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📈 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 22 535 名订阅者,在 教育 类别中位列第 8 857,并在 印度 地区排名第 19 476

📊 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 22 535 名订阅者。

根据 15 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 -434,过去 24 小时变化为 -13,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 6.43%。内容发布后 24 小时内通常能获得 1.32% 的反应,占订阅者总量。
  • 帖子覆盖: 每篇帖子平均可获得 1 450 次浏览,首日通常累积 297 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 16 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。

22 535
订阅者
-1324 小时
-887
-43430
帖子存档
🅾️Free off cost 👉ALL IT companies (materials, exams questions & Answers paper) 👉ALL software courses with certification 🅾️Free off cost Join now 👇 @Allcodingsolution @Allcodingsolution @Allcodingsolution ✅Please send with your friends and college groups

string getLongestRegex(string a, string b, string c) {   const size_t n = a.size();   int idx = -1;   for (int
+1
string getLongestRegex(string a, string b, string c) {   const size_t n = a.size();   int idx = -1;   for (int i = 0; i < n; i++) {     if (c[i] != a[i] && c[i] != b[i]) { idx = i; }   }   if (idx == -1) return "-1";   string res;   for (int i = 0; i < n; i++) {         if (i == idx) {       string cur = "[";       for (int j = 'A'; j <= 'Z'; j++)  if (j != c[i]) cur += j;       cur += "]";       res += cur;     } else {       res += "[ABCDEFGHIJKLMNOPQRSTUVWXYZ]";     }   }   return res; } C++ Amazon

def minimumCost(price): n = len(price) price.append(0) ans = float('inf') last = {price[0]: 0} for i in range(1, n): v = pric
def minimumCost(price):     n = len(price)     price.append(0)     ans = float('inf')     last = {price[0]: 0}     for i in range(1, n):         v = price[i]         price[i] += price[i - 1]         if last.get(v) != None:             ans = min(ans, price[i] - price[last[v]] + v)         last[v] = i     return ans if ans != float('inf') else -1 n = int(input()) price = list(map(int,input().split(' '))) print(minimumCost(price)) Python 3 Amazon

#include <bits/stdc++.h> using namespace std; bool checkbit(int n,int i){     return n&(1<<i); } bool cmp(const pair<int,int> &a,const pair<int,int>&b){     return a.second<b.second; } int main() {     int n;cin>>n;   int a[n];   for(int i=0;i<n;i++) cin>>a[i];   vector<pair<int,int>> v;   for(int i=0;i<n;i++){       int c=0;       for(int j=0;j<31;j++){           if(checkbit(a[i],j)) c++;       }       v.push_back({a[i],c});   }   int idx = 0;   sort(v.begin(),v.end(),cmp);   for(auto it : v){       a[idx++] = it.first;       cout<<it.second<<" ";   }   for(int i=0;i<n;i++){       cout<<a[i]<<" ";   }   return 0; } C++ Telegram:- @allcoding1

def cardinalitySort(nums): return sorted(nums, key=lambda num: [bin(num).count('1'), num]) Cardinality sorting
def cardinalitySort(nums):     return sorted(nums, key=lambda num: [bin(num).count('1'), num]) Cardinality sorting

def gameWinner(colors):     currPlayer = "wendy"     prevPlayer = ""     winner = ""     while True:         moveMade = False         if currPlayer == "wendy":             whiteIndex = colors.find("www")             if whiteIndex = -1:                 # 3 consecutive whites found, remove the middle one                 colorsBuilder = list(colors)                 colorsBuilder.pop(whiteIndex + 1)                 colors = "".join(colorsBuilder)                 moveMade = True                 prevPlayer = currPlayer                 currPlayer = "bob"         else:             blackIndex = colors.find("bbb")             if blackIndex != -1:                 # 3 consecutive blacks found, remove the middle one                 colorsBuilder = list(colors)                 colorsBuilder.pop(blackIndex + 1)                 colors = "".join(colorsBuilder)                 moveMade = True                 prevPlayer = currPlayer                 currPlayer = "wendy"         # if no moves possible break         if not moveMade:             winner = prevPlayer             break     return winner print(gameWinner("wwwbb")) Python 3 Game Winner JP Morgan

Distinct digital number Python 3 JP Morgan
Distinct digital number Python 3 JP Morgan

Amazon seller anybody there in our group

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HCL Recruitment(allcoding1_official).pdf2.71 MB

🎯HCL Off Campus Hiring for Freshers as Engineer Trainee | 3-5LPA Job Role:- Engineer Trainee Education:- Any Degree Batch:- Any Batch CTC/Salary:- 3-5 LPA Apply Now:- www.allcoding1.com Telegram:- @allcoding1 Video👇 🅾️Instagram:- https://instagram.com/allcoding1_official?utm_source=qr&igshid=NGExMmI2YTkyZg%3D%3D ⛔YouTube :- https://youtube.com/shorts/TMaL15W0p3g?feature=share

#Artham Ayyindha Raja😎.... @allcoding1

Repost from allcoding1
import requests import json def getCaptialCity(country): api_request = requests.get(''//@allcoding1 name='+country) data = ap
import requests import json def getCaptialCity(country): api_request = requests.get(''//@allcoding1 name='+country) data = api_request.json()['data'] if(len(data[0]['capital'])==0) return -1 return data[0]['capital'] country = input() print(getCapitalCity(country)) Python IBM exam Ans Telegram:- @allcoding1

Python IBM exam Ans Telegram:- @allcoding1_official
+1
Python IBM exam Ans Telegram:- @allcoding1_official

import requests import json def getCaptialCity(country): api_request = requests.get(''//@allcoding1 name='+country) data = ap
import requests import json def getCaptialCity(country): api_request = requests.get(''//@allcoding1 name='+country) data = api_request.json()['data'] if(len(data[0]['capital'])==0) return -1 return data[0]['capital'] country = input() print(getCapitalCity(country)) Python Telegram:- @allcoding1_official

Both as a queue and as a stack IBM ANS Telegram:- @allcoding1
Both as a queue and as a stack IBM ANS Telegram:- @allcoding1

O(n) IBM ANS Telegram:- @allcoding1
O(n) IBM ANS Telegram:- @allcoding1

allcoding1 - Telegram 频道 @allcoding1 的统计与分析