ACCENTURE EXAM HELP ! CISCO EXAM !

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Ko'proq ko'rsatish

Hindiston31 490 Taʼlim15 294...

📈 Telegram kanali ACCENTURE EXAM HELP ! CISCO EXAM ! analitikasi

ACCENTURE EXAM HELP ! CISCO EXAM ! (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 13 208 obunachidan iborat bo'lib, Taʼlim toifasida 15 294-o'rinni va Hindiston mintaqasida 31 490-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 13 208 obunachiga ega bo‘ldi.

30 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -144 ga, so‘nggi 24 soatda esa 6 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

Tasdiqlash holati: Tasdiqlanmagan
Jalb etish (ER): Auditoriya o‘rtacha 3.08% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.29% ini tashkil etuvchi reaksiyalarni to‘playdi.
Post qamrovi: Har bir post o‘rtacha 407 marta ko‘riladi; birinchi sutkada odatda 170 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 01 Iyul, 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.

13 208

Obunachilar

+624 soatlar

-127 kunlar

-14430 kunlar

407

Post ko'rishlar

~ 17024 soatlar

~ 24048 soatlar

3.08%

Muloqot nisbati

~ 3

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X from Y Code C++ Language ✅ Done TCS Codevita #include <bits/stdc++.h> using namespace std; typedef long long ll; #define codeingarea vector<int> #define all(a) (a).begin() , (a).end() #define sz(a) ((int)((a).size())) const ll INF = 4e18; vector<codeingarea> dp(1e4+1 , codeingaarea(1e4+1 , -1)); int f(int i , int j , int s , int r , string &x , string &y , string &y1) { if(i>j)return 0; if(dp[i][j] != -1)return dp[i][j]; int mini = 1e9; for(int k=i; k<=j; k++) { if(y1.find(x.substr(i , k-i+1)) != string::npos)mini = min(mini , r+f(k+1 , j , s , r , x , y , y1)); if(y.find(x.substr(i , k-i+1)) != string::npos)mini = min(mini , s+f(k+1 , j , s , r , x , y , y1)); } return mini; } int solve(int s , int r , string x, string y) { string y1 = y; reverse(all(y1)); return f(0 , sz(x)-1 , s, r , x , y , y1); } int32_t main(){ ios_base::sync_with_stdio(false); cin.tie(NULL); string x , y; cin >> x >> y; int s, r; cin >> s >> r; int ans = solve(s , r, x , y); cout << ans; return 0; } X from Y C++ Language 100% Working ✅ TCS Codevita

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def min_string_factor(X, Y, S, R): len_X, len_Y = len(X), len(Y) dp = [[float('inf')] * (len_Y + 1) for _ in range(len_X + 1)] dp[0][0] = 0 for i in range(1, len_X + 1): for j in range(1, len_Y + 1): # Select substring from Y if X[i - 1] == Y[j - 1]: dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + S) # Select substring from reversed Y dp[i][j] = min(dp[i][j], dp[i][j - 1] + R) result = min(dp[len_X]) return result if result != float('inf') else "Impossible" # Reading input X = input().strip() Y = input().strip() S, R = map(int, input().strip().split()) # Output result print(min_string_factor(X, Y, S, R)) @codeing_area

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