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allcoding1_official

allcoding1_official

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📈 Telegram 频道 allcoding1_official 的分析概览

频道 allcoding1_official (@allcoding1_official) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 78 794 名订阅者,在 技术与应用 类别中位列第 1 627,并在 印度 地区排名第 3 772 位。

📊 受众指标与增长动态

自 невідомо 创建以来,项目保持高速增长,吸引了 78 794 名订阅者。

根据 29 十月, 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 -1 610,过去 24 小时变化为 -51,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 2.18%。内容发布后 24 小时内通常能获得 0.61% 的反应,占订阅者总量。
  • 帖子覆盖: 每篇帖子平均可获得 1 718 次浏览,首日通常累积 478 次浏览。
  • 互动与反馈: 受众积极参与,单帖平均反应数为 0。
  • 主题关注点: 内容集中在 dsa, stack, namaste, javascript, dev 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 30 十月, 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 技术与应用 类别中的关键影响点。

78 794
订阅者
-5124 小时
-3677 天
-1 61030 天
帖子存档
All this 300 rupees only Contact:- @meterials_available
+8
All this 300 rupees only Contact:- @meterials_available

if not intervals:         return 0         intervals.sort(ke
if not intervals:
        return 0
    
    intervals.sort(key=lambda x: x[0])  # Sort intervals based on start points
    merged_intervals = [intervals[0]]  # Initialize merged intervals with the first interval
    
    for interval in intervals[1:]:
        prev_interval = merged_intervals[-1]
        
        if interval[0] <= prev_interval[1]:  # If overlapping with previous interval
            prev_interval[1] = max(prev_interval[1], interval[1])  # Merge intervals
        else:
            merged_intervals.append(interval)  # Add non-overlapping interval
    
    total_length = sum(end - start for start, end in merged_intervals)
    return total_length

# Example usage:
intervals = [(1, 3), (2, 6), (8, 10), (15, 18)]
total_length_covered = merge_intervals(intervals)
print("Total length covered after merging overlapping intervals:", total_length_covered)

if not intervals: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; return 0 &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; intervals.sort(ke
if not intervals:         return 0         intervals.sort(key=lambda x: x[0])  # Sort intervals based on start points     merged_intervals = [intervals[0]]  # Initialize merged intervals with the first interval         for interval in intervals[1:]:         prev_interval = merged_intervals[-1]                 if interval[0] <= prev_interval[1]:  # If overlapping with previous interval             prev_interval[1] = max(prev_interval[1], interval[1])  # Merge intervals         else:             merged_intervals.append(interval)  # Add non-overlapping interval         total_length = sum(end - start for start, end in merged_intervals)     return total_length # Example usage: intervals = [(1, 3), (2, 6), (8, 10), (15, 18)] total_length_covered = merge_intervals(intervals) print("Total length covered after merging overlapping intervals:", total_length_covered)

int main() { ll N; cin >> N; vector<ll> u(N); for(ll i = 0; i < N; ++i) { cin >> u[i]; } ll M, X; cin >> M >> X; vector<vector<ll>> a(N); for(ll i = 0; i < M; ++i) { ll x, y; cin >> x >> y; a[x].push_back(y); a[y].push_back(x); } vector<bool> v(N, false); queue<ll> q; q.push(0); v[0] = true; ll m = 0; while(!q.empty()) { ll s = q.size(); ll us = 0; for(ll i = 0; i < s; ++i) { ll n = q.front(); q.pop(); us += u[n]; for(ll nb : a[n]) { if(!v[nb]) { q.push(nb); v[nb] = true; } } } m = max(m, us); } cout << m << endl; return 0; } apartment @allcoding1_official

int main() { ll N; cin &gt;&gt; N; vector w(N); for(ll i = 0; i &lt; N; ++i) { cin &gt;&gt; w[i]; } vector s(N, 15); for(ll i
int main() { ll N; cin >> N; vector w(N); for(ll i = 0; i < N; ++i) { cin >> w[i]; } vector s(N, 15); for(ll i = N-2; i >= 0; --i) { for(ll j = i+1; j < N; ++j) { if(w[i] < w[j]) { s[i] = 10; for(ll k = j+1; k < N; ++k) { if(w[j] > w[k]) { s[i] = 5; break; } } break; } } } ll g = 0; for(ll sc : s) { g += sc; } cout << g << endl; return 0; } chair manufacture

class TreeNode:     def init(self, val):         self.val = val         self.children = [] def build_tree(units, relationships):     nodes = {i: TreeNode(units[i]) for i in range(len(units))}     for rel in relationships:         parent, child = rel         nodes[parent].children.append(nodes[child])     return nodes[0]  def max_electricity_per_floor(root):     max_electricity = 0     floor_electricity = {}         def dfs(node, level):         nonlocal max_electricity         if level not in floor_electricity:             floor_electricity[level] = 0         floor_electricity[level] += node.val         max_electricity = max(max_electricity, floor_electricity[level])         for child in node.children:             dfs(child, level + 1)         dfs(root, 0)     return max_electricity num = int(input()) units = list(map(int, input().split())) numRel, memConnect = map(int, input().split()) relationships = [tuple(map(int, input().split())) for _ in range(numRel)] root = build_tree(units, relationships) result = max_electricity_per_floor(root) print(result) Apartment one

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All this 300 rupees only Contact:- @meterials_available
+8
All this 300 rupees only Contact:- @meterials_available

pallindrome code
pallindrome code

All this 300 rupees only Contact:- @meterials_available
+8
All this 300 rupees only Contact:- @meterials_available

X(1), (2) int count; // Line 1 Antint i) { count 1) int getcount () {return count) // Line 5 public class Book public static void main(String[] args) System.out.println(AB.X.count + AB.Y.count); // Line 10 Ans)Compile time error at "// Line 10" @allcoding1_official

Which of the statement(s) given below is/are correct? 1)Routers can be static or dynamic 2)Routers connect two or more separate networks, Ans) both 1 and 2 @allcoding1_official

37 38 In MS Word, how can a document be split into multiple parts such that each part can have different formats and layouts Ans)By creating sections @allcoding1_official

How many pairs of letters are there in the word "DYNASTY" which has as many letters between them (from both the sides) as in the alphabets from A to Z? A) 2 @allcoding1_official

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