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INFOSYS EXAM SOLUTIONS

INFOSYS EXAM SOLUTIONS

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🔥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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📈 Análisis del canal de Telegram INFOSYS EXAM SOLUTIONS

El canal INFOSYS EXAM SOLUTIONS (@coding_are) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 14 191 suscriptores, ocupando la posición 14 646 en la categoría Educación y el puesto 30 070 en la región India.

📊 Métricas de audiencia y dinámica

Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 14 191 suscriptores.

Según los últimos datos del 01 agosto, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de 987, y en las últimas 24 horas de 552, conservando un alto alcance.

  • Estado de verificación: No verificado
  • Tasa de interacción (ER): El promedio de interacción de la audiencia es 7.48%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 2.55% de reacciones respecto al total de suscriptores.
  • Alcance de las publicaciones: Cada publicación recibe en promedio 1 061 visualizaciones. En el primer día suele acumular 362 visualizaciones.
  • Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 3.
  • Intereses temáticos: El contenido se centra en temas clave como placement, gaurntee, suree, capgemini, infosy.

📝 Descripción y política de contenido

El autor describe el recurso como un espacio para expresar opiniones subjetivas:
🔥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...

Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 02 agosto, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Educación.

14 191
Suscriptores
+55224 horas
+8087 días
+98730 días
Archivo de publicaciones
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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)

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include using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int N, M; cin >> N >> M; vector grid(N); for (int i = 0; i < N; i++) { grid[i].resize(M); for (int j = 0; j < M; j++) { cin >> grid[i][j]; } } vector horizontal_rods, vertical_rods; for (int i = 0; i < N; i++) { if (all_of(grid[i].begin(), grid[i].end(), [](char c){ return c != '.'; })) horizontal_rods.push_back(i); } for (int j = 0; j < M; j++) { bool full = true; for (int i = 0; i < N; i++) if (grid[i][j] == '.') full = false; if (full) vertical_rods.push_back(j); } vector> is_cross(N, vector(M, false)); for (int c : vertical_rods) { for (int i = 0; i < N; i++) { int left = c - 1, right = c + 1; if (left >= 0 && right < M && grid[i][left] == 'C' && grid[i][right] == 'C') is_cross[i][c] = true; } } for (int r : horizontal_rods) { for (int j = 0; j < M; j++) { int up = r - 1, down = r + 1; if (up >= 0 && down < N && grid[up][j] == 'C' && grid[down][j] == 'C') is_cross[r][j] = true; } } vector> cable(N, vector(M, false)); for (int i = 0; i < N; i++) for (int j = 0; j < M; j++) if (grid[i][j] == 'C' || is_cross[i][j]) cable[i][j] = true; vector> adj(N * M); int di[4] = {-1, 0, 1, 0}; int dj[4] = {0, 1, 0, -1}; for (int i = 0; i < N; i++) { for (int j = 0; j < M; j++) { if (!cable[i][j]) continue; int id = i * M + j; for (int d = 0; d < 4; d++) { int ni = i + di[d], nj = j + dj[d]; if (ni >= 0 && ni < N && nj >= 0 && nj < M && cable[ni][nj]) adj[id].push_back(ni * M + nj); } } } int start = -1; for (int i = 0; i < N && start == -1; i++) for (int j = 0; j < M; j++) if (cable[i][j] && adj[i * M + j].size() == 1) { start = i * M + j; break; } vector visited(N * M, false); vector sum_h(N, 0), sum_v(M, 0); int curr = start, prevv = -1; visited[curr] = true; while (true) { int cr = curr / M, cc = curr % M; int nextt = -1; for (int nb : adj[curr]) if (nb != prevv && !visited[nb]) { nextt = nb; break; } if (is_cross[cr][cc] && prevv != -1) { int pr = prevv / M, pc = prevv % M; int sign = (grid[cr][cc] == 'C') ? 1 : -1; if (pr == cr) sum_v[cc] += ((pc < cc) ? 1 : -1) * sign; else sum_h[cr] += ((pr < cr) ? 1 : -1) * sign; } if (nextt == -1) break; prevv = curr; curr = nextt; visited[curr] = true; } long long answer = 0; for (int r : horizontal_rods) answer += abs(sum_h[r]) / 2; for (int c : vertical_rods) answer += abs(sum_v[c]) / 2; cout << answer; return 0; } Cable (c++)

Zoobin cpp

include using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int a; cin >> a; vector> b(a), c(a); set d; for (int i = 0; i < a; i++) { cin >> b[i].first >> b[i].second; d.insert(b[i].first); d.insert(b[i].second); } for (int i = 0; i < a; i++) cin >> c[i].first >> c[i].second; vector e(d.begin(), d.end()); auto f = [](vector> g) { for (auto& [h, i] : g) if (h > i) swap(h, i); sort(g.begin(), g.end()); return g; }; auto j = [](const vector>& k) { string l; for (auto [m, n] : k) l += to_string(m) + "-" + to_string(n) + ","; return l; }; vector> o = f(c); string p = j(o); vector> q = f(b); string r = j(q); if (r == p) { cout << 0; return 0; } map s; queue>,int>> t; t.push({q,0}); s[r] = 0; while (!t.empty()) { auto [u,v] = t.front(); t.pop(); map> w; for (auto [x,y] : u) { w[x].push_back(y); w[y].push_back(x); } set> z; for (int aa : e) { function&,set&)> ab = [&](int ac,int ad,vector& ae,set& af) { ae.push_back(ac); af.insert(ac); for (int ag : w[ac]) { if (ag == ad) continue; if (af.count(ag)) { auto ah = find(ae.begin(), ae.end(), ag); if (ah != ae.end()) { vector ai(ah, ae.end()); if (ai.size() >= 3) { int aj = min_element(ai.begin(), ai.end()) - ai.begin(); rotate(ai.begin(), ai.begin() + aj, ai.end()); z.insert(ai); } } } else if (ae.size() < e.size()) ab(ag, ac, ae, af); } ae.pop_back(); af.erase(ac); }; vector ak; set al; ab(aa, -1, ak, al); } for (const auto& am : z) { map an; for (int ao : e) an[ao] = ao; int ap = am.size(); for (int aq = 0; aq < ap; aq++) an[am[aq]] = am[(aq + 1) % ap]; vector> ar; for (auto [as, at] : u) ar.push_back({an[as], an[at]}); ar = f(ar); string au = j(ar); if (au == p) { cout << v + 1; return 0; } if (!s.count(au)) { s[au] = v + 1; t.push({ar, v + 1}); } } } cout << -1; return 0; }