3 minutes
Leetcode 36
The main idea of this solution is to find the rows, columns, and squares of the numbers. By squares, I mean the nine squares in the board:
The code goes through every position, and if the value is not a "."
, we check whether the row, column, or square contains the number. If so, we can return false.
How to Calculate The Rows And Columns
We check whether the row or column contains the number by using primes. We assign every row and column a prime number:
Multiple prime numbers multiplyed together can only be divide by those same prime numbers, If we find a 5
in row 1
we multiply row[5]
by primes[1]
, and primes[1] = 3
, and if we find another 5
in row 1
we can check whether row[5] % primes[1] == 0
which equals 3 % 3 = 0
. But let us say that we have one 5
in row 1
and another 5
in row 2
, the code wont return false because row[5] = primes[1] = 3
, and row[5] % primes[2] == 0 = 3 % 5 != 0
.
How to Calculate The Squares:
We calculate the numbers in the squares by also using primes, but they are used differently. We us the primes for the values of sudoko, "1" : 2, "2" : 3, "3" : 5, "4" : 7, "5" : 11, "6" : 13, "7" : 17, "8" : 19, "9" : 23
.
And we find which square each position goes into by doing squares[i / 3][j / 3]
. squares
is a jagged array of size 3 by 3, each position in squares
represents a blue square in the first image. When we do i / 3
and round it down to the nearest integer and j / 3
rounded down to the nearest integer, we get:
The Code:
func isValidSudoku(board [][]byte) bool {
primes := []int{2, 3, 5, 7, 11, 13, 17, 19, 23}
rows := make(map[int]int)
cols := make(map[int]int)
squares := make([][]int, 3)
squares[0] = []int{1, 1, 1}; squares[1] = []int{1, 1, 1}; squares[2] = []int{1, 1, 1}
for i := 0; i < len(board); i++ {
for j := 0; j < len(board[0]); j++ {
if board[i][j] != '.' {
boardPosition, _ := strconv.Atoi(string(board[i][j]))
if rows[boardPosition] == 0 {
rows[boardPosition] = 1
}
if cols[boardPosition] == 0 {
cols[boardPosition] = 1
}
if rows[boardPosition] % primes[i] == 0 ||
cols[boardPosition] % primes[j] == 0 ||
squares[i / 3][j / 3] % primes[boardPosition - 1] == 0 {
return false
}
cols[boardPosition] *= primes[j]
rows[boardPosition] *= primes[i]
squares[i / 3][j / 3] *= primes[boardPosition - 1]
}
}
}
return true
}