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allcoding1_official

allcoding1_official

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📈 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 85 687 subscribers, ranking 1 509 in the Technologies & Applications category and 3 512 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 85 687 subscribers.

According to the latest data from 20 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -1 460 over the last 30 days and by -39 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 3.36%. Within the first 24 hours after publication, content typically collects 0.73% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 2 882 views. Within the first day, a publication typically gains 625 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 1.
  • 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 21 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 Technologies & Applications category.

85 687
Subscribers
-3924 hours
-3267 days
-1 46030 days
Posts Archive
#include <iostream> #include <vector> #include <algorithm> using namespace std; void dfs(int node, int parent, const vector<vector<int>>& tree, const vector<int>& A, int depth, int& maxDepth) { maxDepth = max(maxDepth, depth); for (int child : tree[node]) { if (child != parent) { if ((A[node] ^ A[child]) < A[node] && (A[node] ^ A[child]) < A[child]) { dfs(child, node, tree, A, depth + 1, maxDepth); } } } } int main() { int N; cin >> N; vector<int> A(N + 1); vector<int> P(N + 1); vector<vector<int>> tree(N + 1); // Read values array for (int i = 1; i <= N; ++i) { cin >> A[i]; } // Read parent array and buildthe tree for (int i = 2; i <= N; ++i) { // P[1] is root with P[1] = 0, so start from 2 cin >> P[i]; tree[P[i]].push_back(i); tree[i].push_back(P[i]); } int maxDepth = 0; dfs(1, 0, tree, A, 1, maxDepth); cout << maxDepth << endl; return 0; }

#include <iostream> #include <vector> #include <algorithm> #include <cmath> using namespace std; const int MOD = 1000000007; int minLampsToLightRoad(int num_positions, int num_lamps, vector<int>& lamp_positions, vector<int>& left_reach, vector<int>& right_reach, vector<pair<int, int>>& queries) { // Step 1: Create intervals for each lamp vector<pair<int, int>> intervals; for (int i = 0; i < num_lamps; ++i) { intervals.push_back({lamp_positions[i] - left_reach[i], lamp_positions[i] + right_reach[i]}); } // Step 2: Sort intervals based on starting position sort(intervals.begin(), intervals.end()); // Precompute the farthest reach for each starting point vector<pair<int, int>> max_reach_from_start; int current_max_reach = -1; for (const auto& interval : intervals) { int start = interval.first; int end = interval.second; if (max_reach_from_start.empty() start > max_reach_from_start.back().first) { max_reach_from_start.push_back({start, end}); } current_max_reach = max(current_max_reach, end); max_reach_from_start.back().second = current_max_reach; } auto min_lamps_needed = [&](int query_left, int query_right) { int count = 0; int max_reach = query_left; while (max_reach <= query_right) { auto it = upper_bound(max_reach_from_start.begin(), max_reach_from_start.end(), make_pair(max_reach, INT_MAX)); if (it == max_reach_from_start.begin() prev(it)->first > max_reach) { return -1; } int next_max_reach = prev(it)->second; if (next_max_reach <= max_reach) { return -1; } max_reach = next_max_reach + 1; count++; if (max_reach > query_right) { break; } } return max_reach > query_right ? count : -1; }; // Step 3: Process each query and sum up the results int result_sum = 0; for (const auto& query : queries) { int result = min_lamps_needed(query.first, query.second); if (result != -1) { result_sum += result; result_sum %= MOD; } } return result_sum; } lightning lamp code , all cases are passing

#include <iostream> #include <vector> #include <unordered_map> #include <unordered_set> #include <queue> #include <algorithm> using namespace std; int maxTreeScore(int node_count, int edge_count, vector<pair<int, int>>& edges, vector<int>& colors) { // Step 1: Parse input and create adjacency list unordered_map<int, vector<int>> adjacency_list; for (const auto& edge : edges) { int start = edge.first; int end = edge.second; adjacency_list[start].push_back(end); adjacency_list[end].push_back(start); } // Step 2: Calculate depth of each node using BFS vector<int> node_depth(node_count + 1, -1); node_depth[1] = 0; queue<int> bfs_queue; bfs_queue.push(1); while (!bfs_queue.empty()) { int current_node = bfs_queue.front(); bfs_queue.pop(); int current_depth = node_depth[current_node]; for (int neighbor : adjacency_list[current_node]) { if (node_depth[neighbor] == -1) { // unvisited node_depth[neighbor] = current_depth + 1; bfs_queue.push(neighbor); } } } // Step 3: Group nodes by depth unordered_map<int, vector<int>> nodes_grouped_by_depth; for (int node = 1; node <= node_count; ++node) { nodes_grouped_by_depth[node_depth[node]].push_back(node); } // Step 4: Calculate distinct colors per depth unordered_map<int, int> distinct_colors_at_depth; for (const auto& pair : nodes_grouped_by_depth) { int depth = pair.first; const vector<int>& nodes = pair.second; unordered_set<int> unique_colors; for (int node : nodes) { unique_colors.insert(colors[node - 1]); } distinct_colors_at_depth[depth] = unique_colors.size(); } // Step 5: Dynamic programming to calculate max score int max_depth = max_element(nodes_grouped_by_depth.begin(), nodes_grouped_by_depth.end(), [](const auto& a, const auto& b) { return a.first < b.first; })->first; vector<int> dp_score(max_depth + 2, 0); for (int depth = max_depth; depth >= 0; --depth) { // Option 1: Move to the next depth without adding score dp_score[depth] = dp_score[depth + 1]; // Option 2: Add the distinct colors to score and move to depth depth + unique_colors_count if (distinct_colors_at_depth.find(depth) != distinct_colors_at_depth.end()) { int unique_colors_count = distinct_colors_at_depth[depth]; if (depth + unique_colors_count <= max_depth) { dp_score[depth] = max(dp_score[depth], dp_score[depth + unique_colors_count] + unique_colors_count); } else { dp_score[depth] = max(dp_score[depth], unique_colors_count); } } } return dp_score[0]; } diving in a tree , all test cases are passing

All codes are available To easily find out ur codes Once check it 👇👇 https://www.instagram.com/allcoding1_official?igsh=ZHJpNXdpeWh1d2No

Coins game
Coins game

#include <iostream> #include <vector> #include <algorithm> using namespace std; void dfs(int node, int parent, const vector<vector<int>> &adj, const vector<int> &A, int length, int &maxLength) { maxLength = max(maxLength, length); for (int neighbor : adj[node]) { if (neighbor == parent) continue; if ((A[node] ^ A[neighbor]) < min(A[node], A[neighbor])) { dfs(neighbor, node, adj, A, length + 1, maxLength); } } } int main() { int N; cin >> N; vector<int> A(N), P(N); for (int i = 0; i < N; ++i) cin >> A[i]; for (int i = 1; i < N; ++i) cin >> P[i]; vector<vector<int>> adj(N); for (int i = 1; i < N; ++i) { int parent = P[i]; adj[parent].push_back(i); adj[i].push_back(parent); } int maxLength = 0; dfs(0, -1, adj, A, 1, maxLength); cout << maxLength << endl; return 0; } Nodes

Shortest string code in python Infosys
Shortest string code in python Infosys

def get_answer(N, K, A, S):     from collections import Counter     def min_deletions_to_palindrome(piece):         count = Counter(piece)         odd_count = sum(1 for freq in count.values() if freq % 2 == 1)         return max(0, odd_count - 1)        start = 0     total_deletions = 0          for length in A:         piece = S[start:start + length]         total_deletions += min_deletions_to_palindrome(piece)         start += length          return total_deletions MINIMAL PALINDROME

Lighting lamp
Lighting lamp

Dividing array
Dividing array

Python periodic strings
Python periodic strings

All codes are available To easily find out ur codes Once check it 👇👇 https://www.instagram.com/allcoding1_official?igsh=ZHJpNXdpeWh1d2No

XR Forming code in python
XR Forming code in python

Python
Python

Code maximum python

///find largest swuare linked in public static int findLargestSquareSize(int[][] samples) { if (samples == null || samples.length == 0) return 0; int n = samples.length; int maxSize = 0; int[][] dp = new int[n][n]; for (int i = 0; i < n; i++) { dp[i][0] = samples[i][0]; dp[0][i] = samples[0][i]; maxSize = Math.max(maxSize, dp[i][0]); maxSize = Math.max(maxSize, dp[0][i]); } for (int i = 1; i < n; i++) { for (int j = 1; j < n; j++) { if (samples[i][j] == 1) { dp[i][j] = Math.min(Math.min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]) + 1; maxSize = Math.max(maxSize, dp[i][j]); } else { dp[i][j] = 0; } } } return maxSize; }

//Equilibrium path #include <bits/stdc++.h> using namespace std; string trim(string str) { str.erase(0, str.find_first_not_of(' ')); str.erase(str.find_last_not_of(' ') + 1); return str; } int solve(int N, vector<int> A) { int total_sum = accumulate(A.begin(), A.end(), 0); int left_sum = 0; int equilibrium_count = 0; for (int i = 0; i < N; ++i) { int right_sum = total_sum - left_sum - A[i]; if (left_sum == right_sum) { equilibrium_count++; } left_sum += A[i]; } return equilibrium_count; } int main() { ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); string inputline; getline(cin, inputline); int N = stoi(trim(inputline)); vector<int> A(N); for (int j = 0; j < N; j++) { getline(cin, inputline); A[j] = stoi(trim(inputline)); } int result = solve(N, A); cout << result << endl; return 0; }

// coins game import java.util.*; class HelloWorld { static int helper(int a, int b, int i ,int[]x,int n){ if(i==x.length) return 0; int left=(x[i]>n)?0:Math.abs(x[i]-a)+helper(x[i],b,i+1,x,n); int right=(x[i]>n)?0:Math.abs(x[i]-b)+helper(a,x[i],i+1,x,n); return Math.min(left,right); } public static void main(String[] args) { Scanner sc=new Scanner(System.in); int N=sc.nextInt(); int Q=sc.nextInt(); int A=sc.nextInt(); int B=sc.nextInt(); int[]x=new int[Q]; for(int i=0;i<Q;i++){ x[i]=sc.nextInt(); } System.out.println(helper(A,B,0,x,N)); } }

All codes are available Once check it 👇👇 https://www.instagram.com/allcoding1_official?igsh=ZHJpNXdpeWh1d2No