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ACCENTURE EXAM HELP ! CISCO EXAM !

ACCENTURE EXAM HELP ! CISCO EXAM !

الذهاب إلى القناة على Telegram

🔥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

إظهار المزيد

📈 نظرة تحليلية على قناة تيليجرام ACCENTURE EXAM HELP ! CISCO EXAM !

تُعد قناة ACCENTURE EXAM HELP ! CISCO EXAM ! (@coding_are) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 13 208 مشتركاً، محتلاً المرتبة 15 294 في فئة التعليم والمرتبة 31 490 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 13 208 مشتركاً.

بحسب آخر البيانات بتاريخ 30 يونيو, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -144، وفي آخر 24 ساعة بمقدار 6، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 3.08‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 1.29‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 407 مشاهدة. وخلال اليوم الأول يجمع عادةً 170 مشاهدة.
  • التفاعلات والاستجابة: يتفاعل الجمهور بانتظام؛ متوسط التفاعلات لكل منشور يبلغ 2.
  • الاهتمامات الموضوعية: يركز المحتوى على مواضيع رئيسية مثل placement, gaurntee, suree, capgemini, infosy.

📝 الوصف وسياسة المحتوى

يصف المؤلف القناة بأنها مساحة للتعبير عن الآراء الذاتية:
🔥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...

بفضل وتيرة التحديث المرتفعة (أحدث البيانات بتاريخ 01 يوليو, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التعليم.

13 208
المشتركون
+624 ساعات
-127 أيام
-14430 أيام
أرشيف المشاركات
def mirror_reflection(digits, side): images = { '0': 'com.tcs.cv.automata.ei.middleware.DocxToHtmlConverter@a7e2d9d:image1.png', '1': 'com.tcs.cv.automata.ei.middleware.DocxToHtmlConverter@a7e2d9d:image2.png', '2': 'com.tcs.cv.automata.ei.middleware.DocxToHtmlConverter@a7e2d9d:image3.png', '3': 'com.tcs.cv.automata.ei.middleware.DocxToHtmlConverter@a7e2d9d:image4.png', '4': 'com.tcs.cv.automata.ei.middleware.DocxToHtmlConverter@a7e2d9d:image5.png', '5': 'com.tcs.cv.automata.ei.middleware.DocxToHtmlConverter@a7e2d9d:image6.png', '6': 'com.tcs.cv.automata.ei.middleware.DocxToHtmlConverter@a7e2d9d:image7.png', '7': 'com.tcs.cv.automata.ei.middleware.DocxToHtmlConverter@a7e2d9d:image8.png', '8': 'com.tcs.cv.automata.ei.middleware.DocxToHtmlConverter@a7e2d9d:image9.png', '9': 'com.tcs.cv.automata.ei.middleware.DocxToHtmlConverter@a7e2d9d:image0.png', } result = [] for d, s in zip(digits, side): if s == 'L': result.append(images[d]) else: # For U, D, R, and S, print the seven segment display result.append(d) return ''.join(result) # Input parsing digits = input().strip() side = input().strip() # Output print(mirror_reflection(digits, side)) Mirror reflection code .... python✅✅✅✅ Full passes ✅✅✅

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import numpy as np def find_widest_valley(n, A, B): def surface_equation(x): return sum(np.sin(a*x + b) for a, b in zip(A, B)) def find_local_maxima(): maxima = [] for i in range(1, n-1): if surface_equation(i-1) < surface_equation(i) > surface_equation(i+1): maxima.append(i) return maxima def find_valley_width(left, right): return right - left maxima = find_local_maxima() widest_valley_width = 0 widest_valley_index = 0 for i in range(len(maxima)-1): left_maxima = maxima[i] right_maxima = maxima[i+1] current_valley_width = find_valley_width(left_maxima, right_maxima) if current_valley_width > widest_valley_width: widest_valley_width = current_valley_width widest_valley_index = left_maxima + 1 return widest_valley_index # Input parsing n = int(input()) A = list(map(int, input().split())) B = list(map(int, input().split())) # Output print(find_widest_valley(n, A, B))

Sending answer

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def convert(expression): stack = [] tokens = expression.split() for token in reversed(tokens): if is_operand(token): stack.append(int(token)) elif is_operator(token): if len(stack) < 2: return float('-inf') operand2 = stack.pop() operand1 = stack.pop() if token == "+": stack.append(operand1 + operand2) elif token == "-": stack.append(operand1 - operand2) elif token == "*": stack.append(operand1 * operand2) elif token == "/": stack.append(operand1 / operand2) elif token == "^": stack.append(operand1 ** operand2) elif token == "%": stack.append(operand1 % operand2) if len(stack) != 1: return float('-inf') return stack.pop() def is_operand(token): return token.lstrip('-').isdigit() def is_operator(token): return token in {"+", "-", "*", "/", "^", "%"} def main(): prefix = input() word_map = { "zero": "0", "one": "1", "two": "2", "three": "3", "four": "4", "five": "5", "six": "6", "seven": "7", "eight": "8", "nine": "9", "mul": "*", "add": "+", "sub": "-", "div": "/", "rem": "%", "pow": "^" } words = prefix.split() real_prefix = "" for word in words: if word in word_map: real_prefix += word_map[word] + " " continue small_words = word.split("c") for s in small_words: if s: if s not in word_map: print("expression evaluation stopped invalid words present") return real_prefix += word_map[s] real_prefix += " " infix = convert(real_prefix) if infix == float('-inf'): print("expression is not complete or invalid") else: print(int(infix)) if name == "main": main() No more suffle✅✅✅✅✅(python All passsd ✅✅✅ Share everyone ✅✅✅

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#include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<tuple<int, int, int>> stocks; for (int i = 0; i < n; ++i) { int a, b, c; cin >> a >> b >> c; stocks.push_back(make_tuple(a, b, c)); } int m; cin >> m; int real = 0; int unreal = 0; vector<vector<int>> prices; for (int i = 0; i < n; ++i) { vector<int> arr(m); for (int j = 0; j < m; ++j) { cin >> arr[j]; } prices.push_back(arr); } int day; cin >> day; for (int i = 0; i < n; ++i) { int a, b, c; tie(a, b, c) = stocks[i]; if (b > day) { continue; } else if (c > day || !c) { unreal += a * (prices[i][day - 1] - prices[i][b - 1]); } else { real += a * (prices[i][c - 1] - prices[i][b - 1]); } } cout << real << endl; cout << unreal << endl; return 0; } Roi code fully passsd ✅✅✅✅✅ Share @codeing_area

def toggle(num): result = "" for bit in num: if bit == "0": result += "1" else: result += "0" return result def get_max_sum(arr, k): max_val = max(arr) max_index = arr.index(max_val) left = max(0, max_index-k) right = min(len(arr)-1, max_index+k) selected = arr[left:max_index+1] + arr[max_index:right+1] arr = [x for x in arr if x not in selected] return sum(selected), arr # Input arr = list(map(int, input().split())) a1, b1 = input().split() a2, b2 = input().split() k = int(input()) sum1, sum2 = 0, 0 while arr: s1, arr = get_max_sum(arr, k) sum1 += s1 if not arr: break s2, arr = get_max_sum(arr, k) sum2 += s2 if sum1 > sum2: a1 = toggle(a1) b2 = toggle(b2) else: a2 = toggle(a2) b1 = toggle(b1) xor1 = int(a1, 2) ^ int(b1, 2) xor2 = int(a2, 2) ^ int(b2, 2) if xor1 > xor2: print("Rahul",end="") elif xor2 > xor1: print("Rupesh",end="") else: print("both",end="") Who is Lucky

def hamming_distance(str1, str2): return sum(c1 != c2 for c1, c2 in zip(str1, str2)) def minimize_cost_and_print_distance(binary_string, a, b): # Count occurrences of "01" and "10" count_01 = binary_string.count("01") count_10 = binary_string.count("10") # Calculate costs cost_01 = count_01 * a cost_10 = count_10 * b # Determine rearranged string for minimum cost if cost_01 < cost_10: rearranged_string = "01" * count_01 + "0" * (len(binary_string) - 2 * count_01) else: rearranged_string = "10" * count_10 + "1" * (len(binary_string) - 2 * count_10) # Calculate and print hamming distance distance = hamming_distance(binary_string, rearranged_string) print(distance) # Input reading and processing T = int(input("Enter the number of test cases: ")) for _ in range(T): binary_string = input("Enter the binary string: ") a, b = map(int, input("Enter A and B separated by space: ").split()) minimize_cost_and_print_distance(binary_string, a, b) Hamming distance Share @codeing_area

Tcs code vitaaa passed successfully ✅✅✅ Share everyone i will post soon
Tcs code vitaaa passed successfully ✅✅✅ Share everyone i will post soon

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ROI 100% Working ✅ Python TCS Codevita @codeing_area n = int(input()) stocks = [] for _ in range(n): a, b, c = map(int, input().split()) stocks.append((a, b, c)) m = int(input()) real = 0 unreal = 0 placementlelo = [] for i in range(n): arr = [int(i) for i in input().split()] placementlelo.append(arr) day = int(input()) for i in range(n): a, b, c = stocks[i] if b > day: continue elif c > day or not c: unreal += a * (placementlelo[i][day - 1] - placementlelo[i][b - 1]) else: real += a * (placementlelo[i][c - 1] - placementlelo[i][b - 1]) print(real) print(unreal) ROI 100% Working ✅ Python TCS Codevita @codeing_area

def toggle(num): result = "" for bit in num: if bit == "0": result += "1" else: result += "0" return result def get_max_sum(arr, k): max_val = max(arr) max_index = arr.index(max_val) left = max(0, max_index-k) right = min(len(arr)-1, max_index+k) selected = arr[left:max_index+1] + arr[max_index:right+1] arr = [x for x in arr if x not in selected] return sum(selected), arr # Input arr = list(map(int, input().split())) a1, b1 = input().split() a2, b2 = input().split() k = int(input()) sum1, sum2 = 0, 0 while arr: s1, arr = get_max_sum(arr, k) sum1 += s1 if not arr: break s2, arr = get_max_sum(arr, k) sum2 += s2 if sum1 > sum2: a1 = toggle(a1) b2 = toggle(b2) else: a2 = toggle(a2) b1 = toggle(b1) xor1 = int(a1, 2) ^ int(b1, 2) xor2 = int(a2, 2) ^ int(b2, 2) if xor1 > xor2: print("Rahul",end="") elif xor2 > xor1: print("Rupesh",end="") else: print("both",end="") Who is Lucky

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Join Fast will share One more code soon👇👇👇👇 https://t.me/ibmsolution