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allcoding1_official

allcoding1_official

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📈 Analytical overview of Telegram channel allcoding1_official

Channel allcoding1_official (@allcoding1_official) in the English language segment is an active participant. Currently, the community unites 82 254 subscribers, ranking 1 512 in the Technologies & Applications category and 3 707 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 82 254 subscribers.

According to the latest data from 25 August, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -1 572 over the last 30 days and by -40 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 3.02%. Within the first 24 hours after publication, content typically collects 0.71% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 2 481 views. Within the first day, a publication typically gains 588 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 0.
  • Thematic interests: Content is focused on key topics such as dsa, stack, namaste, javascript, dev.

📝 Description and content policy

Channel description not provided.

Thanks to the high frequency of updates (latest data received on 26 August, 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 Technologies & Applications category.

82 254
Subscribers
-4024 hours
-2997 days
-1 57230 days
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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; }

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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]; }

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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

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

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& 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; }

Car code int solve(int T, int capacity, vector>& trips) { map 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; }