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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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📈 Analytical overview of Telegram channel MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

Channel MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer (@coding_are) in the English language segment is an active participant. Currently, the community unites 13 241 subscribers, ranking 15 362 in the Education category and 32 092 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 13 241 subscribers.

According to the latest data from 18 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -138 over the last 30 days and by -2 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 2.93%. Within the first 24 hours after publication, content typically collects 1.11% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 388 views. Within the first day, a publication typically gains 147 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 2.
  • Thematic interests: Content is focused on key topics such as placement, gaurntee, suree, capgemini, infosy.

📝 Description and content policy

The author describes the resource as a platform for expressing subjective opinions:
🔥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...

Thanks to the high frequency of updates (latest data received on 19 June, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.

13 241
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Block Extraction C++ TCS CodeVita Zone 2 #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; } Block Extraction C++ TCS CodeVita Zone 2 @codeing_are

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#include <iostream> #include <vector> #include <string> #include <climits> using namespace std; int minCostToFormString(int n, vector<pair<string, int>> &substrings, string mainString) {     int m = mainString.length();     vector<int> dp(m + 1, INT_MAX);     dp[0] = 0;     for (int i = 0; i <= m; ++i) {         if (dp[i] == INT_MAX) continue;         for (auto &sub : substrings) {             string substring = sub.first;             int cost = sub.second;             int subLen = substring.length();                  if (i + subLen <= m && mainString.substr(i, subLen) == substring) {                 dp[i + subLen] = min(dp[i + subLen], dp[i] + cost);             for (int k = 1; k < subLen; ++k) {                 if (i + k <= m && mainString.substr(i, k) == substring.substr(0, k)) {                     dp[i + k] = min(dp[i + k], dp[i] + cost);                 }             }             }              }     }     return (dp[m] == INT_MAX) ? -1 : dp[m]; } int main() {     int n;     cin >> n;     vector<pair<string, int>> substrings;     for (int i = 0; i < n; ++i) {         string substring;         int cost;         cin >> substring >> cost;         substrings.push_back({substring, cost});     }     string mainString;     cin >> mainString;     int result = minCostToFormString(n, substrings, mainString);     if (result == -1) {         cout << "Impossible" ;     } else {         cout << result ;     }     return 0; } form the string c++ code

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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 ✅✅✅

#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

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