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allcoding1_official

Open in Telegram

India3 842 Technologies & Applications1 602...

📈 Analytical overview of Telegram channel allcoding1_official

Channel allcoding1_official (@allcoding1_official) in the English language segment is an active participant. Currently, the community unites 79 172 subscribers, ranking 1 602 in the Technologies & Applications category and 3 842 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 79 172 subscribers.

According to the latest data from 22 October, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -1 680 over the last 30 days and by -35 over the last 24 hours, overall reach remains high.

Verification status: Not verified
Engagement rate (ER): The average audience engagement rate is 1.46%. Within the first 24 hours after publication, content typically collects 0.77% reactions from the total number of subscribers.
Post reach: On average, each post receives 1 153 views. Within the first day, a publication typically gains 608 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, dev.

📝 Description and content policy

Channel description not provided.

Thanks to the high frequency of updates (latest data received on 23 October, 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 Technologies & Applications category.

79 172

Subscribers

-3524 hours

-3517 days

-1 68030 days

1 153

Post views

~ 60824 hours

~ 78848 hours

1.46%

Engagement rate

No data

Posts per day

Ads index

beta

Posts Archive

79 172

Wipro

79 172

5.B 6.D 7.C

79 172

n = int(input()) s = input() consonants = 0 for i in range(n-1,-1,-1): if s[i] not in 'aeiouAEIOU': consonants += 1 if consonants == 3: print(s[i]) break

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#include<stdio.h> int main() { int n; scanf("%d", &n); char str[n + 1]; scanf("%s", str); int count = 0; char thirdLastConsonant; for(int i = n - 1; i >= 0; i--) { if (str[i] != 'a' && str[i] != 'e' && str[i] != 'i' && str[i] != 'o' && str[i] != 'u' && str[i] != 'A' && str[i] != 'E' && str[i] != 'I' && str[i] != 'O' && str[i] != 'U') { count++; if (count == 3) { thirdLastConsonant = str[i]; break; } } } printf("%c\n", thirdLastConsonant); return 0; }

79 172

import java.util.Scanner; public class ChangeCase { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); char c = scanner.next().charAt(0); if (Character.isLowerCase(c)) { System.out.println(Character.toUpperCase(c)); } else { System.out.println(Character.toLowerCase(c)); } scanner.close(); } }

79 172

Exam

79 172

Wipro

79 172

32.C 33.B

79 172

38.C 39.D 40.A

79 172

33.B 34.D 35.C

79 172

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