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India32 337 Educación15 343...

📈 Análisis del canal de Telegram Tcs exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

El canal Tcs exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer (@coding_are) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 13 273 suscriptores, ocupando la posición 15 343 en la categoría Educación y el puesto 32 337 en la región India.

📊 Métricas de audiencia y dinámica

Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 13 273 suscriptores.

Según los últimos datos del 13 junio, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de 114, y en las últimas 24 horas de -6, conservando un alto alcance.

Estado de verificación: No verificado
Tasa de interacción (ER): El promedio de interacción de la audiencia es 2.87%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 1.19% de reacciones respecto al total de suscriptores.
Alcance de las publicaciones: Cada publicación recibe en promedio 381 visualizaciones. En el primer día suele acumular 158 visualizaciones.
Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 1.
Intereses temáticos: El contenido se centra en temas clave como placement, gaurntee, suree, capgemini, infosy.

📝 Descripción y política de contenido

El autor describe el recurso como un espacio para expresar opiniones subjetivas:
“🔥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...”

Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 14 junio, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Educación.

13 273

Suscriptores

-624 horas

-347 días

+11430 días

381

Visitas de la publicación

~ 15824 horas

~ 21048 horas

2.87%

Tasa de compromiso

~ 5

Mensajes por día

Ads index

beta

Archivo de publicaciones

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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 ✅

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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 ✅

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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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