package engine

import (
	"gitea.iliadenisov.ru/developer/scrabble-solver/board"
	"gitea.iliadenisov.ru/developer/scrabble-solver/scrabble"
)

// resolveDirection infers a play's orientation from the placed tiles and the
// board, so a caller need not declare it (docs/ARCHITECTURE.md §5). Two or more
// tiles fix the orientation by the line they share: a common row reads
// horizontally, otherwise vertically (a non-linear placement is left for
// Evaluate to reject). A single tile is ambiguous on its own — it may extend a
// word down a column or across a row — so the orientation is the axis along
// which it abuts existing tiles, preferring the axis that yields the longer word
// and horizontal on a tie. A tile that abuts nothing falls back to horizontal
// and is rejected downstream as disconnected (or, on the first move, as too
// short).
func resolveDirection(b *board.Board, placements []scrabble.Placement) scrabble.Direction {
	if len(placements) >= 2 {
		row := placements[0].Row
		for _, p := range placements[1:] {
			if p.Row != row {
				return scrabble.Vertical
			}
		}
		return scrabble.Horizontal
	}
	if len(placements) == 1 {
		p := placements[0]
		h := runLength(b, p.Row, p.Col, scrabble.Horizontal)
		v := runLength(b, p.Row, p.Col, scrabble.Vertical)
		if v >= 2 && v > h {
			return scrabble.Vertical
		}
		if h >= 2 {
			return scrabble.Horizontal
		}
		if v >= 2 {
			return scrabble.Vertical
		}
	}
	return scrabble.Horizontal
}

// runLength returns how many cells the word through (row, col) along dir would
// span once a tile is placed on the empty target square: the square itself plus
// the runs of filled cells immediately before and after it along dir. A result
// below two means the tile forms no word on that axis. Filled treats
// off-board coordinates as empty, so the walks stop at the board edge.
func runLength(b *board.Board, row, col int, dir scrabble.Direction) int {
	dr, dc := 0, 1
	if dir == scrabble.Vertical {
		dr, dc = 1, 0
	}
	n := 1
	for r, c := row-dr, col-dc; b.Filled(r, c); r, c = r-dr, c-dc {
		n++
	}
	for r, c := row+dr, col+dc; b.Filled(r, c); r, c = r+dr, c+dc {
		n++
	}
	return n
}
