ACCENTURE 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

India28 065 Education14 101...

📈 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 14 133 subscribers, ranking 14 101 in the Education category and 28 065 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 14 133 subscribers.

According to the latest data from 25 September, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -111 over the last 30 days and by -3 over the last 24 hours, overall reach remains high.

Verification status: Not verified
Engagement rate (ER): The average audience engagement rate is 3.69%. Within the first 24 hours after publication, content typically collects 1.55% reactions from the total number of subscribers.
Post reach: On average, each post receives 522 views. Within the first day, a publication typically gains 219 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 26 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.

14 128

Subscribers

-324 hours

-157 days

-11130 days

522

Post views

~ 21924 hours

~ 29048 hours

3.69%

Engagement rate

~ 3

Posts per day

Ads index

beta

Posts Archive

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from collections import deque, defaultdict def Solve(n, r, t, k, s, e): graph = defaultdict(list) for u, v in r: graph[u].append(v) graph[v].append(u) visited_t = set(t) queue_t = deque([(th, 0) for th in t]) while queue_t: current_t, distance_t = queue_t.popleft() if distance_t < k: for neighbor in graph[current_t]: if neighbor not in visited_t: visited_t.add(neighbor) queue_t.append((neighbor, distance_t + 1)) visited_s = set() queue_s = deque([(s, 0)]) while queue_s: current_s, distance_s = queue_s.popleft() if current_s in visited_s or current_s in visited_t: continue if current_s == e: return distance_s visited_s.add(current_s) for neighbor in graph[current_s]: if neighbor not in visited_s and neighbor not in visited_t: queue_s.append((neighbor, distance_s + 1)) return -1 Uber ✅

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typedef long long ll; ll solution(int n, vector a1, vector a2) { vector left(n + 1, 0), right(n + 2, 0), prefixSum(n + 1, 0), maxL(n + 1, 0), maxM(n + 1, LLONG_MIN); for (int i = 1; i <= n; i++) { left[i] = i == 1 ? a1[0] : max(left[i - 1] + (ll)a1[i - 1], (ll)a1[i - 1]); } ll res = LLONG_MIN; for (int i = 1; i <= n; i++) { res = max(res, left[i]); } for (int i = n; i >= 1; i--) { right[i] = i == n ? (ll)a1[n - 1] : max(right[i + 1] + (ll)a1[i - 1], (ll)a1[i - 1]); } for (int i = 1; i <= n; i++) { prefixSum[i] = prefixSum[i - 1] + (ll)a2[i - 1]; } maxL[1] = left[0] - prefixSum[0]; maxM[1] = maxL[1]; for (int i = 2; i <= n; i++) { maxL[i] = left[i - 1] - prefixSum[i - 1]; maxM[i] = max(maxM[i - 1], maxL[i]); } for (int j = 1; j <= n; j++) { res = max(res, maxM[j] + prefixSum[j] + right[j + 1]); } return res; } Tichnas sir and bonds mam Uber ✅

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#include <iostream> #include <string> #include <unordered_map> #include <queue> using namespace std; string reorganizeString(string s) { unordered_map<char, int> freq_map; for (char c : s) { freq_map[c]++; } priority_queue<pair<int, char>> max_heap; for (auto &[ch, freq] : freq_map) { max_heap.push({freq, ch}); } string res; while (max_heap.size() >= 2) { auto [freq1, char1] = max_heap.top(); max_heap.pop(); auto [freq2, char2] = max_heap.top(); max_heap.pop(); res += char1; res += char2; if (--freq1 > 0) max_heap.push({freq1, char1}); if (--freq2 > 0) max_heap.push({freq2, char2}); } if (!max_heap.empty()) { auto [freq, ch] = max_heap.top(); if (freq > 1) return ""; res += ch; } return res; } int main() { string input; cout << "Enter a string: "; cin >> input; string result = reorganizeString(input); if (result.empty()) { cout << "The string cannot be reorganized to avoid adjacent repeating characters." << endl; } else { cout << "Reorganized string: " << result << endl; } return 0; } IBM ✅

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