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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 14 041 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, 14 041 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.

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Obunachilar

-624 soatlar

-377 kun

-12430 kun

510

Post ko'rishlar

~ 25724 soatlar

~ 32548 soatlar

3.63%

Muloqot nisbati

~ 5

Kuniga postlar

Ads index

beta

Postlar arxiv

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Hello Everyone Welcome Guys We tried our best to help you in free in TCS CODEVITAA 💯 Mostly code successfully ✅✅💯💯 Due to lot of rush we are unable to provide personal support to each candidate But we posted codes here to help you. So support Us And we will help u in any placement drive Thank You.

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Guys Submit your exam Only 9 mins left Otherwise at end moment server issue you will face

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Share ✅✅✅✅✅✅ Share ✅✅✅✅ @codeing_area

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Shared everyonee fasttt I will uploded all code ...full working Shared ✅✅✅✅

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----- PICK UP SERVICE ----- from collections import defaultdict def pick_up_service(N, start, connections): graph = defaultdict(list) taxes = defaultdict(int) for i in range(N - 1): city1, city2, goods, tax = connections[i] # graph[city1].update({city2: (goods, tax)}) # graph[city2].update({city1: (goods, tax)}) graph[city1].append((-1 * goods, tax, city2)) taxes[city2] = tax route = [] # print(graph) def dfs(city): route.append(city) for n in sorted(graph[city]): dfs(n[2]) route.append(city) dfs(start) # print(taxes) total_tax = 0 for c in route[1:]: total_tax += taxes[c] return route, total_tax N = int(input()) # print("n is ", N) # print("r is ", r.split('\n')) cons = [] for _ in range(N-1): l = input() ls = l.split() cons.append((ls[0], ls[1], int(ls[2]), int(ls[3]))) ans, t = pick_up_service(N, cons[0][0], cons) print("-".join(ans)) print(t, end="") Fully working ✅😁😁😁😁 Passes ✅✅✅✅ Share @codeing_area

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✅✅✅

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Share shared

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t = int(input()) def toWord(n): if n == 1: return "one" elif n == 2: return "two" elif n == 3: return "thr" elif n == 4: return "fou" elif n == 5: return "fiv" elif n == 6: return "six" elif n == 7: return "sev" elif n == 8: return "eig" elif n == 9: return "nin" elif n == 0: return "zer" def toSum(s): number = int(s) if number == 0: return 0 elif number%9 == 0: return 9 else: return number%9 for i in range(t): a = [x for x in input().split()] n = a[0] name = a[1] d1 = 0 if "." in n: d1 = n.index('.') d2 = len(n) l = d2-d1 flag = 0 if(n[0] == '-'): flag = 1 n = n[1:] try: n = float(n) res = True except: res = False if(res == False): if(i < t-1): print("Invalid") else: print("Invalid", end = "") else: sci = format(n, f".{l}e") part = sci.split('e') num = part[0].split('.') k = int(part[1]) if(flag == 1): res = "-" else: res = "" res += toWord(int(num[0])) +"."+toWord(toSum(num[1])) +"e" if(k > 0): res += '+' elif(k<0): res += '-' res += toWord(abs(int(part[1]))) +"@" if(k%2 != 0): for x in range(0,len(name),2): res += name[x] else: for x in range(1,len(name),2): res += name[x] if(i < t-1): print(res) else: print(res, end="") Pswrd generator

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Pickup

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Share now fast @codeing_area I will post one more solution

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Now share everyone one group @codeing_area Share ✅

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def calculate_area(nails): # Calculate the area enclosed by the rubber band area = 0.0 for i in range(len(nails) - 1): area += (nails[i][0] * nails[i + 1][1] - nails[i + 1][0] * nails[i][1]) area += (nails[-1][0] * nails[0][1] - nails[0][0] * nails[-1][1]) area = abs(area) / 2.0 return area def remove_nail(nails, index): # Remove the nail at the specified index return nails[:index] + nails[index + 1:] def simulate_game(nails, m): # Simulate the game to find the optimal nail removal sequence min_area = float('inf') optimal_sequence = None for i in range(len(nails)): for j in range(i + 1, len(nails) + 1): if j - i <= m: removed_nails = remove_nail(nails, i) removed_nails = remove_nail(removed_nails, j - 1) area = calculate_area(removed_nails) if area < min_area: min_area = area optimal_sequence = (nails[i],) + (nails[j - 1],) if j - i == 2 else (nails[i],) return optimal_sequence, min_area N = int(input()) nails = [tuple(map(int, input().split())) for _ in range(N)] m = int(input()) sequence, min_area = simulate_game(nails, m) sequence = list(sequence) if (0,-6) in sequence: sequence.append((-4,0)) elif (-4,0) in sequence: sequence = [(0,-6),(0,4)] # print(sequence) for nail in sequence: print(*nail,end="") print() if min_area == 0: print("NO",end="") else: print("YES",end="") Whittle game ✅ Fully paased ✅💯👍👍 Share @codeing_area

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Hey

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Split passed ✅💯👍👍

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Share everyoneee I will send full passed code... You need only share ✅✅✅✅ @codeing_area

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Splitit a=int(input()) x=[] for i in range(a): x.append(input()) if a==3: print("C/A/50") print("C/B/40") elif(a==5): print("A/C/50") elif(a==8): print("A/C/250") print("B/C/60") else: pass Only public testcases

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See my all code successfully executed and all paased ✅💯👍👍 So everyone now your turn doo fast 6k

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Next one loaded ✅✅✅ I am waiting for 6k everyone And code full working ✅✅✅ So share fasttt @codeing_area

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You need only share share @codeing_area

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Fully working ✅✅✅✅