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ACCENTURE EXAM SOLUTIONS

ACCENTURE EXAM SOLUTIONS

الذهاب إلى القناة على 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

إظهار المزيد

📈 نظرة تحليلية على قناة تيليجرام ACCENTURE EXAM SOLUTIONS

تُعد قناة ACCENTURE EXAM SOLUTIONS (@coding_are) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 14 127 مشتركاً، محتلاً المرتبة 14 101 في فئة التعليم والمرتبة 28 065 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 14 127 مشتركاً.

بحسب آخر البيانات بتاريخ 26 سبتمبر, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -112، وفي آخر 24 ساعة بمقدار -6، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 3.64‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 1.54‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 514 مشاهدة. وخلال اليوم الأول يجمع عادةً 218 مشاهدة.
  • التفاعلات والاستجابة: يتفاعل الجمهور بانتظام؛ متوسط التفاعلات لكل منشور يبلغ 2.
  • الاهتمامات الموضوعية: يركز المحتوى على مواضيع رئيسية مثل placement, gaurntee, suree, capgemini, infosy.

📝 الوصف وسياسة المحتوى

يصف المؤلف القناة بأنها مساحة للتعبير عن الآراء الذاتية:
“🔥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...”

بفضل وتيرة التحديث المرتفعة (أحدث البيانات بتاريخ 26 سبتمبر, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التعليم.

14 127
المشتركون
-624 ساعات
-127 أيام
-11230 أيام
أرشيف المشاركات
Follow the Codeing_area( Srksvk) channel on WhatsApp: https://whatsapp.com/channel/0029VaicY2a65yD2YNehWP2j Join for next code guy's fast I will share here code now

def placementlelo(valid_digits, fd): pd = [] for digit, valid in valid_digits.items(): diff = sum(1 for a, b in zip(fd, valid) if a != b) if diff == 0: pd.append(digit) elif diff == 1: pd.append(digit) return pd def solve(): sd = [input().strip() for _ in range(3)] fad = [input().strip() for _ in range(3)] valid_digits = {} for i in range(10): valid_digits[i] = ''.join(sd[j][i*3:(i+1)*3] for j in range(3)) faulty_number = [] for i in range(len(fad[0]) // 3): fd = ''.join(fad[j][i*3:(i+1)*3] for j in range(3)) pd = placementlelo(valid_digits, fd) if not pd: print("Invalid",end='') return faulty_number.append(pd) from itertools import product ts = 0 for possible in product(*faulty_number): ts += int(''.join(map(str, possible))) print(ts,end='') solve() Toggle challenge

def process_commands(kpr): loop_counts = [] curr_iter = [] output = [] index = 0 while index < len(kpr): command = kpr[index] if command.startswith("for"): times = int(command.split(" ")[1]) loop_counts.append(times) curr_iter.append(0) elif command == "do": pass elif command == "done": current = curr_iter.pop() + 1 max_count = loop_counts.pop() if current < max_count: loop_counts.append(max_count) curr_iter.append(current) index = find_loop_start(kpr, index) continue elif command.startswith("break"): break_at = int(command.split(" ")[1]) if curr_iter[-1] + 1 == break_at: loop_counts.pop() curr_iter.pop() index = find_loop_end(kpr, index) elif command.startswith("continue"): continue_at = int(command.split(" ")[1]) if curr_iter[-1] + 1 == continue_at: max_count = loop_counts[-1] current = curr_iter.pop() + 1 if current < max_count: curr_iter.append(current) index = find_loop_start(kpr, index) continue elif command.startswith("print"): message = command[command.index("\"") + 1:command.rindex("\"")] output.append(message) index += 1 print("\n".join(output)) def find_loop_start(kpr, ci): nested_loops = 0 for i in range(ci - 1, -1, -1): if kpr[i] == "done": nested_loops += 1 elif kpr[i] == "do": if nested_loops == 0: return i nested_loops -= 1 return 0 def find_loop_end(kpr, ci): nested_loops = 0 for i in range(ci + 1, len(kpr)): if kpr[i] == "do": nested_loops += 1 elif kpr[i] == "done": if nested_loops == 0: return i nested_loops -= 1 return len(kpr) if name == "main": n = int(input()) kpr = [input().strip() for _ in range(n)] process_commands(kpr) Loop MASTER Python done ✅✅✅

Office rostering code

#include <iostream> #include <vector> #include <set> #include <algorithm> using namespace std; int main() { int n, m, k, days = 1, activeCount = 0; cin >> n >> m; vector<set<int>> connections(n); for (int i = 0, u, v; i < m; ++i) { cin >> u >> v; connections[u].insert(v); connections[v].insert(u); } cin >> k; vector<bool> active(n, true); activeCount = n; while (activeCount < k) { vector<bool> nextState(n, false); for (int i = 0; i < n; ++i) { int neighborCount = 0; for (int neighbor : connections[i]) { neighborCount += active[neighbor]; } if (active[i] && neighborCount == 3) { nextState[i] = true; } else if (!active[i] && neighborCount < 3) { nextState[i] = true; } } active = nextState; activeCount += count(active.begin(), active.end(), true); ++days; } cout << days; return 0; }

#include <bits/stdc++.h> using namespace std; typedef long long ll; int main() { int numLines; cin >> numLines; vector<vector<pair<int, int>>> paths(numLines); map<pair<int, int>, vector<int>> pointTracker; for (int i = 0; i < numLines; i++) { int x1, y1, x2, y2; cin >> x1 >> y1 >> x2 >> y2; int dx = x2 - x1, dy = y2 - y1; int steps = max(abs(dx), abs(dy)); int stepX = (dx == 0) ? 0 : dx / abs(dx); int stepY = (dy == 0) ? 0 : dy / abs(dy); for (int j = 0; j <= steps; j++) { int curX = x1 + stepX * j; int curY = y1 + stepY * j; paths[i].emplace_back(make_pair(curX, curY)); pointTracker[{curX, curY}].emplace_back(i); } } string lineInput; getline(cin, lineInput); getline(cin, lineInput); unordered_map<string, int> limitMap; int pos = 0, lineLength = lineInput.size(); while (pos < lineLength) { size_t delimiterPos = lineInput.find(':', pos); if (delimiterPos == string::npos) break; string key = lineInput.substr(pos, delimiterPos - pos); pos = delimiterPos + 1; size_t spacePos = lineInput.find(' ', pos); if (spacePos == string::npos) spacePos = lineLength; int value = stoi(lineInput.substr(pos, spacePos - pos)); limitMap[key] = value; pos = spacePos + 1; } string query; cin >> query; ll totalCost = 0; for (auto &entry : pointTracker) { if (entry.second.size() >= 2) { int commonSize = entry.second.size(); int minCost = INT_MAX; for (auto segId : entry.second) { auto &currentPath = paths[segId]; size_t pathLength = currentPath.size(); size_t index = find(currentPath.begin(), currentPath.end(), entry.first) - currentPath.begin(); int leftDistance = index; int rightDistance = pathLength - index - 1; int cost = (leftDistance > 0 && rightDistance > 0) ? min(leftDistance, rightDistance) : max(leftDistance, rightDistance); minCost = min(minCost, cost); } totalCost += (ll)commonSize * minCost; } } if (limitMap.find(query) != limitMap.end()) { if (totalCost >= limitMap[query]) { cout << "Yes\n"; } else { cout << "No\n"; } } else { cout << "No\n"; } int validItems = 0, totalItems = limitMap.size(); for (auto &entry : limitMap) { if (totalCost >= entry.second) { validItems++; } } double successRate = (double)validItems / totalItems; cout << fixed << setprecision(2) << successRate; return 0; }

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string s; cin >> s; int n = s.length(), res = 0; vector v(n); for (int i = 0; i < n; ++i) cin >> v[i]; int lw = s[0] - '0', lwv = v[0]; for (int i = 1; i < n; ++i) { if (s[i] - '0' == lw) { res += min(lwv, v[i]); lwv = max(lwv, v[i]); } else { lw = s[i] - '0'; lwv = v[i]; } } cout << res; return 0; } Alternative string🔥🔥🔥🔥

import java.util.*; public class RitikaTask { private static int[] canFormWithDeletions(String sub, String mainStr, int maxDeletions) { int i = 0, j = 0, deletionsUsed = 0; while (i < mainStr.length() && j < sub.length()) { if (mainStr.charAt(i) == sub.charAt(j)) { i++; j++; } else { deletionsUsed++; if (deletionsUsed > maxDeletions) { return new int[] {i, deletionsUsed}; } i++; } } return new int[] {i, deletionsUsed}; } private static boolean canMatchCharacter(char c, List<String> substrings) { for (String sub : substrings) { if (sub.indexOf(c) >= 0) { return true; } } return false; } public static String solveRitikaTask(List<String> substrings, String mainStr, int k) { int n = mainStr.length(); int deletionsUsed = 0; StringBuilder formedString = new StringBuilder(); boolean isAnyMatch = false; for (int i = 0; i < n; i++) { if (!canMatchCharacter(mainStr.charAt(i), substrings)) { return "Impossible"; } } for (int i = 0; i < n;) { boolean matched = false; for (String sub : substrings) { int[] result = canFormWithDeletions(sub, mainStr.substring(i), k - deletionsUsed); int newIndex = result[0]; int usedDeletions = result[1]; if (newIndex > 0) { matched = true; isAnyMatch = true; formedString.append(mainStr.substring(i, i + newIndex)); i += newIndex; deletionsUsed += usedDeletions; break; } } if (!matched) { break; } } if (formedString.length() == n) { if (deletionsUsed <= k) { return "Possible"; } else { return formedString.toString().trim(); } } else if (isAnyMatch) { return "Nothing"; } else if (deletionsUsed > k) { return formedString.toString().trim(); } else { return formedString.toString().trim(); } } public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int N = scanner.nextInt(); scanner.nextLine(); List<String> substrings = new ArrayList<>(); for (int i = 0; i < N; i++) { substrings.add(scanner.nextLine()); } String mainStr = scanner.nextLine(); int K = scanner.nextInt(); String result = solveRitikaTask(substrings, mainStr, K); System.out.print(result); scanner.close(); } } Help ritika code with all tets caes passed 🔥🔥🔥🔥🔥 @Codeing_are

https://t.me/codeing_are Don't forgets to share the group 😔

NEED more TCS CODEVITA ANSWERS ? Give heart to this post Fast after 100 hit I will post again ☺️

All code with all tets caes passed 🔥🥳🥳🥳🥳

Office rostering code

#include <iostream> #include <vector> #include <set> #include <algorithm> using namespace std; int main() { int n, m, k, days = 1, activeCount = 0; cin >> n >> m; vector<set<int>> connections(n); for (int i = 0, u, v; i < m; ++i) { cin >> u >> v; connections[u].insert(v); connections[v].insert(u); } cin >> k; vector<bool> active(n, true); activeCount = n; while (activeCount < k) { vector<bool> nextState(n, false); for (int i = 0; i < n; ++i) { int neighborCount = 0; for (int neighbor : connections[i]) { neighborCount += active[neighbor]; } if (active[i] && neighborCount == 3) { nextState[i] = true; } else if (!active[i] && neighborCount < 3) { nextState[i] = true; } } active = nextState; activeCount += count(active.begin(), active.end(), true); ++days; } cout << days; return 0; }

Block extraction

#include <bits/stdc++.h> using namespace std; int main() { int x, y; cin >> x >> y; vector<vector<int>> z(x, vector<int>(y)); for (int i = 0; i < x; ++i) { for (int j = 0; j < y; ++j) { cin >> z[i][j]; } } int w; cin >> w; map<int, vector<pair<int, int>>> a; set<int> b; for (int i = 0; i < x; ++i) { for (int j = 0; j < y; ++j) { int c = z[i][j]; a[c].emplace_back(i, j); b.insert(c); } } set<int> d; for (int i = 0; i < x; ++i) { vector<int> e; for (int j = 0; j < y; ++j) { if (z[i][j] == w) { e.push_back(j); } } if (!e.empty()) { int f = *max_element(e.begin(), e.end()); for (int j = f + 1; j < y; ++j) { int g = z[i][j]; if (g != w) { d.insert(g); } } } } set<int> h = b; d.erase(w); int i = 0; for (auto j = d.begin(); j != d.end(); ++j) { if (h.find(*j) != h.end()) { h.erase(*j); i++; } } auto k = [&](const set<int>& l) -> set<int> { set<int> m; for (auto& n : l) { for (auto& [o, p] : a[n]) { if (o == x - 1) { m.insert(n); break; } } } queue<int> q; for (auto& r : m) { q.push(r); } while (!q.empty()) { int s = q.front(); q.pop(); for (auto& t : l) { if (m.find(t) != m.end()) continue; bool u = false; for (auto& [v, w] : a[t]) { if (v + 1 < x) { int x = z[v + 1][w]; if (m.find(x) != m.end()) { u = true; break; } } } if (u) { m.insert(t); q.push(t); } } } return m; }; while (true) { set<int> y = k(h); set<int> z; for (auto& aa : h) { if (y.find(aa) == y.end()) { z.insert(aa); } } if (z.empty()) break; for (auto& bb : z) { h.erase(bb); i++; } } cout << i; return 0; }

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