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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 81 791 subscribers, ranking 1 506 in the Technologies & Applications category and 3 654 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 81 791 subscribers.

According to the latest data from 04 September, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -1 495 over the last 30 days and by -42 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 4.34%. 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 3 550 views. Within the first day, a publication typically gains 583 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 2.
  • 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 05 September, 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.

81 791
Subscribers
-4224 hours
-3577 days
-1 49530 days
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n = int(raw_input()) k = int(raw_input()) m = int(raw_input()) intervals = [] for i in range(n): s, e, p = map(int, raw_input
n = int(raw_input()) k = int(raw_input()) m = int(raw_input()) intervals = [] for i in range(n): s, e, p = map(int, raw_input().split()) intervals.append((s, e, p)) # Sort by start time intervals.sort() # dp[c][i] = maximum profit using exactly c intervals, # ending with interval i dp = [[-1] * n for _ in range(k + 1)] ans = 0 for i in range(n): dp[1][i] = intervals[i][2] ans = max(ans, dp[1][i]) for c in range(2, k + 1): for i in range(n): s2, e2, p2 = intervals[i] best = -1 for j in range(i): s1, e1, p1 = intervals[j] if e1 <= s2 and (s2 - e1) % m == 0: if dp[c - 1][j] != -1: best = max(best, dp[c - 1][j] + p2) dp[c][i] = best if best > ans: ans = best print ans

MOD = 1000000007 N = int(raw_input()) K = int(raw_input()) ways = [0] * (N + 1) ans = [0] * (N + 1) prefWays = [0] * (N + 1)
MOD = 1000000007 N = int(raw_input()) K = int(raw_input()) ways = [0] * (N + 1) ans = [0] * (N + 1) prefWays = [0] * (N + 1) prefAns = [0] * (N + 1) ways[0] = 1 prefWays[0] = 1 for i in range(1, N + 1): l = max(0, i - K) r = i - 1 ways[i] = (prefWays[r] - (prefWays[l - 1] if l > 0 else 0)) % MOD ans[i] = ( prefAns[r] - (prefAns[l - 1] if l > 0 else 0) + ways[i] ) % MOD prefWays[i] = (prefWays[i - 1] + ways[i]) % MOD prefAns[i] = (prefAns[i - 1] + ans[i]) % MOD print ans[N] % MOD

import sys input = sys.stdin.readline def digit_sum(n: int) -&gt; int: return sum(int(c) for c in str(n)) def solve(N: int, S
import sys input = sys.stdin.readline def digit_sum(n: int) -> int: return sum(int(c) for c in str(n)) def solve(N: int, S: int) -> list: if S == 0: m = 0 else: num_digits = (S + 8) // 9 # ceil(S/9) first = S - 9 * (num_digits - 1) m = int(str(first) + '9' * (num_digits - 1)) if m <= N: return [1, m] else: return [0, -1] if name == "main": try: N = int(input()) S = int(input()) result = solve(N, S) print(" ".join(map(str, result))) except (EOFError, ValueError): pass

#include <bits/stdc++.h> using namespace std; long long solve(int N, int M, long long K, long long C, vector<vector<int>>& g) { vector<vector<long long>> dist(N, vector<long long>(M, LLONG_MAX)); dist[0][0] = g[0][0]; priority_queue<tuple<long long,int,int>, vector<tuple<long long,int,int>>, greater<>> pq; pq.push({dist[0][0], 0, 0}); while (!pq.empty()) { auto [d, r, c] = pq.top(); pq.pop(); if (d > dist[r][c]) continue; if (r == N-1 && c == M-1) break; for (int j = c+1; j < M; j++) { long long diff = llabs((long long)g[r][j] - g[r][c]); long long step = j - c; long long cost; if (step == 1) { cost = g[r][j] + (diff > K ? C : 0); } else { if (diff > K) continue; cost = g[r][j] + C * (step - 1); } long long nd = d + cost; if (nd < dist[r][j]) { dist[r][j] = nd; pq.push({nd, r, j}); } } for (int i = r+1; i < N; i++) { long long diff = llabs((long long)g[i][c] - g[r][c]); long long step = i - r; long long cost; if (step == 1) { cost = g[i][c] + (diff > K ? C : 0); } else { if (diff > K) continue; cost = g[i][c] + C * (step - 1); } long long nd = d + cost; if (nd < dist[i][c]) { dist[i][c] = nd; pq.push({nd, i, c}); } } } return dist[N-1][M-1]; } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int N; cin >> N; int M; cin >> M; long long K; cin >> K; long long C; cin >> C; vector<vector<int>> g(N, vector<int>(M)); for (int i = 0; i < N; i++) for (int j = 0; j < M; j++) cin >> g[i][j]; auto result = solve(N, M, K, C, g); cout << result << endl; return 0; }

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def solve(N: int, t: list) -&gt; list: if all(t[k] &lt;= t[k+1] for k in range(N - 1)): return [0, sum(t)] p = next(k for k i
def solve(N: int, t: list) -> list: if all(t[k] <= t[k+1] for k in range(N - 1)): return [0, sum(t)] p = next(k for k in range(N - 1) if t[k] > t[k+1]) M = [0] * N M[N-1] = t[N-1] for i in range(N - 2, -1, -1): M[i] = min(t[i], M[i+1]) prefix_sum = [0] * (N + 1) for i in range(N): prefix_sum[i+1] = prefix_sum[i] + t[i] best = -1 for i in range(0, p + 2): if i > 0 and t[i-1] > M[i]: continue total = prefix_sum[i] + M[i] * (N - i) best = max(best, total) return [1, best]

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#include using namespace std; long long solve(int N, vector&amp; target) { long long total = 0; for (int i = 0; i &lt; N; i++
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#include <bits/stdc++.h> using namespace std; long long solve(int N, vector<int>& target) { long long total = 0; for (int i = 0; i < N; i++) total += target[i]; long long sumX = 0; long long carry = 0; for (int i = 0; i + 1 < N; i++) { long long cap = min((long long)target[i] - carry, (long long)target[i + 1]); if (cap < 0) cap = 0; sumX += cap; carry = cap; } return total - sumX; } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int N; cin >> N; vector<int> target(N); for (int i = 0; i < N; i++) cin >> target[i]; auto result = solve(N, target); cout << result << endl; return 0; }

#include using namespace std; long long solve(int N, vector&amp; target) { long long total = 0; for (int i = 0; i &lt; N; i++
#include <bits/stdc++.h> using namespace std; long long solve(int N, vector<int>& target) { long long total = 0; for (int i = 0; i < N; i++) total += target[i]; long long sumX = 0; long long carry = 0; for (int i = 0; i + 1 < N; i++) { long long cap = min((long long)target[i] - carry, (long long)target[i + 1]); if (cap < 0) cap = 0; sumX += cap; carry = cap; } return total - sumX; } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int N; cin >> N; vector<int> target(N); for (int i = 0; i < N; i++) cin >> target[i]; auto result = solve(N, target); cout << result << endl; return 0; }

#include using namespace std; int solve(int N, int K, vector&gt;&amp; cost, vector&amp; fatigue) { const long long INF = 4e18
#include <bits/stdc++.h> using namespace std; int solve(int N, int K, vector<vector<int>>& cost, vector<int>& fatigue) { const long long INF = 4e18; vector<vector<long long>> prev(K, vector<long long>(2, INF)); vector<vector<long long>> cur(K, vector<long long>(2, INF)); for (int c = 0; c < K; c++) prev[c][1] = cost[0][c]; for (int i = 1; i < N; i++) { for (int c = 0; c < K; c++) { cur[c][0] = cur[c][1] = INF; } for (int last = 0; last < K; last++) { for (int streak = 1; streak <= 2; streak++) { if (prev[last][streak - 1] == INF) continue; for (int now = 0; now < K; now++) { if (now == last) { if (streak == 2) continue; cur[now][1] = min( cur[now][1], prev[last][streak - 1] + cost[i][now] + fatigue[now] );

#include using namespace std; const int MOD = 1000000007; int solve(int N) { vector&gt; dp(N + 1); // last = 0 -&gt; 1 // las
#include <bits/stdc++.h> using namespace std; const int MOD = 1000000007; int solve(int N) { vector<array<long long, 3>> dp(N + 1); // last = 0 -> 1 // last = 1 -> 3 // last = 2 -> 4 for (int s = 1; s <= N; s++) { if (s == 1) { dp[s][0] = 1; } else { long long ways = 0; if (s >= 1) { ways = (dp[s - 1][1] + dp[s - 1][2]) % MOD; } dp[s][0] = ways; } if (s == 3) { dp[s][1] = (dp[s][1] + 1) % MOD; } if (s > 3) { dp[s][1] = (dp[s - 3][0] + dp[s - 3][2]) % MOD; } if (s == 4) { dp[s][2] = (dp[s][2] + 1) % MOD; } if (s > 4) { dp[s][2] = (dp[s - 4][0] + dp[s - 4][1]) % MOD; } } return (dp[N][0] + dp[N][1] + dp[N][2]) % MOD; } int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int N; cin >> N; cout << solve(N) << '\n'; return 0; }

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from collections import defaultdict n = int(raw_input()) L = int(raw_input()) K = int(raw_input()) pts = list(map(int, raw_in
from collections import defaultdict n = int(raw_input()) L = int(raw_input()) K = int(raw_input()) pts = list(map(int, raw_input().split())) groups = defaultdict(list) for x in pts: groups[x % K].append(x) ans = 0 for arr in groups.values(): arr.sort() left = 0 for right in range(len(arr)): while arr[right] - arr[left] > L: left += 1 ans = max(ans, right - left + 1) print code All test cases pass

from collections import defaultdict n = int(raw_input()) L = int(raw_input()) K = int(raw_input()) pts = list(map(int, raw_input().split())) groups = defaultdict(list) for x in pts: groups[x % K].append(x) ans = 0 for arr in groups.values(): arr.sort() left = 0 for right in range(len(arr)): while arr[right] - arr[left] > L: left += 1 ans = max(ans, right - left + 1) print ans

#include <bits/stdc++.h> using namespace std; long long solve(int N, int S, vector<int>& M) { unordered_map<long long, long long> freq; for (int x : M) : long long ans = 0; for (auto &[x, cnt] : freq) { long long y = (long long)S - 2LL * x; if (y < x) continue; auto it = freq.find(y); if (it == freq.end()) continue; if (x == y) ans += cnt * (cnt - 1) / 2; else ans += cnt * it->second; } return ans; } Barycentric Tether calibration ✅ C++

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