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allcoding1_official

allcoding1_official

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📈 Аналитический обзор Telegram-канала allcoding1_official

Канал allcoding1_official (@allcoding1_official) языкового сегмента Английский является активным участником. Сейчас сообщество объединяет 79 104 подписчиков, занимая 1 602 место в категории Технологии и приложения и 3 841 место в регионе Индия.

📊 Показатели аудитории и динамика

С момента создания невідомо проект демонстрирует стремительный рост, собрав аудиторию из 79 104 подписчиков.

Согласно последним данным от 23 октября, 2026, канал показывает стабильную активность. За последние 30 дней изменение числа участников составило -1 688, а за последние 24 часа — -73, при этом общий охват остаётся высоким.

  • Статус верификации: Не верифицирован
  • Уровень вовлечённости (ER): Средний показатель вовлечённости аудитории составляет 1.52%. В первые 24 часа после публикации контент обычно набирает 0.72% реакций от общего числа подписчиков.
  • Охват публикаций: В среднем каждый пост получает 1 202 просмотров. В течение первых суток публикация набирает 567 просмотров.
  • Реакции и взаимодействия: Аудитория активно поддерживает контент: среднее количество реакций на один пост — 0.
  • Тематические интересы: Контент сосредоточен на ключевых темах, таких как dsa, stack, namaste, javascript, dev.

📝 Описание и контентная политика

Описание канала не предоставлено.

Благодаря высокой частоте обновлений (последние данные получены 24 октября, 2026) канал поддерживает актуальность и высокий уровень охвата публикаций. Аналитика показывает, что аудитория активно взаимодействует с контентом, что делает его важной точкой влияния в категории Технологии и приложения.

79 104
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-4047 дней
-1 68830 дней
Архив постов
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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