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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

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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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📈 Аналітичний огляд Telegram-каналу Tcs exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

Канал Tcs exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer (@coding_are) у мовному сегменті Англійська є активним учасником. На даний момент спільнота об'єднує 13 276 підписників, посідаючи 15 335 місце в категорії Освіта та 32 351 місце у регіоні Індія.

📊 Показники аудиторії та динаміка

З моменту свого створення невідомо, проект продемонстрував стрімке зростання, зібравши аудиторію у 13 276 підписників.

За останніми даними від 12 червня, 2026, канал демонструє стабільну активність. Хоча за останні 30 днів спостерігається зміна кількості учасників на 111, а за останні 24 години на -9, загальне охоплення залишається високим.

  • Статус верифікації: Не верифікований
  • Рівень залученості (ER): Середній показник залученості аудиторії становить 2.86%. Протягом перших 24 годин після публікації контент зазвичай збирає 1.13% реакцій від загальної кількості підписників.
  • Охоплення публікацій: В середньому кожен допис отримує 380 переглядів. Протягом першої доби публікація в середньому набирає 150 переглядів.
  • Реакції та взаємодія: Аудиторія активно підтримує контент: середня кількість реакцій на один пост – 1.
  • Тематичні інтереси: Контент зосереджений навколо ключових тем, таких як placement, gaurntee, suree, capgemini, infosy.

📝 Опис та контентна політика

Автор описує ресурс як майданчик для висловлення суб'єктивної думки:
🔥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...

Завдяки високій частоті оновлень (останні дані отримано 13 червня, 2026), канал підтримує актуальність та високий рівень охоплення публікацій. Аналітика показує, що аудиторія активно взаємодіє з контентом, що робить його важливою точкою впливу в категорії Освіта.

13 276
Підписники
-924 години
-397 днів
+11130 день
Архів дописів
from collections import deque from typing import List, Dict, Tuple, Set class InputData: def init(self): self.connections: List[Tuple[int, int]] = [] self.costs: Dict[int, int] = {} self.values: Dict[int, int] = {} self.budget: int = 0 self.depth: int = 0 self.seed_order: List[int] = [] def read_input() -> InputData: data = InputData() n = int(input()) num_connections = int(input()) for _ in range(num_connections): u, v = map(int, input().split()) data.connections.append((u, v)) num_seeds = int(input()) for _ in range(num_seeds): user_id, cost = map(int, input().split()) data.costs[user_id] = cost data.seed_order.append(user_id) for i in range(1, n + 1): value = int(input()) data.values[i] = value data.budget, data.depth = map(int, input().split()) return data def bfs_reach(start: int, adj: List[List[int]], d: int) -> List[int]: visited: Set[int] = set() q = deque([(start, 0)]) visited.add(start) reach = [] while q: u, dist = q.popleft() reach.append(u) if dist == d: continue for v in adj[u]: if v not in visited: visited.add(v) q.append((v, dist + 1)) return reach def solve_campaign( connections: List[Tuple[int, int]], costs: Dict[int, int], values: Dict[int, int], seed_order: List[int], budget: int, d: int ) -> Tuple[List[int], int, int]: n = len(values) adj = [[] for _ in range(n + 1)] for u, v in connections: adj[u].append(v) adj[v].append(u) reach_map: Dict[int, List[int]] = {} for s in costs: reach_map[s] = bfs_reach(s, adj, d) chosen: List[int] = [] total_value = 0 total_cost = 0 covered: Set[int] = set() remaining = seed_order.copy() while True: best_eff = -1 best_seed = -1 best_gain = 0 best_cost = 0 for s in remaining: c = costs[s] if total_cost + c > budget: continue gain = 0 for u in reach_map[s]: if u not in covered: gain += values[u] if gain == 0: continue eff = gain / c if ( eff > best_eff or (eff == best_eff and seed_order.index(s) < seed_order.index(best_seed)) ): best_eff = eff best_seed = s best_gain = gain best_cost = c if best_seed == -1: break chosen.append(best_seed) total_cost += best_cost total_value += best_gain for u in reach_map[best_seed]: covered.add(u) remaining.remove(best_seed) return chosen, total_value, total_cost def main(): data = read_input() chosen, total_value, total_cost = solve_campaign( data.connections, data.costs, data.values, data.seed_order, data.budget, data.depth, ) print(" ".join(map(str, chosen))) print(total_value) print(total_cost) if name == "main": main()

Puzzle Python 3 All tests caes passed ✅

from typing import List, Set class Trie: def init(self): self.end = False self.next = {} def insert_word(root: Trie, word: str) -> None: cur = root for c in word: if c not in cur.next: cur.next[c] = Trie() cur = cur.next[c] cur.end = True dirs = [(1,0), (-1,0), (0,1), (0,-1), (1,1), (1,-1), (-1,1), (-1,-1)] def dfs(board: List[List[str]], x: int, y: int, node: Trie, path: List[str], res: Set[str]): c = board[x][y] if c not in node.next: return node = node.next[c] path.append(c) if node.end: res.add("".join(path)) board[x][y] = "#" for dx, dy in dirs: nx, ny = x + dx, y + dy if 0 <= nx < len(board) and 0 <= ny < len(board[0]) and board[nx][ny] != "#": dfs(board, nx, ny, node, path, res) board[x][y] = c path.pop() def solve_problem(rows: int, cols: int, board: List[List[str]], words: List[str]) -> None: root = Trie() for w in words: insert_word(root, w) found = set() path = [] for i in range(rows): for j in range(cols): dfs(board, i, j, root, path, found) if not found: print(0) return ans = sorted(found) print(len(ans)) for w in ans: print(w) if name == "main": R, C = map(int, input().split()) board = [list(input().strip().replace(" ", "")) for _ in range(R)] n = int(input()) words = [input().strip() for _ in range(n)] solve_problem(R, C, board, words)

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