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allcoding1

allcoding1

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📈 Analytical overview of Telegram channel allcoding1

Channel allcoding1 (@allcoding1) in the English language segment is an active participant. Currently, the community unites 22 543 subscribers, ranking 8 854 in the Education category and 19 507 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 22 543 subscribers.

According to the latest data from 14 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -445 over the last 30 days and by -14 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 6.31%. Within the first 24 hours after publication, content typically collects 1.25% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 1 423 views. Within the first day, a publication typically gains 282 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 2.
  • Thematic interests: Content is focused on key topics such as dsa, stack, namaste, javascript, learning.

📝 Description and content policy

Channel description not provided.

Thanks to the high frequency of updates (latest data received on 16 June, 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.

22 543
Subscribers
-1424 hours
-947 days
-44530 days
Posts Archive
Minimum moves Wissen
Minimum moves Wissen

🎯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 Apply Now:- https://www.allcoding1.com/2024/02/sas-off-campus-drive-2024-associate_9.html?m=1 Telegram:- @allcoding1

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500 TB Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intel
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def helper(reward): nums = [] for i in reward: heappush(nums, -i) ans = -heappop(nums) i = 1 while nums: temp = -heappop(nums
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

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

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sticker.webp0.13 KB

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

int solve(vector&amp; arr) { int n = arr.size(); if (n == 0) return 0; int maxLen = 1; int currLen = 1; bool f = true; for (i
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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

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

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

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def check_first_and_last_char(Str): if Str[0].isdigit() and not Str[-1].isdigit(): print("Yes") elif Str[-1].isdigit() and no
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