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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 79 557 subscribers, ranking 1 588 in the Technologies & Applications category and 3 855 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 79 557 subscribers.

According to the latest data from 14 October, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -1 647 over the last 30 days and by 0 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 0.88%. Within the first 24 hours after publication, content typically collects 0.65% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 699 views. Within the first day, a publication typically gains 517 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 15 October, 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.

79 557
Subscribers
No data24 hours
-3057 days
-1 64730 days
Posts Archive
Infosys Exam 2PM BE READY I WILL SHARE ANSWERS Loading..... Please share with your friends @allcoding1_official

Infosys Exam 2PM BE READY I WILL SHARE ANSWERS Please share with your friends

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import java.util.*; class Main { public static long solve(int N, int M, int G, int[] A, int[] B) { if (N == 0 || M == 0) return 0; int maxL = 1; for (int offset = -(M - 1); offset < N; offset++) { int startA = Math.max(0, offset); int endA = Math.min(N - 1, offset + M - 1); int len = endA - startA + 1; if (len <= maxL) continue; int[] diffs = new int[len]; for (int k = 0; k < len; k++) { int idxA = startA + k; int idxB = idxA - offset; diffs[k] = A[idxA] - B[idxB]; } Map> shiftMap = new HashMap<>(); for (int k = 0; k < len; k++) { shiftMap.computeIfAbsent(diffs[k], x -> new ArrayList<>()).add(k); } for (int d : shiftMap.keySet()) { List exactIndices = shiftMap.get(d); int nExact = exactIndices.size(); for (int e = 0; e < nExact; e++) { int left = exactIndices.get(e); int right = exactIndices.get(e); int graceCount = 0; int curLen = 0; for (int k = left; k < len; k++) { if (diffs[k] == d) { curLen++; } else if (Math.abs(diffs[k] - d) <= G && graceCount == 0) { graceCount++; curLen++; } else { break; } } for (int k = left - 1; k >= 0; k--) { if (diffs[k] == d) { curLen++; } else if (Math.abs(diffs[k] - d) <= G && graceCount == 0) { graceCount++; curLen++; } else { break; } } maxL = Math.max(maxL, curLen); } } } return maxL; }

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int solve(int n, int index, int maxSum) { auto getSum = [](long long peak, long long count, long long diff) { long long k = m
int solve(int n, int index, int maxSum) { auto getSum = [](long long peak, long long count, long long diff) { long long k = min(count, (peak - 1) / diff); long long sum = k * peak - diff * k * (k + 1) / 2; sum += (count - k); return sum; }; auto possible = [&](long long peak) { long long left = getSum(peak, index, 1); long long right = getSum(peak, n - index - 1, 2); long long total = peak + left + right; return total <= maxSum; }; long long low = 1; long long high = maxSum; long long ans = 1; while (low <= high) { long long mid = low + (high - low) / 2; if (possible(mid)) { ans = mid; low = mid + 1; } else { high = mid - 1; } } return (int)ans; }

import java.util.*; class Main { static final int MOD = 1000000007; public static int solve(int N, int F, int T) { int m = N - 2; // dp[s] = ways to obtain sum s using processed nodes long[] dp = new long[T + 1]; dp[0] = 1; for (int i = 0; i < m; i++) { long[] ndp = new long[T + 1]; for (int sum = 0; sum <= T; sum++) { if (dp[sum] == 0) continue; for (int v = 1; v <= F && sum + v <= T; v++) { ndp[sum + v] += dp[sum]; if (ndp[sum + v] >= MOD) ndp[sum + v] %= MOD; } } dp = ndp; } long waysPerPosition = 0; for (int a = 1; a <= F; a++) { for (int b = 1; b <= F; b++) { int need = T - a * b; if (need >= 0 && need <= T) { waysPerPosition += dp[need]; waysPerPosition %= MOD; } } } return (int) ((waysPerPosition * (N - 1)) % MOD); }

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import java.util.*; class Main { public static int solve(int N, int[] arr) { int maxVal = 0; for (int x : arr) { if (x &gt; m
import java.util.*; class Main { public static int solve(int N, int[] arr) { int maxVal = 0; for (int x : arr) { if (x > maxVal) maxVal = x; } int[] freq = new int[maxVal + 1]; for (int x : arr) { freq[x]++; } int need = N / 2; int left = 1; int count = 0; int ans = Integer.MAX_VALUE; for (int right = 1; right <= maxVal; right++) { count += freq[right]; while (count >= need) { ans = Math.min(ans, right - left + 1); count -= freq[left]; left++; } } return ans; }

#include <bits/stdc++.h> using namespace std; long long sideSum(long long v, long long count, long long step) { if (count <= 0) return 0; long long full = min(count, (v - 1) / step); long long sum = full * v - step * full * (full + 1) / 2; sum += (count - full) * 1LL; return sum; } bool feasible(long long v, long long n, long long index, long long maxSum) { long long left = index; long long right = n - 1 - index; long long total = v + sideSum(v, left, 1) + sideSum(v, right, 2); return total <= maxSum; } int solve(int n, int index, int maxSum) { long long lo = 1, hi = maxSum, ans = 1; while (lo <= hi) { long long mid = lo + (hi - lo) / 2; if (feasible(mid, n, index, maxSum)) { ans = mid; lo = mid + 1; } else { hi = mid - 1; } } return (int)ans; } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int n; cin >> n; int index; cin >> index; int maxSum; cin >>

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import java.util.*; class Main { public static int solve(int P, int M, int k, int[][] piles) { List best = new ArrayList<>(); for (int i = 0; i < P; i++) { ArrayList a = new ArrayList<>(); for (int j = 0; j < M; j++) { if (piles[i][j] != -1) a.add(piles[i][j]); } int n = a.size(); int[] mx = new int[n + 1]; Arrays.fill(mx, Integer.MIN_VALUE); mx[0] = 0; if (n > 0) { int[] pre = new int[n + 1]; for (int j = 0; j < n; j++) pre[j + 1] = pre[j] + a.get(j); for (int len = 1; len <= n; len++) { int bestSum = Integer.MIN_VALUE; for (int s = 0; s + len <= n; s++) { bestSum = Math.max(bestSum, pre[s + len] - pre[s]); } mx[len] = bestSum; } } best.add(mx); } int NEG = -1000000000; int[] dp = new int[k + 1]; Arrays.fill(dp, NEG); dp[0] = 0; for (int[] cur : best) { int[] ndp = new int[k + 1]; Arrays.fill(ndp, NEG); int lim = cur.length - 1; for (int used = 0; used <= k; used++) { if (dp[used] == NEG) continue; for (int take = 0; take <= lim && used + take <= k; take++) { ndp[used + take] = Math.max( ndp[used + take], dp[used] + cur[take] ); } } dp = ndp; } return dp[k]; }

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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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allcoding1_official - Statistics & analytics of Telegram channel @allcoding1_official