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

India28 010 Education14 044...

📈 Analytical overview of Telegram channel ACCENTURE EXAM SOLUTIONS

Channel ACCENTURE EXAM SOLUTIONS (@coding_are) in the English language segment is an active participant. Currently, the community unites 14 139 subscribers, ranking 14 044 in the Education category and 28 010 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 14 139 subscribers.

According to the latest data from 21 September, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -136 over the last 30 days and by 1 over the last 24 hours, overall reach remains high.

Verification status: Not verified
Engagement rate (ER): The average audience engagement rate is 4.15%. Within the first 24 hours after publication, content typically collects 1.53% reactions from the total number of subscribers.
Post reach: On average, each post receives 586 views. Within the first day, a publication typically gains 216 views.
Reactions and interaction: The audience actively supports content: the average number of reactions per post is 1.
Thematic interests: Content is focused on key topics such as placement, gaurntee, suree, capgemini, infosy.

📝 Description and content policy

The author describes the resource as a platform for expressing subjective opinions:
“🔥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...”

Thanks to the high frequency of updates (latest data received on 22 September, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.

14 139

Subscribers

+124 hours

-367 days

-13630 days

586

Post views

~ 21624 hours

~ 29048 hours

4.15%

Engagement rate

~ 3

Posts per day

Ads index

beta

Posts Archive

14 136

Follow the Codeing_area( Srksvk) channel on WhatsApp: https://whatsapp.com/channel/0029VaicY2a65yD2YNehWP2j Fully done go and check ✅✅

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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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*3PM CODES START FROM HERE 👇* *We will Share Codes Solutions Here* Start Exam around 3:15 PM https://whatsapp.com/channel/0029VaicY2a65yD2YNehWP2j

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Infosys 10am exam successfully completed by remote access 👍👍👍👍👍 3/3 hard code done with all tests caes passed ✅✅ Contact for placement exam @srksvk

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Infosys 10am exam successfully completed by remote access 👍👍👍👍👍 3/3 hard code done with all tests caes passed ✅✅ Contact for placement exam @srksvk

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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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