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

Ko'proq ko'rsatish

Hindiston28 152 Taʼlim14 133...

📈 Telegram kanali ACCENTURE EXAM SOLUTIONS analitikasi

ACCENTURE EXAM SOLUTIONS (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 14 076 obunachidan iborat bo'lib, Taʼlim toifasida 14 133-o'rinni va Hindiston mintaqasida 28 152-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 14 076 obunachiga ega bo‘ldi.

06 Oktabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -120 ga, so‘nggi 24 soatda esa -10 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

Tasdiqlash holati: Tasdiqlanmagan
Jalb etish (ER): Auditoriya o‘rtacha 3.44% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.82% ini tashkil etuvchi reaksiyalarni to‘playdi.
Post qamrovi: Har bir post o‘rtacha 485 marta ko‘riladi; birinchi sutkada odatda 256 ta ko‘rish yig‘iladi.
Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 2 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 07 Oktabr, 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.

14 083

Obunachilar

-1024 soatlar

-337 kun

-12030 kun

485

Post ko'rishlar

~ 25624 soatlar

~ 32748 soatlar

3.44%

Muloqot nisbati

~ 4

Kuniga postlar

Ads index

beta

Postlar arxiv

14 076

int findTotal(string s){ unordered_setst; for(int i=0;i

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After 12k I willl post all

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Fully passed

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#include <iostream> #include <vector> #include <algorithm> long long solve(int N, std::vector<int>& A) { std::sort(A.begin(), A.end()); long long minSum = A[0]; int prev = A[0]; for (int i = 1; i < N; ++i) { if (A[i] <= prev) { A[i] = prev + 1; } prev = A[i]; minSum += A[i]; } return minSum; } .. Minimum unique sum 🙃🙃

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Make it 12k everyone Then I will share all codes ☺️ Share @codeing_area

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int n = S.length(); unordered_map leftFreq, rightFreq; unordered_set leftSet, rightSet; // Initialize the rightFreq map and rightSet with the entire string S for (char c : S) { rightFreq[c]++; rightSet.insert(c); } int maxX = 0; // Traverse the string and adjust the left and right sets and maps for (int i = 0; i < n - 1; ++i) { char c = S[i]; leftFreq[c]++; rightFreq[c]--; if (rightFreq[c] == 0) { rightSet.erase(c); } leftSet.insert(c); int currentSum = leftSet.size() + rightSet.size(); maxX = max(maxX, currentSum); } return n - maxX; Infosys ✅ Fully passed Split string ✅✅

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Shortest string

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Done first ✅✅✅✅

14 076

Make it 12k

14 076

Gys share FAstt

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long long solve(int n,vector<int>v){ long long ans=0; sort(v.begin(),v.end()); for(int i=1;i<n;i++){ if(v[i]<=v[i-1]){ v[i]=v[i-1]+1; } } ans=accumulate(v.begin(),v.end(),0); return ans; } Minimum unique sum✅ Infosys ( Full passed ✅✅ Share @codeing_area

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Guys, Make it 12k .. after 12 k I will send sloution

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Next sloution After 12k I will uploaded ✅✅

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#include <bits/stdc++.h> using namespace std; int equalzeroandone(vector<int>v){ int n=v.size(); for(int i=0;i<n;i++){ if(v[i]==0){ v[i]=-1; } } int sum=0; int ans=-1; map<int,int>mp; for(int i=0;i<n;i++){ sum+=v[i]; if(sum==0){ ans=i+1; } if(mp.find(sum)!=mp.end()){ ans=max(ans,i-mp[sum]); } else{ mp[sum]=i; } } return ans; } int main() { int n; cin>>n; vector<int>v(n); for(int i=0;i<n;i++){ cin>>v[i]; } cout<<equalzeroandone(v); } Equal no of zero and one(Infosys) Full passed ✅✅ Shared @codeing_area

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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

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