# Challenge 126

## Task 1 - Count Numbers

You are given a positive integer \$N.

Write a script to print count of numbers from 1 to \$N that donâ€™t contain digit 1.

Example:

Input: \$N = 15
Output: 8

There are 8 numbers between 1 and 15 that don't contain digit 1.
2, 3, 4, 5, 6, 7, 8, 9.

Input: \$N = 25
Output: 13

There are 13 numbers between 1 and 25 that don't contain digit 1.
2, 3, 4, 5, 6, 7, 8, 9, 20, 22, 23, 24, 25.

### Raku

Loop over 1..\$N and grep for numbers that don't have 1. hyper hints the compiler that this can be run in parallel.

print .join: ', ' with (1..\$N).hyper.grep(*.comb.grep(1).elems == 0);
put '.';

### C

argv holds the input & argc holds the number of inputs. The input should be a single integer so argc should be equal to 2. After checking for that, we check if valid integer was passed.

After checks, we loop over all numbers from 1 to num (passed value) and check if they contain digit 1. The check is performed by converting num to a string and matching for '1' in the string. If we find it then we set the flag to false and don't print the number, otherwise we print it.

if (argc != 2) {
puts("Usage: ./ch-1 <number>");
exit(0);
}

const char *errstr;
int num = strtonum(argv[1], 1, INT_MAX, &errstr);
if (errstr != NULL)
errx(1, "number is %s: %s", errstr, argv[1]);

for (int idx = 1; idx <= num; idx++) {
bool take = true;

int str_size = 1 + snprintf(NULL, 0, "%d", idx);
char *num_str = calloc(str_size, sizeof(char));
snprintf(num_str, str_size, "%d", idx);

for (int x = 0; num_str[x] != '\0'; x++)
if (num_str[x] == '1') take = false;
free(num_str);

if (take == true) printf("%d ", idx);
}
printf("\n");

## Task 2 - Minesweeper Game

You are given a rectangle with points marked with either x or *. Please consider the x as a land mine.

Write a script to print a rectangle with numbers and x as in the Minesweeper game.

A number in a square of the minesweeper game indicates the number of mines within the neighbouring squares (usually 8), also implies that there are no bombs on that square.

Example:

Input:
x * * * x * x x x x
* * * * * * * * * x
* * * * x * x * x *
* * * x x * * * * *
x * * * x * * * * x

Output:
x 1 0 1 x 2 x x x x
1 1 0 2 2 4 3 5 5 x
0 0 1 3 x 3 x 2 x 2
1 1 1 x x 4 1 2 2 2
x 1 1 3 x 2 0 0 1 x

### Raku

@rect holds the challenge grids and @grids holds the solution. We loop over every cell and if it's not a land mine then we loop over it's neighbors and find the number of neighboring land mines and set it's value.

The neighbors is returned by neighbors subroutine which is adapted from Octans::Neighbors (projects/octans) which was adapted from my 2020 AoC day 11's solution.

my @rect = \$input.IO.lines.map(*.words.cache);
die "Not rectangle" unless [==] @rect.map(*.elems);

my @grid;
for 0 .. @rect.end -> \$r {
for 0 .. @rect[\$r].end -> \$c {
given @rect[\$r][\$c] {
when "x" { @grid[\$r][\$c] = @rect[\$r][\$c] }
when "*" {
@grid[\$r][\$c] = 0;
for neighbors(@rect, \$r, \$c).List -> \$pos {
@grid[\$r][\$c]++ if @rect[\$pos[0]][\$pos[1]] eq "x";
}
}
}
}
}
.put for @grid;

Andinus / 2021-08-19 / Modified: 2022-10-04 Tue 21:36 Emacs 27.2 (Org mode 9.4.4)