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

Ko'proq ko'rsatish

Hindiston28 155 Taʼlim14 131...

📈 Telegram kanali ACCENTURE EXAM SOLUTIONS analitikasi

ACCENTURE EXAM SOLUTIONS (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 14 088 obunachidan iborat bo'lib, Taʼlim toifasida 14 131-o'rinni va Hindiston mintaqasida 28 155-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 14 088 obunachiga ega bo‘ldi.

05 Oktabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -120 ga, so‘nggi 24 soatda esa -12 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

Tasdiqlash holati: Tasdiqlanmagan
Jalb etish (ER): Auditoriya o‘rtacha 3.51% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.81% ini tashkil etuvchi reaksiyalarni to‘playdi.
Post qamrovi: Har bir post o‘rtacha 495 marta ko‘riladi; birinchi sutkada odatda 255 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 06 Oktabr, 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.

14 088

Obunachilar

-1224 soatlar

-327 kun

-12030 kun

495

Post ko'rishlar

~ 25524 soatlar

~ 31648 soatlar

3.51%

Muloqot nisbati

~ 4

Kuniga postlar

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14 088

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#include <vector> const int MOD = 1000000007; int countPrettyPartitions(int N, int L, int R, std::vector<int>& A) { std::vector<int> dp(N + 1, 0); dp[0] = 1; for (int i = 1; i <= N; ++i) { int xor_value = 0; for (int j = i; j >= 1; --j) { xor_value ^= A[j - 1]; if (L <= xor_value && xor_value <= R) { dp[i] = (dp[i] + dp[j - 1]) % MOD; } } } return dp[N]; } Dividing array

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int findTotal(string s){ unordered_setst; for(int i=0;i

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After 12k I willl post all

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Fully passed

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#include <iostream> #include <vector> #include <algorithm> long long solve(int N, std::vector<int>& A) { std::sort(A.begin(), A.end()); long long minSum = A[0]; int prev = A[0]; for (int i = 1; i < N; ++i) { if (A[i] <= prev) { A[i] = prev + 1; } prev = A[i]; minSum += A[i]; } return minSum; } .. Minimum unique sum 🙃🙃

14 088

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