Machine Learning šŸ”¹Deep Learning šŸ”¹Ethical Hacking šŸ”¹SPSS šŸ”¹Statistics šŸ”¹Data Base šŸ”¹Learning language resources English , šŸ‡«šŸ‡· All courses (100 rupees) Contact:- @meterials_available

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Advanced SubArray Problem
Advanced SubArray Problem

Elections code
Elections code

int winning_party(int voters, int votes[]) { &nbsp;&nbsp;&nbsp; unordered_map vote_counts; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nb
int winning_party(int voters, int votes[]) {     unordered_map vote_counts;         for (int i = 0; i < voters; ++i) {         vote_counts[votes[i]]++;     }         for (auto it = vote_counts.begin(); it != vote_counts.end(); ++it) {         if (it->second > voters / 2) {             return it->first;         }     }         return -1; } Elections

def is_prime(n): if n <= 1: return False if n <= 3: return True if n % 2 == 0 or n % 3 == 0: return False i = 5 while i * i <= n: if n % i == 0 or n % (i + 2) == 0: return False i += 6 return True def next_prime(N): if N <= 1: return 2 prime = N + 1 while True: if is_prime(prime): return prime prime += 1 # Test the function N = 4 print(next_prime(N)) # Output: 5

encoded_value = "" for digit in str(input1): encoded_value += str(int(digit) ** 2) return int(encoded_value) Minimum array sum