#include <ctype.h>
#include <stdbool.h>
#include <stddef.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "aocdailies.h"
#include "aocutils.h"

/* === aoc202408 =======================================================
   Oh! another square of text!
   Idea: for all points p with an antenna find all points q>p with
a corresponding frequency. For each such pair calculate the 2 antinodes
and add the resulting points (if not there already) to an array.
   The answer to Part One is the number of elements in the array
   TODO: I don't like my `linearize2d` thing! Thinking of a replacement
===================================================================== */

void aoc202408(char *data, size_t len) {
    (void)len; // unused argument
    unsigned anti[256][2]; // locations of antinodes
    unsigned nanti = 0;
    unsigned size = 1 + strchr(data, '\n') - data;
    for (unsigned row = 0; row < size-1; row++) {
        for (unsigned col = 0; col < size-1; col++) {
            char *p = data + linearize2d(size, row, col);
            if (*p != '.') {
                char *q = strchr(p + 1, *p);
                while (q) {
                    printf("found '%c'('%c')\n", *p, *q);
                    ptrdiff_t delta = q - p;
                    #if 0
    int row_pos = (int)(linear_pos / size);
    int col_pos = (int)(linear_pos % size);
                    #endif
                    if (p >= data + delta) {
                        // antinode
                        anti[nanti][0] = row(p - delta);
                        anti[nanti++][1] = col(p - delta);
                    }
                    if (q < "end of data" - delta) {
                        // antinode
                        anti[nanti][0] = row(q + delta);
                        anti[nanti++][1] = col(q + delta);
                    }
                    q = strchr(q + 1, *p);
                }
            }
        }
    }
    printf("There are %u antinodes within the map.\n", nanti);
}

/* === aoc202407 =======================================================
   Part one looks easy
   Part two also easy
===================================================================== */

static unsigned long long concat(unsigned long long a, unsigned long long b) {
    unsigned long long r = a, t = b;

/* === aoc202406 =======================================================
   Hmmm, I've done something like this already this year... on day 4
   I need rcmap() moved to aocutils ... done; now it's called `linearize2d`
Part Two: brute force? It likely is faster than coming up with an algo
   Let's go: save map, for every available position put in an obstacle,
   loop around until some place is visited 4? times or area is exited
   TODO remove linearize2d
===================================================================== */

static bool validpos(unsigned s, int r, int c) {
    if (r < 0) return false;
    if (c < 0) return false;
    if ((unsigned)r >= s - 1) return false;
    if ((unsigned)c >= s - 1) return false;

        accumsum += accum;
    } while ((line = strtok(NULL, "\n")) != NULL);
    printf("The sum of middle update numbers is {%u}.\n", accumsum);
    printf("The sum of middle update numbers for newly ordered updates is {%u}.\n", accumsum2);
}

/* === aoc202404 =======================================================
   TODO remove linearize2d
===================================================================== */

static bool masat(char *data, unsigned size, unsigned row, unsigned col, int drow, int dcol) {
    int maxrow = (int)row + 3*drow;
    int maxcol = (int)col + 3*dcol;
    if ((0 <= maxrow) && (maxrow < (int)size) && (0 <= maxcol) && (maxcol < (int)size)) {
        if (data[linearize2d(size+1, (unsigned)((int)row + drow), (unsigned)((int)col + dcol))] != 'M') return false;