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ACCENTURE EXAM SOLUTIONS

Kanalga Telegram’da o‘tish

🔥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 436 Taʼlim14 513...

📈 Telegram kanali ACCENTURE EXAM SOLUTIONS analitikasi

ACCENTURE EXAM SOLUTIONS (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 14 042 obunachidan iborat bo'lib, Taʼlim toifasida 14 513-o'rinni va Hindiston mintaqasida 28 436-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 14 042 obunachiga ega bo‘ldi.

13 Oktabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -124 ga, so‘nggi 24 soatda esa -6 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

Tasdiqlash holati: Tasdiqlanmagan
Jalb etish (ER): Auditoriya o‘rtacha 3.63% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.83% ini tashkil etuvchi reaksiyalarni to‘playdi.
Post qamrovi: Har bir post o‘rtacha 510 marta ko‘riladi; birinchi sutkada odatda 257 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 14 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.

14 042

Obunachilar

-624 soatlar

-377 kun

-12430 kun

510

Post ko'rishlar

~ 25724 soatlar

~ 32548 soatlar

3.63%

Muloqot nisbati

~ 5

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int f(int i, int j, int flag, vector& arr, int sum, vector>>& dp) { if (i >= j) return 0; if (dp[i][j][flag] != -1) return dp[i][j][flag]; int firsttwo = 0; int lasttwo = 0; int firstlast = 0; if (i + 1 < arr.size() && arr[i] + arr[i + 1] == sum) { firsttwo = 1 + f(i + 2, j, flag, arr, sum, dp); } if (arr[i] + arr[j] == sum) { firstlast = 1 + f(i + 1, j - 1, flag, arr, sum, dp); } if (j >= 0 && arr[j] + arr[j - 1] == sum) { lasttwo = 1 + f(i, j - 2, flag, arr, sum, dp); } return dp[i][j][flag] = max({firsttwo, lasttwo, firstlast}); } int solution(vector& arr) { int n = arr.size(); vector>> dp(n, vector>(n, vector(3, -1))); int sum1 = arr[0] + arr[n - 1]; int sum2 = arr[0] + arr[1]; int sum3 = arr[n - 1] + arr[n - 2]; int ans1 = f(0, n - 1, 0, arr, sum1, dp); int ans2 = f(0, n - 1, 1, arr, sum2, dp); int ans3 = f(0, n - 1, 2, arr, sum3, dp); int res = max(ans1, max(ans2, ans3)); return res; }

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#include <bits/stdc++.h> using namespace std; int solution(vector<int>& A) { sort(A.begin(), A.end()); if (A.size() <= 2) { return A.size(); } int maxCount = 0; unordered_map<int, int> differences[A.size()]; for (int i = 1; i < A.size(); i++) { for (int j = 0; j < i; j++) { int currentDiff = A[i] - A[j]; int currentCnt = 1; if (differences[j].count(currentDiff) != 0) { currentCnt += differences[j][currentDiff]; differences[i][currentDiff] = currentCnt; } else { differences[i][currentDiff] = 1 + currentCnt; } maxCount = max(maxCount, differences[i][currentDiff]); } } return maxCount; } task 2

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