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
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ACCENTURE EXAM SOLUTIONS (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 14 133 obunachidan iborat bo'lib, Taʼlim toifasida 14 101-o'rinni va Hindiston mintaqasida 28 065-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 14 133 obunachiga ega bo‘ldi.
25 Sentabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -111 ga, so‘nggi 24 soatda esa -3 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 3.69% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.55% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 522 marta ko‘riladi; birinchi sutkada odatda 219 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 2 ta reaksiya keladi.
- Tematik yo‘nalishlar: Kontent placement, gaurntee, suree, capgemini, infosy kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“🔥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...”
Yuqori yangilanish chastotasi (oxirgi ma’lumot 26 Sentabr, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.
14 133
Obunachilar
-324 soatlar
-157 kun
-11130 kun
Postlar arxiv
14 133
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14 133
Ab inbev exam help available
Contact fast and book your slots
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14 133
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14 133
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 ✅
14 133
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;
}
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14 133
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14 133
#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;
}
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Final call,
IBM 11am exam slot available
Contact fast and book your slots l
Contact @srksvk
200% suree clearance grauntee 🔥
Remote slot also availableNote:- one slot available
14 133
Final call,
IBM 11am exam slot available
Contact fast and book your slots l
Contact @srksvk
200% suree clearance grauntee 🔥
Remote slot also available
Note:- one slot available
14 133
IBM 11am exam slot available
Contact fast and book your slots
Contact @srksvk
200% suree clearance grauntee 🔥
Remote slot also available
14 133
IBM 11am exam slot available
Contact fast and book your slots
Contact @srksvk
200% suree clearance grauntee 🔥
Remote slot also available
14 133
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14 133
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Contact @srksvk
200% suree clearance grauntee 🔥Remote slot also available
