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Tcs exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

Tcs exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

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📈 نظرة تحليلية على قناة تيليجرام Tcs exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

تُعد قناة Tcs exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer (@coding_are) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 13 273 مشتركاً، محتلاً المرتبة 15 343 في فئة التعليم والمرتبة 32 337 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 13 273 مشتركاً.

بحسب آخر البيانات بتاريخ 13 يونيو, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار 114، وفي آخر 24 ساعة بمقدار -6، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 2.87‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 1.19‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 381 مشاهدة. وخلال اليوم الأول يجمع عادةً 158 مشاهدة.
  • التفاعلات والاستجابة: يتفاعل الجمهور بانتظام؛ متوسط التفاعلات لكل منشور يبلغ 1.
  • 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 14 يونيو, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التعليم.

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int gen_ans(int N, int M, int K, vector<vector<int>> A) { vector<vector<int>> col(N+1, vector<int>(M)); for(int j=0; j<M; j++) for(int i=0; i<N; i++) col[i+1][j] = col[i][j] + (A[i][j] == K); long long Tot = 0, Freq = 0; int MaxB = 0; vector<int> r(M), v(M); for(int i=0; i<N; i++) for(int k=i+1; k<N; k++){ int h = k - i + 1; for(int j=0; j<M; j++){ r[j] = (A[i][j] == K && A[k][j] == K); v[j] = (col[k+1][j] - col[i][j] == h); } for(int j=0; j<M; ){ if(!r[j]) { j++; continue; } int s = j; while(j<M && r[j]) j++; int e = j-1, cnt = 0, f = -1, l = -1; for(int t=s; t<=e; t++) if(v[t]){ if(f<0) f = t; l = t; cnt++; } if(cnt >= 2){ Tot += 1LL*cnt*(cnt-1)/2; int w = l - f + 1; int b = 2*(h + w) - 4; if(b > MaxB){ MaxB = b; Freq = 1; } else if(b == MaxB) Freq++; } } } if(!Tot) return 0; return Tot ^ MaxB ^ Freq; } All passed ✅

def a(s): b = len(s) c = [[0] * b for _ in range(b)] for d in range(b): c[d][d] = 1 for f in range(2, b + 1): for g in range(b - f + 1): h = g + f - 1 if s[g] == s[h] and f == 2: c[g][h] = 2 elif s[g] == s[h]: c[g][h] = c[g + 1][h - 1] + 2 else: c[g][h] = max(c[g][h - 1], c[g + 1][h]) print(c[0][b - 1]) a(input()) https://t.me/coding_are //Infosys LPS - Full pass ✅

One more Done ✅✅ from collections import defaultdict def count(n, c, m, a): mod = 10**9 + 7 start = tuple(a) dp = [defaultdict(int) for _ in range(m+1)] dp[0][start] = 1 for i in range(m): ndp = defaultdict(int) for s in dp[i]: cnt = dp[i][s] ndp[s] = (ndp[s] + cnt) % mod lst = list(s) for j in range(n): if lst[j] == 0: continue if j > 0 and lst[j-1] == 0: new = lst.copy() new[j-1], new[j] = new[j], new[j-1] ndp[tuple(new)] = (ndp[tuple(new)] + cnt) % mod if j < n-1 and lst[j+1] == 0: new = lst.copy() new[j+1], new[j] = new[j], new[j+1] ndp[tuple(new)] = (ndp[tuple(new)] + cnt) % mod dp[i+1] = ndp total = sum(dp[m].values()) % mod return total Language pyhton ✅

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