ACCENTURE EXAM HELP ! CISCO EXAM !

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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

Hindiston31 462 Taʼlim15 244...

📈 Telegram kanali ACCENTURE EXAM HELP ! CISCO EXAM ! analitikasi

ACCENTURE EXAM HELP ! CISCO EXAM ! (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 13 202 obunachidan iborat bo'lib, Taʼlim toifasida 15 244-o'rinni va Hindiston mintaqasida 31 462-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 13 202 obunachiga ega bo‘ldi.

29 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -155 ga, so‘nggi 24 soatda esa -4 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

Tasdiqlash holati: Tasdiqlanmagan
Jalb etish (ER): Auditoriya o‘rtacha 3.03% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.29% ini tashkil etuvchi reaksiyalarni to‘playdi.
Post qamrovi: Har bir post o‘rtacha 400 marta ko‘riladi; birinchi sutkada odatda 170 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 30 Iyun, 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.

13 202

Obunachilar

-424 soatlar

-177 kunlar

-15530 kunlar

400

Post ko'rishlar

~ 17024 soatlar

~ 24048 soatlar

3.03%

Muloqot nisbati

~ 3

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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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