Endi mavjud! Telegram Tadqiqoti 2025 — yilning asosiy insaytlari 
Accenture exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer
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Ko'proq ko'rsatish📈 Telegram kanali Accenture exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer analitikasi
Accenture exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 13 224 obunachidan iborat bo'lib, Taʼlim toifasida 15 290-o'rinni va Hindiston mintaqasida 31 740-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 13 224 obunachiga ega bo‘ldi.
24 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -147 ga, so‘nggi 24 soatda esa 5 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 2.63% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.55% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 348 marta ko‘riladi; birinchi sutkada odatda 73 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 25 Iyun, 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.
13 224
Obunachilar
+524 soatlar
-257 kunlar
-14730 kunlar
Postlar arxiv
#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 ✅✅
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https://t.me/codeing_area/7643?single
https://t.me/codeing_area/8059?single
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.
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1) @codeing_area
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3) @codeing_area
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Guys we are upload all code passed .. but no one sharing group
#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll solve(string s1, string s2) {
ll m = s1.length();
ll n = s2.length();
vector<vector<ll>>dp(m+1,vector<ll>(n + 1));
for (ll i=1;i<= m;i++) {
dp[i][0]=i;
}
for (ll j = 1; j <= n;j++) {
dp[0][j] = j;
}
for (ll i = 1; i <= m; ++i)
{
for (ll j = 1; j <= n; ++j)
{
if (s1[i - 1] == s2[j - 1])
{
dp[i][j] = dp[i - 1][j - 1];
}
else
{
dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1]);
if (i > 1 && j > 1 && s1[i - 1] == s2[j - 2])
{
dp[i][j] = min(dp[i][j], 1 + dp[i - 1][j - 2]);
}
if (i > 1 && j > 1 && s1[i - 2] == s2[j - 1])
{
dp[i][j] = min(dp[i][j], 1 + dp[i - 2][j - 1]);
}
}
}
}
return dp[m][n];
}
signed main()
{
string s1,s2; cin>>s1>>s2;
cout<<solve(s1,s2)<<endl;
return 0;
}
String operation.
Just paste it
Guys we are upload all code passed pic ... but no one sharing group
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@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
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#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
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@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,✅✅✅✅
