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allcoding1_official

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تُعد قناة allcoding1_official (@allcoding1_official) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 84 584 مشتركاً، محتلاً المرتبة 1 497 في فئة التكنولوجيات والتطبيقات والمرتبة 3 527 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 84 584 مشتركاً.

بحسب آخر البيانات بتاريخ 10 يوليو, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -1 556، وفي آخر 24 ساعة بمقدار -30، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 2.01‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 0.85‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 1 701 مشاهدة. وخلال اليوم الأول يجمع عادةً 723 مشاهدة.
  • التفاعلات والاستجابة: يتفاعل الجمهور بانتظام؛ متوسط التفاعلات لكل منشور يبلغ 1.
  • الاهتمامات الموضوعية: يركز المحتوى على مواضيع رئيسية مثل dsa, stack, namaste, javascript, dev.

📝 الوصف وسياسة المحتوى

وصف القناة غير متوفر.

بفضل وتيرة التحديث المرتفعة (أحدث البيانات بتاريخ 11 يوليو, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التكنولوجيات والتطبيقات.

84 584
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-3024 ساعات
-4257 أيام
-1 55630 أيام
أرشيف المشاركات
Performing strange task code Python 3
Performing strange task code Python 3

Good Subset code in python
Good Subset code in python

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

Line segments code in python Telegram:-@allcoding1
Line segments code in python Telegram:-@allcoding1

// 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}, {3, 6}}; int N = sizeof(vertex) / sizeof(vertex[0]); cout << getMinSubtreeSumDifference(vertex, edges, N); return 0; } C++ Delete Edge to minimize subtree sum difference Telegram - @allcoding1

class Solution: def countSubstrings(self, s: str) -> int: n = len(s) c=0 for i in range(0,n-1): #i=0 print('i:',i) for j in range(i+1,n+1): #j=1 print('j',j) temp = s[i:j] if len(temp) == 1: c+=1 # if the len of substring is even, # check if the reverse of the string is same as the string elif(len(temp)%2 == 0): if (temp == temp[::-1]): c+=1 print("c",c) else: # create a dict to check how many times # each value has occurred d = {} for l in range(len(temp)): if temp[l] in d: d[temp[l]] = d[temp[l]]+1 else: d[temp[l]] = 1 print(d) return c op = Solution() op.countSubstrings('aabb') Python Wedding code Telegram:- @allcoding1

int ans = 0; for(int i = 0; i < N; ){ int num = A.get(i); int j; for(j = 0; j < N; j++){ if(i == j) continue; if(num % A.get(j) == 0) break; } if(j == N) ans++; i++; } return ans; Java ANOTHER DIVISOR PROBLEM Code Telegram - @allcoding1

Java Divide into Minimum Sequences Telegram - @allcoding1
+1
Java Divide into Minimum Sequences Telegram - @allcoding1

x=[] for i in range(N): x.append(1) a=1 while(a<N): //allcoding1 for i in range(a,N,a+1): x[i]=x[i]+1 a=a+1 print(x.count(K)) Python Performing strange task code 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

def flip( ch): return '1' if (ch == '0') else '0' # Utility method to get minimum flips when # alternate string starts with expected char def getFlipWithStartingCharcter(str, expected): flipCount = 0 for i in range(len( str)): # if current character is not expected, # increase flip count if (str[i] != expected): flipCount += 1 # flip expected character each time expected = flip(expected) return flipCount # method return minimum flip to make binary # string alternate def minFlipToMakeStringAlternate(str): # return minimum of following two # 1) flips when alternate string starts with 0 # 2) flips when alternate string starts with 1 return min(getFlipWithStartingCharcter(str, '0'), //allcoding1 getFlipWithStartingCharcter(str, '1')) //allcoding1 # Driver code to test above method if name == "main": str = "0001010111" print(minFlipToMakeStringAlternate) Flip or swap code in Python Telegram:-@allcoding1

// Java program to find the XOR of // all elements in the array class GFG { // Function to find the XOR of // all elements in the array static int xorOfArray(int arr[], int n) { // Resultant variable int xor_arr = 0; // Iterating through every element in // the array for (int i = 0; i < n; i++) { // Find XOR with the result xor_arr = xor_arr ^ arr[i]; } // Return the XOR return xor_arr; } // Driver Code public static void main (String[] args) { int arr[] = { 3, 9, 12, 13, 15 }; int n = arr.length; // allcoding1 System.out.println(xorOfArray(arr, n)); } } Prifix XOR code in Java

#include using namespace std; #define ll long long examcell ll qdSum(ll n){ if (n==0){ return 0; } return ((n%10)*(n%10)*(n%1
#include <iostream> using namespace std; #define ll long long examcell ll qdSum(ll n){ if (n==0){ return 0; } return ((n%10)*(n%10)*(n%10)*(n%10)) + qdSum(n/10); } ll getProduct(ll n){ // Base Case if(n == 0){ return 1 ; } // get the last digit and multiply it with remaining digits return (n%10) * getProduct(n/10) ; } examcell ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); } bool isSpecial(ll N){ if(gcd(qdSum(N),getProduct(N)) > 1){ return 1; } return 0; } int main() { int T; cin>>T; while(T--){ ll N; cin>>N; int count=0; while(N != 1){ if(isSpecial(N)){ count++; } N--; } cout<<count<<endl; } } C++

#include <iostream> using namespace std; class gas { public: int gas; int distance; }; int findStartIndex(gas stationQueue[], int n) { int start_point = 0; int end_point = 1; int curr_gas = stationQueue [start_point].gas - stationQueue [start_point].distance; while (end_point != start_point || curr_gas < 0) { while (curr_gas < 0 && start_point != end_point) { curr_gas -= stationQueue[start_point].gas - stationQueue //allcoding1 [start_point].distance; start_point = (start_point + 1) % n; if (start_point == 0) return -1; } curr_gas += stationQueue[end_point].gas - stationQueue [end_point].distance; end_point = (end_point + 1) % n; } return start_point; } int main() { gas gasArray[] = {{4, 6}, {6, 5}, {7, 3}, {4, 5}}; int n = sizeof(gasArray)/sizeof(gasArray [0]); int start = findStartIndex(gasArray, n); if(start == -1) cout<<"No solution"; else cout<<"Index of first gas station : "<<start; } GAS STATION C++

Python Make palindrome Code Telegram - @allcoding1
Python Make palindrome Code Telegram - @allcoding1

INFOSYS, TATA STEEL EXAM ANSWERS (23/4/22) 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

def flip( ch): return '1' if (ch == '0') else '0' # Utility method to get minimum flips when # alternate string starts with e
def flip( ch): return '1' if (ch == '0') else '0' # Utility method to get minimum flips when # alternate string starts with expected char def getFlipWithStartingCharcter(str, expected): flipCount = 0 for i in range(len( str)): # if current character is not expected, # increase flip count if (str[i] != expected): flipCount += 1 # flip expected character each time expected = flip(expected) return flipCount # method return minimum flip to make binary # string alternate def minFlipToMakeStringAlternate(str): # return minimum of following two # 1) flips when alternate string starts with 0 # 2) flips when alternate string starts with 1 return min(getFlipWithStartingCharcter(str, '0'), //allcoding1 getFlipWithStartingCharcter(str, '1')) # Driver code to test above method if name == "main": str = "0001010111" print(minFlipToMakeStringAlternate) Python Telegram - @allcoding1

INFOSYS EXAM ANS once check all codes are available http://www.joboffersadda.com/2022/01/infosys-exam-answer.html

Repost from allcoding1
Both i and 2 follow TATA steel exam Ans Telegram - @allcoding_1 🔔Unmute this channel to never miss any updates
Both i and 2 follow TATA steel exam Ans Telegram - @allcoding_1 🔔Unmute this channel to never miss any updates

Repost from allcoding1
Immedate left of D TATA steel exam Ans Telegram - @allcoding_1 🔔Unmute this channel to never miss any updates
Immedate left of D TATA steel exam Ans Telegram - @allcoding_1 🔔Unmute this channel to never miss any updates