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کانال allcoding1_official (@allcoding1_official) در بخش زبانی انگلیسی بازیگری فعال است. در حال حاضر جامعه شامل 84 536 مشترک است و جایگاه 1 499 را در دسته فناوری و برنامه‌ها و رتبه 3 531 را در منطقه الهند دارد.

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از زمان ایجاد در невідомо، پروژه رشد سریعی داشته و 84 536 مشترک جذب کرده است.

بر اساس آخرین داده‌ها در تاریخ 11 ژوئیه, 2026، کانال فعالیت پایداری دارد. در ۳۰ روز گذشته تغییر اعضا برابر -1 593 و در ۲۴ ساعت گذشته برابر -70 بوده و همچنان دسترسی گسترده‌ای حفظ شده است.

  • وضعیت تأیید: تأیید نشده
  • نرخ تعامل (ER): میانگین تعامل مخاطب 2.09% است و در ۲۴ ساعت نخست پس از انتشار، محتوا معمولاً 0.84% واکنش نسبت به کل مشترکان کسب می‌کند.
  • دسترسی پست‌ها: هر پست به طور میانگین 1 769 بازدید دریافت می‌کند. در اولین روز معمولاً 712 بازدید جمع‌آوری می‌شود.
  • واکنش‌ها و تعامل: مخاطبان به‌طور فعال حمایت می‌کنند؛ میانگین واکنش به هر پست 1 است.
  • علایق موضوعی: محتوا بر موضوعات کلیدی مانند dsa, stack, namaste, javascript, dev تمرکز دارد.

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به لطف به‌روزرسانی‌های پرتکرار (آخرین داده در تاریخ 12 ژوئیه, 2026)، کانال همواره به‌روز و دارای دسترسی بالاست. تحلیل‌ها نشان می‌دهد مخاطبان به‌طور فعال با محتوا تعامل دارند و آن را به نقطه اثرگذاری مهم در دسته فناوری و برنامه‌ها تبدیل کرده‌اند.

84 536
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-4307 روز
-1 59330 روز
آرشیو پست ها
Performing strange task code Python 3
Performing strange task code Python 3

Good Subset code in python
Good Subset code in python

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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

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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

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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