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ACCENTURE EXAM SOLUTIONS

ACCENTURE EXAM SOLUTIONS

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πŸ”₯Guys plz Stop fearing for daily exams πŸ“ πŸ‘¨β€πŸ’» @srksvk is here to help you all at lowest cost possible.πŸ’ͺ πŸŒ€ ” Our Only Aim Is To Let Get Placed To You In A Reputed Company πŸ”₯Effort from our side = πŸ’― πŸ“±Main Channel: @coding_are πŸ“±Tel I'd : @srksvk

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πŸ“ˆ Analytical overview of Telegram channel ACCENTURE EXAM SOLUTIONS

Channel ACCENTURE EXAM SOLUTIONS (@coding_are) in the English language segment is an active participant. Currently, the community unites 14 128 subscribers, ranking 14 097 in the Education category and 28 067 in the India region.

πŸ“Š Audience metrics and dynamics

Since its creation on Π½Π΅Π²Ρ–Π΄ΠΎΠΌΠΎ, the project has demonstrated rapid growth, gathering an audience of 14 128 subscribers.

According to the latest data from 26 September, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -112 over the last 30 days and by -6 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 3.64%. Within the first 24 hours after publication, content typically collects 1.54% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 514 views. Within the first day, a publication typically gains 218 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.

πŸ“ Description and content policy

The author describes the resource as a platform for expressing subjective opinions:
β€œπŸ”₯Guys plz Stop fearing for daily exams πŸ“ πŸ‘¨β€πŸ’» @srksvk is here to help you all at lowest cost possible.πŸ’ͺ πŸŒ€ ” Our Only Aim Is To Let Get Placed To You In A Reputed Company πŸ”₯Effort from our side = πŸ’― πŸ“±Main Channel: @coding_are πŸ“±Tel I'd : @srks...”

Thanks to the high frequency of updates (latest data received on 27 September, 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.

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Capgemini exam cleard πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸ”₯πŸ”₯πŸŽ‰ Got next round mail πŸ’ŒπŸ’ŒπŸ’ŒπŸ’ŒπŸ’ŒπŸ’ŒπŸ’Œ Helped proof πŸ‘‡πŸ‘‡
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Capgemini exam cleard πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸŽ‰πŸ”₯πŸ”₯πŸŽ‰ Got next round mail πŸ’ŒπŸ’ŒπŸ’ŒπŸ’ŒπŸ’ŒπŸ’ŒπŸ’Œ Helped proof πŸ‘‡πŸ‘‡πŸ‘‡πŸ‘‡πŸ‘‡πŸ‘‡πŸ‘‡ https://t.me/codeing_are/12204?single Contact for placement exam @srksvk

Capgemini codeing round help successfully completed by Remote access to πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸŽ‰ 2/2 code done with all tets caes passed
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Capgemini codeing round help successfully completed by Remote access to πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸŽ‰ 2/2 code done with all tets caes passed πŸ”₯πŸ”₯πŸ”₯πŸ”₯ Contact for placement exam @srksvk

Capgemini communication help successfully done by Remote access πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯ All answers provided
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Capgemini communication help successfully done by Remote access πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯ All answers provided on time with πŸ’―% Correct answer πŸ”₯πŸ”₯πŸ”₯ Contact for placement exam @srksvk Slot 2

Capgemini communication help successfully done by Remote access πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯ All answers provided
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Capgemini communication help successfully done by Remote access πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯ All answers provided on time with πŸ’―% Correct answer πŸ”₯πŸ”₯πŸ”₯ Contact for placement exam @srksvk

Cognizant communication help successfully done by Remote access πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯ All answers provided on time with πŸ’―
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Cognizant communication help successfully done by Remote access πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯ All answers provided on time with πŸ’― correct answer πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯πŸ”₯ 77/77 questions done βœ…βœ… Contact for placement exam @srksvk

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from collections import deque import itertools def get_shortest_path(grid, N): start = None end = None for i in range(N): for j in range(N): if grid[i][j] == 'S': start = (i, j) elif grid[i][j] == 'D': end = (i, j) queue = deque([(start, 0)]) visited = {start} while queue: (x, y), dist = queue.popleft() if grid[x][y] == 'D': return dist for nx, ny in [(x+1, y), (x-1, y), (x, y+1), (x, y-1)]: if 0 <= nx < N and 0 <= ny < N and (nx, ny) not in visited and grid[nx][ny] != 'T': visited.add((nx, ny)) queue.append(((nx, ny), dist + 1)) return float('inf') def get_sheets(grid, N, M): sheets = [] for i in range(0, N, M): for j in range(0, N, M): sheet = [] for x in range(M): row = [] for y in range(M): row.append(grid[i+x][j+y]) sheet.append(row) sheets.append(sheet) return sheets def make_grid(arrangement, sheets, N, M): grid = [["" for _ in range(N)] for _ in range(N)] num_sheets = N // M for idx, sheet_idx in enumerate(arrangement): sheet = sheets[sheet_idx] base_i = (idx // num_sheets) * M base_j = (idx % num_sheets) * M for i in range(M): for j in range(M): grid[base_i + i][base_j + j] = sheet[i][j] return grid def solve(): N, M = map(int, input().split()) original_grid = [list(input().strip()) for _ in range(N)] sheets = get_sheets(original_grid, N, M) num_sheets = (N // M) ** 2 s_sheet = d_sheet = None for i, sheet in enumerate(sheets): for row in sheet: if 'S' in row: s_sheet = i if 'D' in row: d_sheet = i min_dist = float('inf') nums = list(range(num_sheets)) nums.remove(s_sheet) nums.remove(d_sheet) for middle_perm in itertools.permutations(nums): arrangement = [s_sheet] + list(middle_perm) + [d_sheet] grid = make_grid(arrangement, sheets, N, M) min_dist = min(min_dist, get_shortest_path(grid, N)) return min_dist if name == "main": print(solve()) Arrange Map - Codevita βœ…

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#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; }

Floded are

struct P {     double a, b;     P(double a = 0, double b = 0) : a(a), b(b) {} }; P r(const P &p, double t) {     return P(p.a * cos(t) - p.b * sin(t),              p.a * sin(t) + p.b * cos(t)); } pair g(const vector

&q, double t) {     double x1 = 1e9, x2 = -1e9, y1 = 1e9, y2 = -1e9;     for (const auto &p : q) {         P s = r(p, t);         x1 = min(x1, s.a);         x2 = max(x2, s.a);         y1 = min(y1, s.b);         y2 = max(y2, s.b);     }     return {x2 - x1, y2 - y1}; } int main() {     int n;     cin >> n;     vector

q(n);     for (int i = 0; i < n; ++i) {         cin >> q[i].a >> q[i].b;     }     double m = 1e9, w = 0, h = 0;     for (int i = 0; i < 360; ++i) {         double t = i * M_PI / 180.0;         auto [cw, ch] = g(q, t);         double a = cw * ch;         if (a < m) {             m = a;             w = cw;             h = ch;         }     }     if (w > h) {         swap(w, h);     }     cout << fixed << setprecision(0) << round(w) << " "          << fixed << setprecision(0) << round(h);     return 0; }

Fast

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

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That code shared join fast ⏩

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