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ACCENTURE EXAM SOLUTIONS

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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📈 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 14 088 subscribers, ranking 14 131 in the Education category and 28 155 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 14 088 subscribers.

According to the latest data from 05 October, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -120 over the last 30 days and by -12 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 3.51%. Within the first 24 hours after publication, content typically collects 1.81% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 495 views. Within the first day, a publication typically gains 255 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 2.
  • 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 06 October, 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.

14 088
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Don't forget share the group @codeing_arra

const int MOD 1000000007; int subsetSumCount(const vector& A, int L, int R, int K) { vector dp(K+1, 0); dp[0] = 1; for (int i=L; i<=R; ++i) { for (int j = K; j >= A[1]; --j) { dp[j] = (dp[j] + dp[j- A[i]]) % MOD; } } return dp[K];} int findXOR(int n, int Q, const vector& A, const vector>& B) { int result = 0; for (const auto& query: B) { int L= query [0] - 1; int R= query [1] -1; int K = query[2]; int P=subsetSumCount(A,L,R,K); result ^= P; } return result; } Subset sum Fully passed ✅✅✅✅ Just paste it

Share with your Friends and in Big groups we will definitely post answers here We will Try to share Some answers Here also in our Groups. So must Join 1) @codeing_area 2) @codeing_area 3) @codeing_area

#include <bits/stdc++.h> using namespace std; #define ll long long const ll MOD=1e9+7; struct FenwickTree { vector<ll> bit; ll n; FenwickTree(ll size) { n = size; bit.assign(n + 1, 0); } void update(ll idx, ll val) { for (; idx <= n; idx += idx & -idx) bit[idx] = max(bit[idx], val); } ll query(ll idx) { ll res = 0; for (; idx > 0; idx -= idx & -idx) res = max(res, bit[idx]); return res; } }; ll solve(vector<ll>& a) { ll n=a.size(); vector<pair<ll,ll>>ia(n); for (ll i=0;i<n;i++) { ia[i] = make_pair(a[i], i); } sort(ia.begin(),ia.end(),[](const pair<ll,ll>&p1,const pair<ll,ll>&p2) { return p1.first<p2.first; }); FenwickTree fenwick(n); ll maxBeauty=0; for (ll i=0;i<n;i++) { ll val=ia[i].first; ll og=ia[i].second; ll beauty=0; for (ll j=og-1;j>=0;j--) { if (__gcd(abs(a[j]),abs(a[og]))>1) { beauty=max(beauty,fenwick.query(j+1)+(a[og]-a[j])*(a[og]-a[j]) % MOD); } } fenwick.update(og + 1, beauty); maxBeauty = max(maxBeauty, beauty); } return maxBeauty; } signed main() { ll n; cin>>n; vector<ll>a(n); for (ll i=0;i<n;i++) cin>>a[i]; cout<<solve(a)<<endl; return 0; } subsequence beauty Share group @srksvk

Share guy's @codeing_area

Share everyone for answer 🙏 @codeing_area

dp[i+1][0][0][(j + addv)%k][pv+1] += dp[i][1][0][j][pv]; dp[i+1][0][0][(j + addv)%k][pv+1] %= mod; } else{ dp[i+1][1][0][(j + addv)%k][pv+1] += dp[i][1][0][j][pv]; dp[i+1][1][0][(j + addv)%k][pv+1] %= mod; } // le = 0, re = 1 if(r[i] == '0'){ dp[i+1][0][1][j][pv-1] += dp[i][0][1][j][pv]; dp[i+1][0][1][j][pv-1] %= mod; } else{ dp[i+1][0][1][(j + addv)%k][pv+1] += dp[i][0][1][j][pv]; dp[i+1][0][1][(j + addv)%k][pv+1] %= mod; dp[i+1][0][0][j][pv-1] += dp[i][0][1][j][pv]; dp[i+1][0][0][j][pv-1] %= mod; } // le = 0, re = 0 dp[i+1][0][0][j][pv-1] += dp[i][0][0][j][pv]; dp[i+1][0][0][j][pv-1] %= mod; dp[i+1][0][0][(j + addv)%k][pv+1] += dp[i][0][0][j][pv]; dp[i+1][0][0][(j + addv)%k][pv+1] %= mod; } } } } int ans = 0; for(int i = 0; i <= n; i++){ for(int j = 0; j < 2; j++){ for(int z = 0; z < 2; z++){ ans += dp[n][j][z][0][i]; ans %= mod; } } } return ans; } int32_t main(){ cin.tie(0); cout.tie(0); ios_base::sync_with_stdio(false); int m, n, k; cin>>m>>n>>k; string l, r; cin>>l>>r; cout<

Share the group ✅✅⏩ Fast for next answer

import sys MOD = 1000000007 def getCount(N, M, S):     dp = [[[0] * (M + 1) for _ in range(N + 1)] for _ in range(N + 1)]     dp[0][0][0] = 1         for i in range(N):         for j in range(N + 1):             for k in range(M + 1):                 if dp[i][j][k] == 0:                     continue                                 if j + 1 <= N:                     if k < M and S[k] == '(':                         dp[i + 1][j + 1][k + 1] = (dp[i + 1][j + 1][k + 1] + dp[i][j][k]) % MOD                     else:                         dp[i + 1][j + 1][k] = (dp[i + 1][j + 1][k] + dp[i][j][k]) % MOD                                 if j > 0:                     if k < M and S[k] == ')':                         dp[i + 1][j - 1][k + 1] = (dp[i + 1][j - 1][k + 1] + dp[i][j][k]) % MOD                     else:                         dp[i + 1][j - 1][k] = (dp[i + 1][j - 1][k] + dp[i][j][k]) % MOD         return dp[N][0][M] def main():     N = int(sys.stdin.readline().strip())     M = int(sys.stdin.readline().strip())     S = sys.stdin.readline().strip()         result = getCount(N, M, S)         print(result) if name == "main":     main() Bracket sequence,✅✅✅✅

Pre bookings slot help successfully done ... Now I will upload answer... guy's just share group fast ⏩⏩⏩

String container Fully passed 🔥🔥🔥 Infosys 🔥🔥 All three code done 😎😎😎
String container Fully passed 🔥🔥🔥 Infosys 🔥🔥 All three code done 😎😎😎

Share with your Friends and in Big groups we will definitely post answers here We will Try to share Some answers Here also in our Groups. So must Join 1) @codeing_area 2) @codeing_area 3) @codeing_area

Don't ask these code everyone... Note:: This one prebooking 👍👍👍 So I am not share these code... because plug 👍👍

Count Travel routers 🔥🔥🔥🔥🔥 Fully passed ✅✅✅✅
Count Travel routers 🔥🔥🔥🔥🔥 Fully passed ✅✅✅✅

Infosys 🔥🔥🔥🔥🔥 Cost of string s Fully ✅✅✅
Infosys 🔥🔥🔥🔥🔥 Cost of string s Fully ✅✅✅

Just share group everyone ✅🥺😀 If you want more answers 😜

100% running code

const int MOD = 1000000007; int subsetSumCount(const vector& A, int L, int R, int K) { vector dp(K + 1, 0); dp[0] = 1; for (int i = L; i <= R; ++i) { for (int j = K; j >= A[i]; --j) { dp[j] = (dp[j] + dp[j - A[i]]) % MOD; } } return dp[K]; } int findXOR(int n, int Q, const vector& A, const vector>& B ) { int result = 0; for (const auto& query : B ) { int L = query[0] - 1; int R = query[1] - 1; int K = query[2]; int P = subsetSumCount(A, L, R, K); result ^= P; } return result; } //subarray subset sum

#include <bits/stdc++.h> #define int long long using namespace std; #define ll long long ll solve(ll n,ll k,vector<ll>&a) { k++; unordered_map<ll,ll>freq; map<ll,vector<ll>>mpp; for (ll i=0;i<n;i++) { freq[a[i]]++; mpp[freq[a[i]]].push_back(a[i]); } ll ans=0; for (auto it:mpp) ans+=it.second.size(); return ans; } signed main() { ll n,k; cin>>n>>k; vector<ll>a(n); for(ll i=0;i<n;i++) cin>>a[i]; cout<<solve(n,k,a); return 0; } sequence split (Infosys) 100💯✅ running

So do fastt