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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Hindiston28 083 Taʼlim14 103...

📈 Telegram kanali ACCENTURE EXAM SOLUTIONS analitikasi

ACCENTURE EXAM SOLUTIONS (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 14 134 obunachidan iborat bo'lib, Taʼlim toifasida 14 103-o'rinni va Hindiston mintaqasida 28 083-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 14 134 obunachiga ega bo‘ldi.

22 Sentabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -130 ga, so‘nggi 24 soatda esa 2 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

Tasdiqlash holati: Tasdiqlanmagan
Jalb etish (ER): Auditoriya o‘rtacha 4.14% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.53% ini tashkil etuvchi reaksiyalarni to‘playdi.
Post qamrovi: Har bir post o‘rtacha 586 marta ko‘riladi; birinchi sutkada odatda 217 ta ko‘rish yig‘iladi.
Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 1 ta reaksiya keladi.
Tematik yo‘nalishlar: Kontent placement, gaurntee, suree, capgemini, infosy kabi asosiy mavzularga jamlangan.

📝 Tavsif va kontent siyosati

Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“🔥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...”

Yuqori yangilanish chastotasi (oxirgi ma’lumot 23 Sentabr, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

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Obunachilar

+224 soatlar

-337 kun

-13030 kun

586

Post ko'rishlar

~ 21724 soatlar

~ 28948 soatlar

4.14%

Muloqot nisbati

~ 3

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Taimur and Shadiur Code : ALL PASSED ✅ MOD = 10**9+7 def get_ans(N, Tpos, Shpos): a0 = min(Tpos, Shpos) - 1 b0 = N - max(Tpos, Shpos) dp = [[0]*(b0+1) for _ in range(a0+1)] dp2 = [[0]*(b0+1) for _ in range(a0+1)] dp[a0][b0] = 1 for _ in range(N): for b in range(b0+1): run = 0 for a in range(a0, -1, -1): run = (run + dp[a][b]) % MOD dp2[a][b] = run for a in range(a0+1): run = 0 for b in range(b0, -1, -1): run = (run + dp[a][b]) % MOD dp2[a][b] = (dp2[a][b] + run) % MOD for a in range(a0+1): for b in range(b0+1): dp2[a][b] = (dp2[a][b] - dp[a][b]) % MOD dp, dp2 = dp2, [[0]*(b0+1) for _ in range(a0+1)] return sum(sum(row) for row in dp) % MOD N = int(input()) Tpos = int(input()) Shpos = int(input()) print(get_ans(N, Tpos, Shpos)) *

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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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