CareerCup

Coupla quick solutions:

n squared:

#!perl -l

use strict;
use warnings;

my @output;

my $y = 0;
my @lines = <DATA>;
for my $line (@lines) {
  chomp $line;
  for my $x (0 .. length ($line) - 1) {
    $output[$y][$x] ||= '-';
    if (substr($line, $x, 1) eq 'x') {
      for my $i (0 .. length ($line) - 1) {
        $output[$y][$i] = 'x';
      }
      for my $i (0 .. $#lines) {
        $output[$i][$x] = 'x';
      }
    }
  }
  $y++;
}

print @$_ for @output;
__DATA__
-----
-----
---x-
x----
-----

Linear:

use strict;
use warnings;

my $y = 0;
my @lines = <DATA>;
my (%x, %y);
for my $line (@lines) {
  chomp $line;
  for my $x (0 .. length ($line)) {
    if (substr($line, $x, 1) eq 'x') {
      $x{$x}++;
      $y{$y}++;
    }
  }
  $y++;
}

for my $i (0 .. @lines - 1) {
  for my $j (0 .. length($lines[0]) - 1) {
    print $y{$i} || $x{$j} ? 'x' : '-';
  }
  print "\n";
}
__DATA__
-----
-----
---x-
x----
-----

public class Attempt1{
    boolean[] row = new boolean[n];
    boolean[] column = new boolean[n];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (matrix[i][j] == "x") {
                row[i] = true;
                column[j] = true;
            }
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (row[i] || column[j]) {
                matrix[i][j] = "x";
            }
        }
    }
    System.out.println("Resultant matrix: ");
    printMatrix(matrix, n);
    return matrix;
}

Here is a solution that I can think of. Please provide inputs if there is a better way to solve it. Thank you!

public class StackOver1{
    boolean[] row = new boolean[n];
    boolean[] column = new boolean[n];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (matrix[i][j] == "x") {
                row[i] = true;
                column[j] = true;
            }
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (row[i] || column[j]) {
                matrix[i][j] = "x";
            }
        }
    }
    System.out.println("Resultant matrix: ");
    printMatrix(matrix, n);
    return matrix;
}

Here are my questions: 1. Is there a better way to solve this?
2. Should I consider a better data structure?

public class Attempt1{
    boolean[] row = new boolean[n];
    boolean[] column = new boolean[n];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (matrix[i][j] == "x") {
                row[i] = true;
                column[j] = true;
            }
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (row[i] || column[j]) {
                matrix[i][j] = "x";
            }
        }
    }
    System.out.println("Resultant matrix: ");
    printMatrix(matrix, n);
    return matrix;
}

Here are my questions: 1. Is there a better way to solve this? 2. Should I consider a better data structure?

Thank you.

We could use a HashSet to store (row,column) values where 'x' is found instead of O(n2) space this would require O(k) space where k is the times 'x' appears in the matrix.

traversing a row till the end can be skipped once a 'x' is found in a row.
once a whole row is set with 'x' it could be marked to prevent duplicate 'x' assignment

test