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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 153Taʼlim14 642...

📈 Telegram kanali ACCENTURE EXAM SOLUTIONS analitikasi

ACCENTURE EXAM SOLUTIONS (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 13 954 obunachidan iborat bo'lib, Taʼlim toifasida 14 642-o'rinni va Hindiston mintaqasida 28 153-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 13 954 obunachiga ega bo‘ldi.

02 Noyabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -150 ga, so‘nggi 24 soatda esa -1 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

Tasdiqlash holati: Tasdiqlanmagan
Jalb etish (ER): Auditoriya o‘rtacha 3.64% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.39% ini tashkil etuvchi reaksiyalarni to‘playdi.
Post qamrovi: Har bir post o‘rtacha 508 marta ko‘riladi; birinchi sutkada odatda 194 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 03 Noyabr, 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.

13 954

Obunachilar

-124 soatlar

-317 kun

-15030 kun

508

Post ko'rishlar

~ 19424 soatlar

~ 26748 soatlar

3.64%

Muloqot nisbati

~ 3

Kuniga postlar

Ads index

Postlar arxiv

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Those who have TCS, COGNIZANT, HEXAWARE,TECH MAHINDRA,FEDEX,ZOHO, ACCENTURE OR ANY OTHER EXAM help available Scroll up and check past results 😃 Both Coding and Aptitude help available Contact @srksvkfor help 🙏 💯 Clearance and genuine help Note: it's paid help

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Array A length n code

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C++

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magic Number Code python 3 All test cases passed

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#include <bits/stdc++.h> using namespace std; int count(string s) { int N, i, cnt = 0, ans = 0; N = s.length(); for (i = 0; i < N; i++) { if (s[i] == 'R') cnt++; if (s[i] == 'L') ans += cnt; } return ans; } int main() { string s = "RRLL"; cout << count(s) << endl; return 0; } C++

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Share the group everyone

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if (n > k && n > a) { printf("N is the Biggest number is %d", n); return n; } if (k > n && k > a) { printf("k is the Biggest number is %d", k); return k; } if (a > n && a > k) { printf("a is the Biggest number is %d", a); return a; }

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def getLargestString(s, k): frequency_array = [0] * 26 for i in range(len(s)): frequency_array[ord(s[i]) - ord('a')] += 1 ans = "" i = 25 while i >= 0: if (frequency_array[i] > k): temp = k st = chr( i + ord('a')) while (temp > 0): ans += st temp -= 1 frequency_array[i] -= k j = i - 1 while (frequency_array[j] <= 0 and j >= 0): j -= 1 if (frequency_array[j] > 0 and j >= 0): str1 = chr(j + ord( 'a')) ans += str1 frequency_array[j] -= 1 else: break elif (frequency_array[i] > 0): temp = frequency_array[i] frequency_array[i] -= temp st = chr(i + ord('a')) while (temp > 0): ans += st temp -= 1 else: i -= 1 return ans if name == "main": S = input() k = 3 print (getLargestString(S, k)) Python Bob code

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from itertools import permutations def count(arr): z=[] perm = permutations(arr) for i in list(perm): z.append(list(i)) q=[] for i in range(len(arr)-1): x,y=arr[i],arr[i+1] for j in range(len(z)): if z[j].index(x)!=len(z[j])-1: if z[j][z[j].index(x)+1]==y: q.append(z[j]) for i in range(len(q)): if q[i] in z: z.remove(q[i]) return len(z) a= int(input()) b=list(map(int,input().strip().split())) print(count(b))

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Java

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