Question

in javascript OR JAVA

You are given two numbers n and m representing the dimensions of an n × m rectangular board. The rows of the board are numbered from 1 to n, and the columns are numbered from 1 to m. Each cell has a value equal to the product of its row index and column index (both 1-based); in other words, board[i][j] = (i + 1) * (j + 1).

Initially, all the cells in the board are considered active, though some of them will eventually be deactivated through a sequence of queries - specifically, you will be given an array queries, where each query is of one of the following 3 types:

• [0] - find the minimum value among all remaining active cells on the board.
• [1, i] - deactivate all cells in row i;
• [2, j] - deactivate all cells in column j;

Given the dimensions n, m, and the array of queries, your task is to return an array consisting of calculated values (results of the queries of the 0^th type), in the order in which they were calculated.

Answer 1

Hey , i am solving this problem using JAVA.

Actually I am sorry to say this, The question is not very much clear. But still I tried my best to solve the question. If you feel something has been missed or you want me to change some code , you can ask me in the comment section. I will make changes ASAP.

Assumptions I made:

1. ACTIVE cell means the cell with NON - ZERO value. ( Because as per the question , no cell will be initialized with ZERO value , (i+1)*(j+1) where i and j starts from 0 ).

2. DEACTIVATED cell means , the cell with value ZERO. So I made that cell value as ZERO.

3. Array of Queries:

you will be given an array queries, where each query is of one of the following 3 types:

• [0] - find the minimum value among all remaining active cells on the board.
• [1, i] - deactivate all cells in row i;
• [2, j] - deactivate all cells in column j;

So, What I did is I made a 2 dimensional queries array.

4. your task is to return an array consisting of calculated values (results of the queries of the 0^th type), in the order in which they were calculated.

• So what I assumed is, to return all those minimum values which were deactivated by the query ,   [ 0 ] -- find the minimum value among all remaining active cells on the board.
• So I returned an array of size queries.length ( if you want you can modify that size, but it will have a zero ( 0 ) value in the empty spaces in the array )

Now I am providing the code here:

===========================================================================================================================================================

import java.util.*;

public class Main {
  
// main( ) method:
public static void main(String[] args) {
  
Scanner sc = new Scanner(System.in);
// ask user to enter the no.of rows
System.out.print("Enter the number of rows: ");
int rows = sc.nextInt();
// ask user to enter the no.of columns
System.out.print("\nEnter the number of columns: ");
int cols = sc.nextInt();
  
int[][] board = new int[rows][cols];
  
// initialization of values on board
for(int i=0; i<rows; i++) {
for(int j=0; j<cols;j++) {
board[i][j] = (i+1)*(j+1);
}
}
// end of for loop ( intializiation completed)
  
  
// array of queries
int[][] queries = new int[3][];
queries[0] = new int[] {0};
// [1,i] to deactivate all cells in 'i'th row.
queries[1] = new int[]{1,1};
// [2,j] to deactivate all cells in 'j'th column
queries[2] = new int[]{2,2};
  
int[] res = solve(queries , board);
  
for(int z =0; z< res.length;z++) {
System.out.println("calculated values (results of the queries of the 0th type): "+ res[z]);
}
}
  
  
// ACTIVE CELLS === Cells whose value is NOT zero
// DEACTIVATED CELLS ==== Cells whose value is ZERO.
public static int[] solve(int[][] queries , int[][] board ) {
  
int[] result = new int[queries.length] ;
int z =0;
for(int i = 0; i<queries.length;i++) {
  
// find the minimum value among all remaining active cells on the board.
if(queries[i][0] == 0) {
  
// initialize ' min ' variable with some very high value.
int min = 1000000;
int min_row = 0,min_col = 0 ;
for(int x =0; x < board.length; x++) {
for(int y =0; y < board[i].length; y++) {
  
if(board[x][y] < min ) {
min = board[x][y];
min_col = y;
min_row = x;
}
}
} // for loop END ( found a minimum value) and it's indexes.
  
result[z] = min;z++;
board[min_row][min_col] = 0;
  
  
}
  
// [1,i] === deactivate all cells in row i;
else if(queries[i][0] == 1) {
  
int x = queries[i][1];
for(int k =0; k < board[0].length; k++) {
board[x][k] = 0;
}
}
  
// [2,j] ===== deactivate all cells in column j;
else if(queries[i][0] == 2 ) {
  
int y = queries[i][1];
  
for(int k=0; k < board.length; k++) {
board[k][y] = 0;
}
  
  
}
  
}
return result;
}
  
}

==============================================================

OUTPUT

The first value is 1 , because the minimum value is 1. ( i.e.. i =0, j =0. the value at a[ i ][ j ] = ( i + 1 ) * ( j + 1) = 1

And the other two values are 0 because , the size of array is queries.length ( which is 3 in our case ). So the default value in the array is zero.

But last two values is not because of any query.

I hope you got your answer, if not please do comment I will modify code as per your request. THANK YOU !.