allcoding1_official
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📈 Telegram 频道 allcoding1_official 的分析概览
频道 allcoding1_official (@allcoding1_official) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 79 104 名订阅者,在 技术与应用 类别中位列第 1 602,并在 印度 地区排名第 3 841 位。
📊 受众指标与增长动态
自 невідомо 创建以来,项目保持高速增长,吸引了 79 104 名订阅者。
根据 23 十月, 2026 的最新数据,频道保持稳定运转。过去 30 天订阅人数变化为 -1 688,过去 24 小时变化为 -73,整体触达仍然可观。
- 认证状态: 未认证
- 互动率 (ER): 平均受众互动率为 1.52%。内容发布后 24 小时内通常能获得 0.72% 的反应,占订阅者总量。
- 帖子覆盖: 每篇帖子平均可获得 1 202 次浏览,首日通常累积 567 次浏览。
- 互动与反馈: 受众积极参与,单帖平均反应数为 0。
- 主题关注点: 内容集中在 dsa, stack, namaste, javascript, dev 等核心主题上。
📝 描述与内容策略
尚未提供频道描述。
凭借高频更新(最新数据采集于 24 十月, 2026),频道始终保持新鲜度与高覆盖。分析显示受众积极互动,使其成为 技术与应用 类别中的关键影响点。
79 104
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79 104
1Ans) java code
import java.util.ArrayList;
import java.util.List;
public class Solution {
public static int getAns(int N, List> A) {
// This will store the maximum beauty sum of two non-intersecting squares
int maxSumOfBeauties = 0;
// Precompute the beauty of all sub-grids for quick lookup
int[][] beauty = new int[N][N];
for (int row = 0; row < N; row++) {
for (int col = 0; col < N; col++) {
// Calculate beauty for sub-grid starting at (row, col)
int minVal = Integer.MAX_VALUE;
int maxVal = Integer.MIN_VALUE;
for (int i = row; i < N; i++) {
for (int j = col; j < N; j++) {
minVal = Math.min(minVal, A.get(i).get(j));
maxVal = Math.max(maxVal, A.get(i).get(j));
beauty[i][j] = maxVal - minVal;
}
}
}
}
// Try all possible pairs of non-intersecting sub-grids
for (int r1 = 0; r1 < N; r1++) {
for (int c1 = 0; c1 < N; c1++) {
for (int r2 = 0; r2 < N; r2++) {
for (int c2 = 0; c2 < N; c2++) {
if (r1 != r2 && c1 != c2) {
// Ensure the sub-grids do not share any rows or columns
int beauty1 = beauty[r1][c1];
int beauty2 = beauty[r2][c2];
maxSumOfBeauties = Math.max(maxSumOfBeauties, beauty1 + beauty2);
}
}
}
}
}
return maxSumOfBeauties;
}
public static void main(String[] args) {
// Example usage:
List> grid = new ArrayList<>();
grid.add(List.of(2, 3));
grid.add(List.of(2, 3));
int N = 2;
System.out.println(getAns(N, grid)); // Output should be 0
}
}
79 104
1Ans) java code
import java.util.ArrayList;
import java.util.List;
public class Solution {
public static int getAns(int N, List> A) {
// This will store the maximum beauty sum of two non-intersecting squares
int maxSumOfBeauties = 0;
// Precompute the beauty of all sub-grids for quick lookup
int[][] beauty = new int[N][N];
for (int row = 0; row < N; row++) {
for (int col = 0; col < N; col++) {
// Calculate beauty for sub-grid starting at (row, col)
int minVal = Integer.MAX_VALUE;
int maxVal = Integer.MIN_VALUE;
for (int i = row; i < N; i++) {
for (int j = col; j < N; j++) {
minVal = Math.min(minVal, A.get(i).get(j));
maxVal = Math.max(maxVal, A.get(i).get(j));
beauty[i][j] = maxVal - minVal;
}
}
}
}
// Try all possible pairs of non-intersecting sub-grids
for (int r1 = 0; r1 < N; r1++) {
for (int c1 = 0; c1 < N; c1++) {
for (int r2 = 0; r2 < N; r2++) {
for (int c2 = 0; c2 < N; c2++) {
if (r1 != r2 && c1 != c2) {
// Ensure the sub-grids do not share any rows or columns
int beauty1 = beauty[r1][c1];
int beauty2 = beauty[r2][c2];
maxSumOfBeauties = Math.max(maxSumOfBeauties, beauty1 + beauty2);
}
}
}
}
}
return maxSumOfBeauties;
}
public static void main(String[] args) {
// Example usage:
List> grid = new ArrayList<>();
grid.add(List.of(2, 3));
grid.add(List.of(2, 3));
int N = 2;
System.out.println(getAns(N, grid)); // Output should be 0
}
}
79 104
Q1: Two Square minimax
You are given a square grid A of size N x N.
Your want to choose two non-intersecting square sub-grids from the grid such that they have no common row or column The objective is to maximize the sum of beauties of both square sub-grids.
The beauty of a sub-grid is defined as the difference betwee the maximum value and the minimum value within that sub- grid.
tikqupta2s
Find the maximum possible sum of beauties for the sel two sub-grids.
Input Format
The next line contains an integer, N, denoting the number o columns in A.
Each line i of the N subsequent lines (where 0 ≤ i < N) com space separated integers each describing the row A[i].
Constraints
2 <= N <= 250
1 <= A[i][j] <= 10^5
Sample Test Cases
Case 1
Input:
2
23
23
Output:
0
Explanation:
Given N 2, A = [[2, 3], [2, 3]].
We take cell (1, 2) and cell (2, 1) and the answer will be 0
Public class solution {
Public static int get_ans(int N, List> A}}
79 104
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⚡️ Enumeration
⚡️ Firewalls HIDs Honeypot
⚡️ Malware and Threats
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⚡️ Pentesting
⚡️ Sql Injection
⚡️ System Hacking
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⚡️ Reconnaissance and Footprinting
⚡️ Network Scanning
⚡️ Enumeration
⚡️ Firewalls HIDs Honeypot
⚡️ Malware and Threats
⚡️ Mobile Platform
⚡️ Pentesting
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79 104
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79 104
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⚡️ Reconnaissance and Footprinting
⚡️ Network Scanning
⚡️ Enumeration
⚡️ Firewalls HIDs Honeypot
⚡️ Malware and Threats
⚡️ Mobile Platform
⚡️ Pentesting
⚡️ Sql Injection
⚡️ System Hacking
⚡️ Web Application
⚡️ Wireless Network
⚡️ Cloud Computing
⚡️ Web Server
⚡️ Social Engineering
⚡️ Session Hijacking
⚡️ Sniffing
⚡️ BufferOverflow
⚡️ Cryptography
⚡️ Denial Of Service
All courses (100 rupees)
Contact:- @meterials_available
