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MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

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πŸ“ˆ Analytical overview of Telegram channel MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

Channel MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer (@coding_are) in the English language segment is an active participant. Currently, the community unites 13 252 subscribers, ranking 15 363 in the Education category and 32 287 in the India region.

πŸ“Š Audience metrics and dynamics

Since its creation on Π½Π΅Π²Ρ–Π΄ΠΎΠΌΠΎ, the project has demonstrated rapid growth, gathering an audience of 13 252 subscribers.

According to the latest data from 15 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by 92 over the last 30 days and by -11 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 2.94%. Within the first 24 hours after publication, content typically collects 1.28% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 390 views. Within the first day, a publication typically gains 170 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.

πŸ“ Description and content policy

The author describes the resource as a platform for expressing subjective opinions:
β€œπŸ”₯Guys plz Stop fearing for daily exams πŸ“ πŸ‘¨β€πŸ’» @srksvk is here to help you all at lowest cost possible.πŸ’ͺ πŸŒ€ ” Our Only Aim Is To Let Get Placed To You In A Reputed Company πŸ”₯Effort from our side = πŸ’― πŸ“±Main Channel: @coding_are πŸ“±Tel I'd : @srks...”

Thanks to the high frequency of updates (latest data received on 16 June, 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.

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def GetAnswer(N, A): MOD = 10**9 + 7 table = [[0] * (N+1) for _ in range(N+1)] table[0][0] = 1 nums = [0] + A for i in range(1, N+1): next_table = [[0] * (N+1) for _ in range(N+1)] for x in range(i): for y in range(i): next_table[x][y] = table[x][y] for col in range(i): total = 0 for row in range(i): if row == 0 or nums[row] < nums[i]: total = (total + table[row][col]) % MOD next_table[i][col] = total for row in range(i): total = 0 for col in range(i): if col == 0 or nums[col] > nums[i]: total = (total + table[row][col]) % MOD next_table[row][i] = total table = next_table result = 0 for row in table: result = (result + sum(row)) % MOD return result Share to All Groups πŸ“£ Follow the Codeing_area( Srksvk) channel on WhatsApp: https://whatsapp.com/channel/0029VaicY2a65yD2YNehWP2j

#include <bits/stdc++.h> using namespace std; string trim(string); long calc(int N, int M, int X, vector<int> A, vector<int> B) { const int MAX = 100005; vector<bool> visited(MAX, false); unordered_set<int> corrupted(B.begin(), B.end()); queue<pair<int, int>> q; q.push({0, 0}); visited[0] = true; while (!q.empty()) { auto [step, jumps] = q.front(); q.pop(); for (int jump : A) { int next = step + jump; if (next == X) return jumps + 1; if (next <= X && !visited[next] && corrupted.find(next) == corrupted.end()) { visited[next] = true; q.push({next, jumps + 1}); } } } return -1; } int main() { cout << calc(1, 1, 5, {1}, {6}) << endl; cout << calc(2, 2, 5, {1, 4}, {1, 3}) << endl; cout << calc(4, 2, 4, {2, 3, 4, 1}, {1, 3}) << endl; return 0; } Share To Clg Group, public channel groups βœ… Follow the Codeing_area( Srksvk) channel on WhatsApp: https://whatsapp.com/channel/0029VaicY2a65yD2YNehWP2j

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import sys import heapq from collections import defaultdict def Diff_LCM(N, A): MOD = 10**9 + 7 M = max(abs(x) for x in A) spf = [0] * (M + 1) if M >= 1: spf[1] = 1 for i in range(2, M + 1): if spf[i] == 0: spf[i] = i ii = i * i if ii <= M: for j in range(ii, M + 1, i): if spf[j] == 0: spf[j] = i factors = [[] for _ in range(N)] global_exp = {} for i, v in enumerate(A): x = abs(v) while x > 1: p = spf[x] cnt = 0 while x % p == 0: x //= p cnt += 1 factors[i].append((p, cnt)) if cnt > global_exp.get(p, 0): global_exp[p] = cnt if not global_exp: return 0 contrib = {p: [] for p in global_exp} at_pos = [[] for _ in range(N)] for i, fl in enumerate(factors): for p, cnt in fl: if global_exp[p] == cnt: contrib[p].append(i) at_pos[i].append(p) ptr = {p: 0 for p in global_exp} next_pos = {} heap = [] for p, lst in contrib.items(): pos0 = lst[0] next_pos[p] = pos0 heap.append((-pos0, p)) heapq.heapify(heap) S = [0] * (N + 1) for i in range(N): S[i+1] = S[i] + A[i] size = 1 while size < N + 1: size <<= 1 INF = 10**30 tree = [-INF] * (2 * size) for i in range(N + 1): tree[size + i] = S[i] for i in range(size - 1, 0, -1): tree[i] = max(tree[2*i], tree[2*i + 1]) def rmq(l, r): l += size r += size m = -INF while l <= r: if l & 1: m = max(m, tree[l]) l += 1 if not r & 1: m = max(m, tree[r]) r -= 1 l //= 2; r //= 2 return m ans = 0 for l in range(N): if l > 0: for p in at_pos[l-1]: ptr[p] += 1 idx = ptr[p] lst = contrib[p] newpos = lst[idx] if idx < len(lst) else N next_pos[p] = newpos heapq.heappush(heap, (-newpos, p)) while True: negpos, p = heap[0] pos = -negpos if next_pos[p] != pos: heapq.heappop(heap) else: r_end = pos break lo = l + 1 hi = r_end if lo <= hi: best_prefix = rmq(lo, hi) ans = max(ans, best_prefix - S[l]) return ans % MOD def main(): data = sys.stdin.read().split() N = int(data[0]) A = list(map(int, data[1:])) print(Diff_LCM(N, A)) if name == "main": main()

long Pastry_Contrast(int N,int Q,vector>Pastry,vector>Queries){ vectorF(N),D(N); for(int i=0;i>e; vectorst; for(int i=0;i=D[i]){ int u=st.back();st.pop_back(); e.push_back({i,u,(F[i]-F[u])*(D[i]+D[u])}); } if(!st.empty()){ int u=st.back(); e.push_back({i,u,(F[i]-F[u])*(D[i]+D[u])}); } st.push_back(i); } sort(e.begin(),e.end(),[](auto&a,auto&b){return a[0]t(2*p,LLONG_MAX); auto upd=&{ int x=u+p; t[x]=min(t[x],v); for(x>>=1;x;x>>=1) t[x]=min(t[2*x],t[2*x+1]); }; auto qmn=&{ long s=LLONG_MAX; for(l+=p,r+=p;l<=r;l>>=1,r>>=1){ if(l&1) s=min(s,t[l++]); if(!(r&1)) s=min(s,t[r--]); } return s; }; vector>q(Q); for(int i=0;i

int Tom_Jerry(int N, vector<int> Tom, vector<int> Jerry) { int L = 2*N; vector<vector<int>> G(N+1); for(int i=0;i<N;i++) G[Tom[i]].push_back(Jerry[i]); int B = floor(sqrt(2.0*N)); vector<int> small, large; for(int v=1;v<=N;v++) if(!G[v].empty()) { if(v<=B) small.push_back(v); else large.push_back(v); } vector<vector<pair<int,int>>> comp(B+1); for(int v: small){ auto &g = G[v]; sort(g.begin(),g.end()); auto &c = comp[v]; for(int x: g){ if(c.empty()||c.back().first!=x) c.emplace_back(x,1); else c.back().second++; } } long long ans=0; int M = small.size(); for(int i=0;i<M;i++){ int v = small[i]; auto &cv = comp[v]; long long P = 1LL*v*v; if(P<=L){ int l=0, r=cv.size()-1; while(l<=r){ long long s=cv[l].first+cv[r].first; if(s<P) l++; else if(s>P) r--; else{ if(l==r) ans += 1LL*cv[l].second*(cv[l].second-1)/2; else ans += 1LL*cv[l].second*cv[r].second; l++; r--; } } } for(int j=i+1;j<M;j++){ int u = small[j]; long long P2 = 1LL*v*u; if(P2> L) break; auto &cu = comp[u]; int l=0, r=cu.size()-1; while(l<cv.size() && r>=0){ long long s=cv[l].first+cu[r].first; if(s<P2) l++; else if(s>P2) r--; else{ ans += 1LL*cv[l].second*cu[r].second; l++; r--; } } } } for(int v: large){ auto &gv = G[v]; for(int x: gv){ for(int u: small){ long long P = 1LL*v*u; if(P> L) break; int need = P - x; auto &cu = comp[u]; int lo=0, hi=cu.size(); while(lo<hi){ int m=(lo+hi)/2; if(cu[m].first<need) lo=m+1; else hi=m; } if(lo<cu.size() && cu[lo].first==need) ans += cu[lo].second; } } } return (int)ans; }

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