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allcoding1_official

allcoding1_official

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📈 Análisis del canal de Telegram allcoding1_official

El canal allcoding1_official (@allcoding1_official) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 79 104 suscriptores, ocupando la posición 1 602 en la categoría Tecnologías y Aplicaciones y el puesto 3 841 en la región India.

📊 Métricas de audiencia y dinámica

Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 79 104 suscriptores.

Según los últimos datos del 23 octubre, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -1 688, y en las últimas 24 horas de -73, conservando un alto alcance.

  • Estado de verificación: No verificado
  • Tasa de interacción (ER): El promedio de interacción de la audiencia es 1.52%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 0.72% de reacciones respecto al total de suscriptores.
  • Alcance de las publicaciones: Cada publicación recibe en promedio 1 202 visualizaciones. En el primer día suele acumular 567 visualizaciones.
  • Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 0.
  • Intereses temáticos: El contenido se centra en temas clave como dsa, stack, namaste, javascript, dev.

📝 Descripción y política de contenido

No se ha proporcionado la descripción del canal.

Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 24 octubre, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Tecnologías y Aplicaciones.

79 104
Suscriptores
-7324 horas
-4047 días
-1 68830 días
Archivo de publicaciones
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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