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ACCENTURE EXAM HELP ! CISCO EXAM !

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

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📈 Аналитический обзор Telegram-канала ACCENTURE EXAM HELP ! CISCO EXAM !

Канал ACCENTURE EXAM HELP ! CISCO EXAM ! (@coding_are) языкового сегмента Английский является активным участником. Сейчас сообщество объединяет 13 208 подписчиков, занимая 15 294 место в категории Образование и 31 490 место в регионе Индия.

📊 Показатели аудитории и динамика

С момента создания невідомо проект демонстрирует стремительный рост, собрав аудиторию из 13 208 подписчиков.

Согласно последним данным от 30 июня, 2026, канал показывает стабильную активность. За последние 30 дней изменение числа участников составило -144, а за последние 24 часа — 6, при этом общий охват остаётся высоким.

  • Статус верификации: Не верифицирован
  • Уровень вовлечённости (ER): Средний показатель вовлечённости аудитории составляет 3.08%. В первые 24 часа после публикации контент обычно набирает 1.29% реакций от общего числа подписчиков.
  • Охват публикаций: В среднем каждый пост получает 407 просмотров. В течение первых суток публикация набирает 170 просмотров.
  • Реакции и взаимодействия: Аудитория активно поддерживает контент: среднее количество реакций на один пост — 2.
  • Тематические интересы: Контент сосредоточен на ключевых темах, таких как placement, gaurntee, suree, capgemini, infosy.

📝 Описание и контентная политика

Автор описывает ресурс как площадку для выражения субъективного мнения:
🔥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...

Благодаря высокой частоте обновлений (последние данные получены 01 июля, 2026) канал поддерживает актуальность и высокий уровень охвата публикаций. Аналитика показывает, что аудитория активно взаимодействует с контентом, что делает его важной точкой влияния в категории Образование.

13 208
Подписчики
+624 часа
-127 дней
-14430 день
Архив постов
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

Share our channel One more code soon Need your support guys Sitting from last 3 hours to share Codes in free Share it @codeing_area

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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