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Tcs exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

Tcs exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

前往频道在 Telegram

🔥Guys plz Stop fearing for daily exams 📝 👨‍💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srksvk

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📈 Telegram 频道 Tcs exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer 的分析概览

频道 Tcs exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer (@coding_are) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 13 282 名订阅者,在 教育 类别中位列第 15 335,并在 印度 地区排名第 32 351

📊 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 13 282 名订阅者。

根据 12 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 111,过去 24 小时变化为 -9,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 2.86%。内容发布后 24 小时内通常能获得 1.13% 的反应,占订阅者总量。
  • 帖子覆盖: 每篇帖子平均可获得 380 次浏览,首日通常累积 150 次浏览。
  • 互动与反馈: 受众积极参与,单帖平均反应数为 1
  • 主题关注点: 内容集中在 placement, gaurntee, suree, capgemini, infosy 等核心主题上。

📝 描述与内容策略

作者将该频道定位为表达主观观点的平台:
🔥Guys plz Stop fearing for daily exams 📝 👨‍💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srks...

凭借高频更新(最新数据采集于 13 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。

13 282
订阅者
-924 小时
-397
+11130
帖子存档
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#include <bits/stdc++.h> using namespace std; static int min_ops(const vector<int>& a) { int n = (int)a.size(); string s(n, '\0'), g(n, '\0'); for (int i = 0; i < n; ++i) s[i] = char(a[i]); for (int i = 0; i < n; ++i) g[i] = char(i); if (s == g) return 0; vector<array<int, 3>> mv; mv.reserve(n * n * n); for (int i = 0; i < n; ++i) { for (int j = i + 1; j <= n; ++j) { int ln = j - i; for (int k = 0; k <= n - ln; ++k) { if (k == i) continue; mv.push_back({i, j, k}); } } } deque<string> q1, q2; unordered_map<string, int> d1, d2; q1.push_back(s); d1.emplace(s, 0); q2.push_back(g); d2.emplace(g, 0); auto expand = [&](deque<string>& q, unordered_map<string, int>& dself, unordered_map<string, int>& dother) -> int { int m = (int)q.size(); while (m--) { string x = q.front(); q.pop_front(); int dx = dself[x]; for (auto& t : mv) { int i = t[0], j = t[1], k = t[2]; int ln = j - i; string b = x.substr(i, ln); string r = x.substr(0, i) + x.substr(j); string y = r.substr(0, k) + b + r.substr(k); if (dself.find(y) != dself.end()) continue; int nd = dx + 1; auto it = dother.find(y); if (it != dother.end()) return nd + it->second; dself.emplace(y, nd); q.push_back(y); } } return -1; }; while (!q1.empty() && !q2.empty()) { if (q1.size() <= q2.size()) { int ans = expand(q1, d1, d2); if (ans != -1) return ans; } else { int ans = expand(q2, d2, d1); if (ans != -1) return ans; } } return 0; } int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n; if (!(cin >> n)) return 0; string line; getline(cin, line); // consume endline after n getline(cin, line); // blank line vector<string> sh(n), og(n); for (int i = 0; i < n; ++i) getline(cin, sh[i]); getline(cin, line); // blank line for (int i = 0; i < n; ++i) getline(cin, og[i]); unordered_map<string, int> mp; mp.reserve(n * 2); for (int i = 0; i < n; ++i) mp[og[i]] = i; vector<int> a(n); for (int i = 0; i < n; ++i) a[i] = mp[sh[i]]; cout << min_ops(a) << "\n"; return 0; } Order it (cpp)