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allcoding1_official

allcoding1_official

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Hindiston3 707Texnologiyalar & Aralashmalar1 512

📈 Telegram kanali allcoding1_official analitikasi

allcoding1_official (@allcoding1_official) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 82 254 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 512-o'rinni va Hindiston mintaqasida 3 707-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 82 254 obunachiga ega bo‘ldi.

25 Avgust, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -1 572 ga, so‘nggi 24 soatda esa -40 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 3.02% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.71% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 2 481 marta ko‘riladi; birinchi sutkada odatda 588 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 0 ta reaksiya keladi.
  • Tematik yo‘nalishlar: Kontent dsa, stack, namaste, javascript, dev kabi asosiy mavzularga jamlangan.

📝 Tavsif va kontent siyosati

Kanal uchun tavsif kiritilmagan.

Yuqori yangilanish chastotasi (oxirgi ma’lumot 26 Avgust, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

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allcoding1_official
ACCENTURE IS HIRING – FRESHERS Company: Accenture Position: Custom Software Engineer Experience: 0–2 Years APPLY NOW — ACCENTURE CAREERS: https://www.accenture.com/in-en/careers/jobdetails?id=ATCI-5704900-S2062039_en Share with friends

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import java.util.*; class Main { public static int solve(int N, String colors, int[] neededTime) { long answer = 0; int i = 0; while (i < N) { int j = i; while (j + 1 < N && colors.charAt(j + 1) == colors.charAt(i)) j++; int len = j - i + 1; if (len >= 2) { long keep = 0; long diffuse = neededTime[i]; for (int k = i + 1; k <= j; k++) { long newKeep = diffuse; long newDiffuse = Math.min(keep, diffuse) + neededTime[k]; keep = newKeep; diffuse = newDiffuse; } answer += Math.min(keep, diffuse); } i = j + 1; } return (int) answer; }
3 007
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Matn yo'q...
1 867
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import java.util.*; class Main { public static int solve(String word1, String word2) { int n = word1.length(); int m = word2.length(); final int INF = 1_000_000; int[][][] dp = new int[n + 1][m + 1][2]; for (int i = 0; i <= n; i++) { for (int j = 0; j <= m; j++) { dp[i][j][0] = INF; dp[i][j][1] = INF; } } dp[n][m][0] = 0; dp[n][m][1] = 0; for (int i = n; i >= 0; i--) { for (int j = m; j >= 0; j--) { for (int p = 0; p < 2; p++) { if (i == n && j == m) continue; int ans = INF; if (i < n) { ans = Math.min(ans, 1 + dp[i + 1][j][p ^ 1]); } if (j < m) { ans = Math.min(ans, 1 + dp[i][j + 1][p ^ 1]); } if (i < n && j < m) { char c = word1.charAt(i); if (p == 1) { c = (char) ('a' + (c - 'a' + 13) % 26); } if (c == word2.charAt(j)) ans = Math.min(ans, dp[i + 1][j + 1][p]); else ans = Math.min(ans, 1 + dp[i + 1][j + 1][p]); } dp[i][j][p] = ans; } } } return dp[0][0][0]; }
1 922
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Matn yo'q...
1 085
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AnimatedSticker.tgs
1 545
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def solve(N: int, nums: list) -&gt; int: def max_product(arr): mx = mn = ans = arr[0] for x in arr[1:]: if x &lt; 0: mx, mn =
def solve(N: int, nums: list) -> int: def max_product(arr): mx = mn = ans = arr[0] for x in arr[1:]: if x < 0: mx, mn = mn, mx mx = max(x, mx * x) mn = min(x, mn * x) ans = max(ans, mx) return ans res = -10**18 i = 0 while i < N: j = i while j + 1 < N and abs(nums[j]) < abs(nums[j + 1]): j += 1 res = max(res, max_product(nums[i:j + 1])) i = j + 1 return res
1 832
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def is_vowel(ch): return ch in 'AEIOU' def solve(): data = sys.stdin.read().strip() if not data: return 0 s = data.split()[0]
def is_vowel(ch): return ch in 'AEIOU' def solve(): data = sys.stdin.read().strip() if not data: return 0 s = data.split()[0] n = len(s) dp_vowel = [0] * (n + 1) dp_consonant = [0] * (n + 1) dp_consonant[0] = 1 for i in range(1, n + 1): if s[i-1] != '0': digit = int(s[i-1]) letter = chr(ord('A') + digit - 1) if is_vowel(letter): dp_vowel[i] = (dp_vowel[i] + dp_consonant[i-1]) % MOD else: dp_consonant[i] = (dp_consonant[i] + dp_consonant[i-1] + dp_vowel[i-1]) % MOD if i >= 2 and s[i-2] != '0': two_digit = int(s[i-2:i]) if 10 <= two_digit <= 26: letter = chr(ord('A') + two_digit - 1) if is_vowel(letter): dp_vowel[i] = (dp_vowel[i] + dp_consonant[i-2]) % MOD else: dp_consonant[i] = (dp_consonant[i] + dp_consonant[i-2] +dp_vowel[i-2]) % MOD
1 049
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int solve(int N, int fee, vector&amp; prices) { int max_cash = 0; int bought_even_stock = -1e9; int bought_odd_stock = -1e9;
int solve(int N, int fee, vector<int>& prices) { int max_cash = 0; int bought_even_stock = -1e9; int bought_odd_stock = -1e9; for (int idx = 0; idx < N; idx++) { int current_price = prices[idx]; int updated_cash = max_cash; int updated_even_stock = bought_even_stock; int updated_odd_stock = bought_odd_stock; if (current_price % 2 == 0) { updated_even_stock = max(bought_even_stock, max_cash - current_price); if (bought_odd_stock != -1e9) { updated_cash = max(updated_cash, bought_odd_stock + current_price - fee); } } else { updated_odd_stock = max(bought_odd_stock, max_cash - current_price); if (bought_even_stock != -1e9) { updated_cash = max(updated_cash, bought_even_stock + current_price - fee); } } max_cash = updated_cash; bought_even_stock = updated_odd_stock; } return max_cash; }
1 059
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Car code int solve(int T, int capacity, vector<vector<int>>& trips) { map<int, long long> passenger_diffs; for (int idx = 0; idx < T; idx++) { int count = trips[idx][0]; int start_loc = trips[idx][1]; int end_loc = trips[idx][2]; passenger_diffs[start_loc] += count; passenger_diffs[end_loc] -= count; } long long accumulated_standing_cost = 0; long long active_riders = 0; int last_point = -1; for (auto& entry : passenger_diffs) { int curr_point = entry.first; if (last_point != -1 && curr_point > last_point) { long long dist = curr_point - last_point; if (active_riders > capacity) { accumulated_standing_cost += (active_riders - capacity) * dist; } } active_riders += entry.second; last_point = curr_point; } return accumulated_standing_cost; }
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