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ACCENTURE EXAM SOLUTIONS

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

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📈 Telegram 频道 ACCENTURE EXAM SOLUTIONS 的分析概览

频道 ACCENTURE EXAM SOLUTIONS (@coding_are) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 13 961 名订阅者,在 教育 类别中位列第 14 633,并在 印度 地区排名第 28 223 位。

📊 受众指标与增长动态

自 невідомо 创建以来,项目保持高速增长,吸引了 13 961 名订阅者。

根据 31 十月, 2026 的最新数据,频道保持稳定运转。过去 30 天订阅人数变化为 -145,过去 24 小时变化为 -9,整体触达仍然可观。

  • 认证状态: 未认证
  • 互动率 (ER): 平均受众互动率为 3.54%。内容发布后 24 小时内通常能获得 1.43% 的反应,占订阅者总量。
  • 帖子覆盖: 每篇帖子平均可获得 494 次浏览,首日通常累积 200 次浏览。
  • 互动与反馈: 受众积极参与,单帖平均反应数为 2。
  • 主题关注点: 内容集中在 placement, gaurntee, suree, capgemini, infosy 等核心主题上。

📝 描述与内容策略

作者将该频道定位为表达主观观点的平台:
“🔥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...”

凭借高频更新(最新数据采集于 01 十一月, 2026),频道始终保持新鲜度与高覆盖。分析显示受众积极互动,使其成为 教育 类别中的关键影响点。

13 961
订阅者
-924 小时
-417 天
-14530 天
帖子存档
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fibonacci triangle code python

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maximum product
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maximum product

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int Children = 0,Adult = 0;         for(int i=0;i
int Children = 0,Adult = 0;         for(int i=0;i

def cond1(s):       for i in range(len(s) - 2):         if (
def cond1(s):       for i in range(len(s) - 2):         if (s[i] == s[i + 1] and             s[i + 1] == s[i + 2]):             return True           return False def cond2(s):     for i in range(len(s) - 2):         if ((s[i] < s[i + 1] and              s[i + 1] < s[i + 2]) or             (s[i] > s[i + 1] and              s[i + 1] > s[i + 2])):             return True           return False def cond3(s):     a = [0] * 10     for i in range(len(s)):         a[s[i] - '0'] = a[s[i] - '0'] + 1           for i in range(len(9)):               if (a[i] >= 4):             return True           return False   def isFancy(s):     if (cond1(s) or cond2(s) or cond3(s)):         return True     else:         return False s = input(“ “) if (isFancy(s)):     print("Yes") else:     print("No") Language: Python3

class Solution: def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]: dict1 = dict() for i in nums1: if i not
class Solution:     def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:         dict1 = dict()         for i in nums1:             if i not in dict1:                 dict1[i] = 1             else:                 dict1[i] += 1         ret = []         for i in nums2:             if i in dict1 and dict1[i]>0:                 ret.append(i)                 dict1[i] -= 1         return ret Python3

Those who needs help to clear ACCENTURE,ACCOLITE DIGITAL,ATOS, CAPGEMINI,IBM or any other placement papers can contact us and Book your slots 100% clearance guarant Telegram@srksvk Note: it's paid

Those who needs help to clear KEKA technologies, cognizant, AMAZON,NETCRACKER, CAPGEMINI, WIPRO, or any other placement papers can contact us and Book your slots 100% clearance guarantee Telegram--  @srksvk Note: it's paid Telegram -- https://t.me/rituyr

Find max code HSBC
Find max code HSBC