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لا تقع ضحية للمخادعين! تيليمتريو يكتشف ويُميّز هذه القنوات 👉 إذا كنت تريد رؤية العلامة، اشترك

allcoding1_official

allcoding1_official

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📈 نظرة تحليلية على قناة تيليجرام allcoding1_official

تُعد قناة allcoding1_official (@allcoding1_official) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 79 104 مشتركاً، محتلاً المرتبة 1 602 في فئة التكنولوجيات والتطبيقات والمرتبة 3 841 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 79 104 مشتركاً.

بحسب آخر البيانات بتاريخ 23 أكتوبر, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -1 688، وفي آخر 24 ساعة بمقدار -73، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 1.52‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 0.72‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 1 202 مشاهدة. وخلال اليوم الأول يجمع عادةً 567 مشاهدة.
  • التفاعلات والاستجابة: يتفاعل الجمهور بانتظام؛ متوسط التفاعلات لكل منشور يبلغ 0.
  • الاهتمامات الموضوعية: يركز المحتوى على مواضيع رئيسية مثل dsa, stack, namaste, javascript, dev.

📝 الوصف وسياسة المحتوى

وصف القناة غير متوفر.

بفضل وتيرة التحديث المرتفعة (أحدث البيانات بتاريخ 24 أكتوبر, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التكنولوجيات والتطبيقات.

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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 } }

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 } }

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}}

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