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allcoding1

allcoding1

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📈 نظرة تحليلية على قناة تيليجرام allcoding1

تُعد قناة allcoding1 (@allcoding1) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 22 590 مشتركاً، محتلاً المرتبة 8 822 في فئة التعليم والمرتبة 19 518 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 22 590 مشتركاً.

بحسب آخر البيانات بتاريخ 12 يونيو, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -437، وفي آخر 24 ساعة بمقدار -6، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 5.99‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 1.25‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 1 353 مشاهدة. وخلال اليوم الأول يجمع عادةً 283 مشاهدة.
  • التفاعلات والاستجابة: يتفاعل الجمهور بانتظام؛ متوسط التفاعلات لكل منشور يبلغ 2.
  • الاهتمامات الموضوعية: يركز المحتوى على مواضيع رئيسية مثل dsa, stack, namaste, javascript, learning.

📝 الوصف وسياسة المحتوى

وصف القناة غير متوفر.

بفضل وتيرة التحديث المرتفعة (أحدث البيانات بتاريخ 13 يونيو, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التعليم.

22 590
المشتركون
-624 ساعات
-967 أيام
-43730 أيام
أرشيف المشاركات
def max_sum_of_distinct_characters(S): n = len(S) left_chars = set() right_chars = set() left_count = [0] * n right_count = [0] * n for i in range(n): left_chars.add(S[i]) left_count[i] = len(left_chars) for i in range(n-1, -1, -1): right_chars.add(S[i]) right_count[i] = len(right_char) max_sum = 0 for i in range(n-1): max_sum = max(max_sum, left_count[i] + right_count[i+1]) return n- max_sum Split String code Python 3 All passed

import sys det solve(N, A) for i in range(N): if A[i]=0; A[1] ps=0 M-1 pm={0:-1) for i in range(N): ps+=A[i] if ps in pm: m=max(m,i-pm[ps]) else: pm[ps]=i return m def main(): Nint(sys.stdin.readline().strip()) A-[] for_ in range(N): A.append(int(sys.stdin.readline().strip())) result = solve(N, A) print(result) Largest Subarray with equal number Infosys

def solve(N, A): unique_sums = set() for start in range(N): current_sum = 0 for end in range(start, N): current_sum += A[end] unique_sums.add(current_sum) print(len(unique_sums))

Minimum substring ..
Minimum substring ..

def count_distinct_strings(S): distinct_strings = set() for i in range(len(S) - 1): new_string = S[:i] + S[i+2:] distinct_strings.add(new_string) return len(distinct_strings) # Read input string S = input().strip() # Get the number of distinct strings that can be generated result = count_distinct_strings(S) print(result)

def minimum_unique_sum(A): N = len(A) A.sort() total = A[0] for i in range(1, N): if A[i] <= A[i-1]: A[i] = A[i-1] + 1 total += A[i] return total # Input format N = int(input()) A = [] for i in range(N): A.append(int(input())) # Output result = minimum_unique_sum(A) print(result)

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def split_string_cost(S): # Length of the string S len_S = len(S) # To store the cost of the split parts max_cost = 0 # Set to keep track of distinct characters in the first part distinct_chars_A = set() # List to keep track of the cost for the second part from each split position cost_B = [0] * len_S # Set to keep track of distinct characters in the second part distinct_chars_B = set() # Calculate cost for second part from the end for i in range(len_S - 1, -1, -1): distinct_chars_B.add(S[i]) cost_B[i] = len(distinct_chars_B) # Calculate maximum sum of cost for parts A and B for i in range(len_S - 1): distinct_chars_A.add(S[i]) cost_A = len(distinct_chars_A) cost = cost_A + cost_B[i + 1] max_cost = max(max_cost, cost) # Calculate the result as |S| - X result = len_S - max_cost return result # Example usage S = "aaabbb" print(split_string_cost(S)) # Output: 3

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