Problem Statement
We have to implement an isValidSudoku function that takes a 9x9 matrix (representing a Sudoku grid) as input and returns true if the grid is valid and false otherwise.
Sudoku Grid
A Sudoku grid is valid if
- Every row contains digits from 1 to 9 without repetition.
Rows in a Sudoku Grid
- Every column contains digits from 1 to 9 without repetition.
Columns in a Sudoku Grid
- Every 3x3 sub-matrix contains digits from 1 to 9 without repetition.
Sub-Matrices in a Sudoku Grid
The input Sudoku table could be incomplete where “.” will represent an empty cell.
Optimal Solution
The time complexity of validating a single row or column for repetition using a hashmap (like in containsDuplicate) will be $O(9)$. Repeating this process for every row and column in the input matrix will result in total time complexity of $9 \times O(9) + 9 \times O(9)$.
To validate all 3x3 sub-matrices we have to iterate over every 3rd row and column of the Sudoku grid. On each iteration, we will start from the top-left corner of the 3x3 grid and reach the bottom-right corner using nested loops. The total time complexity of validating each 3x3 submatrix will be $9 \times O(3) \times O(3)$.
Since the total time complexity of this solution will be constant for all inputs (because the Sudoku grid has a constant size of 9x9) this is the optimal solution.
Psuedo code for the Optimal Solution
loop row_index from 0 to 9
row_hashmap = Hashmap()
sudoku_row = sudoku_grid[row_index]
loop column_index from 0 to 9
if row_hashmap[sudoku_row[column_index]]
return false
else
row_hashmap[sudoku_row[column_index]] = 1
loop column_index from 0 to 9
column_hashmap = Hashmap()
loop row_index from 0 to 9
value = sudoku_grid[row_index][column_index]
if column_hashmap[value]
return false
else
column_hashmap[value] = 1
loop row_index from 0 to 6 with step 3
loop column_index from 0 to 6 with step 3
sub_matrix_hashmap = Hashmap()
loop sm_row_index from 0 to 2
loop sm_column_index from 0 to 2
r_value = row_index+sm_row_index
c_value = column_index+sm_column_index
if sub_matrix_hashmap[r_value][c_value]
return false
else
sub_matrix_hashmap[r_value][c_value] = 1
Time Complexity Analysis
Best Case Scenario
The best-case input for the optimal solution will be a completed Sudoku grid. The time taken to return the result will be $9 \times O(9) + 9 \times O(9) + O(81)$ which could be simplified to $O(1)$.
Worst Case Scenario
In the worst-case scenario, the input will be an invalid Sudoku grid. The time taken to produce the result will be the same as the best-case scenario i.e. O(1).
Space Complexity Analysis
The space complexity for the optimal solution will be the sum of memory space required by the row_hashmap, column_hashmap, and sub_matrix_hashmap i.e. $O(9)+O(9)+O(9)$ which could be simplified to $O(1)$.
Code for the Optimal Solution
First, we will implement the validCellValue function to check if the value in a cell is within 1 to 9.
func validCellValue(inputValue byte)(bool){
// Converting byte value to integer
// to validate if inputValue lies between 1-len(board)
intInputValue, _ := strconv.Atoi(string(inputValue))
if ((intInputValue<10) && (intInputValue>0)){
return true
}
return false
}
To validate each row we can create a function called checkRowValidity.
func checkRowValidity(board [][]byte)(bool){
// Iterating over each row
// The length of the board could be hard coded to 9
for rowIndex:=0;rowIndex<len(board);rowIndex++{
// The rowHashMap would be used to check for
// repeating values in the row
rowHashMap:=make(map[byte]int)
// Iterating over each column in the row
for colIndex:=0;colIndex<len(board);colIndex++{
cellValue := board[rowIndex][colIndex]
// If the value of the cell isn't blank i.e. "."
if(string(cellValue) != "."){
// If the cell value is within 1-9
if(validCellValue(cellValue)){
// Check for value in rowHashMap
_, key_exists := rowHashMap[cellValue]
if key_exists{
// Exit the function if we
// encountered a repeating value
// in the current row
return false
} else {
// If value isn't present
// then add it the the hashmap
rowHashMap[cellValue] = 1
}
} else {
// Exit the function if we
// encountered a value that's not
// within range 1-9
return false
}
}
}
}
return true
}
We can implement a similar function checkColumnValidity to validate columns.
func checkColValidity(board [][]byte)(bool){
// Iterating over each column
for colIndex:=0;colIndex<len(board);colIndex++{
// The colHashMap would be used to check for
// repeating values in the column
colHashMap:=make(map[byte]int)
// Iterating over each row in the column
for rowIndex:=0;rowIndex<len(board);rowIndex++{
cellValue := board[rowIndex][colIndex]
// If the value of the cell isn't blank i.e. "."
if(string(cellValue)!="."){
// If the cell value is within 1-9
if(validCellValue(cellValue)){
// Check for value in colHashMap
_, key_exists := colHashMap[cellValue]
if key_exists{
// Exit the function if we
// encountered a repeating value
// in the current column
return false
} else {
// If value isn't present
// then add it the the hashmap
colHashMap[cellValue] = 1
}
} else {
// Exit the function if we
// encountered a value that's not
// within range 1-9
return false
}
}
}
}
return true
}
For validating every 3x3 sub-matrix we can implement checkSubMatrixValidity.
func checkSubMatrixValidity(board [][]byte)(bool){
// Iterating over every 3rd row
for rowIndex:=0;rowIndex<len(board);rowIndex+=3{
// Iterating over every 3rd column
for colIndex:=0;colIndex<len(board);colIndex+=3{
// The subMatrixHashMap would be used to check for
// repeating values in the 3x3 sub-matrix
subMatrixHashMap := make(map[byte]int)
// Iterate over every row in the sub-matrix
for i:=0;i<3;i++{
// Iterate over every column in the sub-matrix
for j:=0;j<3;j++{
cellValue := board[rowIndex+i][colIndex+j]
// If the value of the cell isn't blank i.e. "."
if(string(cellValue)!="."){
// If the cell value is within 1-9
if(validCellValue(cellValue)){
// Check for value in subMatrixHashMap
_, key_exists := subMatrixHashMap[cellValue]
if key_exists{
// Exit the function if we
// encountered a repeating value
// in the current sub-matrix
return false
} else {
// If value isn't present
// then add it the the hashmap
subMatrixHashMap[cellValue] = 1
}
} else {
// Exit the function if we
// encountered a value that's not
// within range 1-9
return false
}
}
}
}
}
}
return true
}
The isvalidSudoku function will validate the board by calling checkRowValidity, checkColumnValidity, and checkSubMatrixValidity and if all three functions return true then the input Sudoku board is valid.
func isValidSudoku(board [][]byte)(bool){
if(checkRowValidity(board) &&
checkColValidity(board) &&
checkSubMatrixValidity(board)){
return true
}
return false
}
Complete Code for the Optimal Solution
package main
import (
"fmt"
"strconv"
)
func validCellValue(inputValue byte)(bool){
// Converting byte value to integer
// to validate if inputValue lies between 1-len(board)
intInputValue, _ := strconv.Atoi(string(inputValue))
if ((intInputValue<10) && (intInputValue>0)){
return true
}
return false
}
func checkRowValidity(board [][]byte)(bool){
// Iterating over each row
// The length of board the could be hard coded to 9
for rowIndex:=0;rowIndex<len(board);rowIndex++{
// The rowHashMap would be used to check for
// repeating values in the row
rowHashMap:=make(map[byte]int)
// Iterating over each column in the row
for colIndex:=0;colIndex<len(board);colIndex++{
cellValue := board[rowIndex][colIndex]
// If the value of the cell isn't blank i.e. "."
if(string(cellValue) != "."){
// If the cell value is within 1-9
if(validCellValue(cellValue)){
// Check for value in rowHashMap
_, key_exists := rowHashMap[cellValue]
if key_exists{
// Exit the function if we
// encountered a repeating value
// in the current row
return false
} else {
// If value isn't present
// then add it the the hashmap
rowHashMap[cellValue] = 1
}
} else {
// Exit the function if we
// encountered a value that's not
// within range 1-9
return false
}
}
}
}
return true
}
func checkColValidity(board [][]byte)(bool){
// Iterating over each column
for colIndex:=0;colIndex<len(board);colIndex++{
// The colHashMap would be used to check for
// repeating values in the column
colHashMap:=make(map[byte]int)
// Iterating over each row in the column
for rowIndex:=0;rowIndex<len(board);rowIndex++{
cellValue := board[rowIndex][colIndex]
// If the value of the cell isn't blank i.e. "."
if(string(cellValue)!="."){
// If the cell value is within 1-9
if(validCellValue(cellValue)){
// Check for value in colHashMap
_, key_exists := colHashMap[cellValue]
if key_exists{
// Exit the function if we
// encountered a repeating value
// in the current column
return false
} else {
// If value isn't present
// then add it the the hashmap
colHashMap[cellValue] = 1
}
} else {
// Exit the function if we
// encountered a value that's not
// within range 1-9
return false
}
}
}
}
return true
}
func checkSubMatrixValidity(board [][]byte)(bool){
// Iterating over every 3rd row
for rowIndex:=0;rowIndex<len(board);rowIndex+=3{
// Iterating over every 3rd column
for colIndex:=0;colIndex<len(board);colIndex+=3{
// The subMatrixHashMap would be used to check for
// repeating values in the 3x3 sub-matrix
subMatrixHashMap := make(map[byte]int)
// Iterate over every row in the sub-matrix
for i:=0;i<3;i++{
// Iterate over every column in the sub-matrix
for j:=0;j<3;j++{
cellValue := board[rowIndex+i][colIndex+j]
// If the value of the cell isn't blank i.e. "."
if(string(cellValue)!="."){
// If the cell value is within 1-9
if(validCellValue(cellValue)){
// Check for value in subMatrixHashMap
_, key_exists := subMatrixHashMap[cellValue]
if key_exists{
// Exit the function if we
// encountered a repeating value
// in the current sub-matrix
return false
} else {
// If value isn't present
// then add it the the hashmap
subMatrixHashMap[cellValue] = 1
}
} else {
// Exit the function if we
// encountered a value that's not
// within range 1-9
return false
}
}
}
}
}
}
return true
}
func isValidSudoku(board [][]byte)(bool){
if(checkRowValidity(board) &&
checkColValidity(board) &&
checkSubMatrixValidity(board)){
return true
}
return false
}
func main(){
// A valid sudoku grid
// inputBoard := [][]string{
// {"5", "3", ".", ".", "7", ".", ".", ".", "."},
// {"6", ".", ".", "1", "9", "5", ".", ".", "."},
// {".", "9", "8", ".", ".", ".", ".", "6", "."},
// {"8", ".", ".", ".", "6", ".", ".", ".", "3"},
// {"4", ".", ".", "8", ".", "3", ".", ".", "1"},
// {"7", ".", ".", ".", "2", ".", ".", ".", "6"},
// {".", "6", ".", ".", ".", ".", "2", "8", "."},
// {".", ".", ".", "4", "1", "9", ".", ".", "5"},
// {".", ".", ".", ".", "8", ".", ".", "7", "9"},
// }
inputBoard := [][]byte{
{53,51,46,46,55,46,46,46,46},
{54,46,46,49,57,53,46,46,46},
{46,57,56,46,46,46,46,54,46},
{56,46,46,46,54,46,46,46,51},
{52,46,46,56,46,51,46,46,49},
{55,46,46,46,50,46,46,46,54},
{46,54,46,46,46,46,50,56,46},
{46,46,46,52,49,57,46,46,53},
{46,46,46,46,56,46,46,55,57}}
fmt.Println("Validitity of sudoku grid:", isValidSudoku(inputBoard))
// An invalid sudoku grid (Repeating 8s in first column)
// inputBoard := [][]string{
// {"8", "3", ".", ".", "7", ".", ".", ".", "."},
// {"6", ".", ".", "1", "9", "5", ".", ".", "."},
// {".", "9", "8", ".", ".", ".", ".", "6", "."},
// {"8", ".", ".", ".", "6", ".", ".", ".", "3"},
// {"4", ".", ".", "8", ".", "3", ".", ".", "1"},
// {"7", ".", ".", ".", "2", ".", ".", ".", "6"},
// {".", "6", ".", ".", ".", ".", "2", "8", "."},
// {".", ".", ".", "4", "1", "9", ".", ".", "5"},
// {".", ".", ".", ".", "8", ".", ".", "7", "9"},
// }
inputBoard = [][]byte{
{56,51,46,46,55,46,46,46,46},
{54,46,46,49,57,53,46,46,46},
{46,57,56,46,46,46,46,54,46},
{56,46,46,46,54,46,46,46,51},
{52,46,46,56,46,51,46,46,49},
{55,46,46,46,50,46,46,46,54},
{46,54,46,46,46,46,50,56,46},
{46,46,46,52,49,57,46,46,53},
{46,46,46,46,56,46,46,55,57}}
fmt.Println("Validitity of sudoku grid:", isValidSudoku(inputBoard))
// An invalid sudoku grid (Repeating 8s in the first 3x3 submatrix)
// inputBoard := [][]string{
// {"5", "3", ".", ".", "7", ".", ".", ".", "."},
// {"6", "8", ".", "1", "9", "5", ".", ".", "."},
// {".", "9", "8", ".", ".", ".", ".", "6", "."},
// {"8", ".", ".", ".", "6", ".", ".", ".", "3"},
// {"4", ".", ".", "8", ".", "3", ".", ".", "1"},
// {"7", ".", ".", ".", "2", ".", ".", ".", "6"},
// {".", "6", ".", ".", ".", ".", "2", "8", "."},
// {".", ".", ".", "4", "1", "9", ".", ".", "5"},
// {".", ".", ".", ".", "8", ".", ".", "7", "9"},
// }
inputBoard = [][]byte{
{53,51,46,46,55,46,46,46,46},
{54,56,46,49,57,53,46,46,46},
{46,57,56,46,46,46,46,54,46},
{56,46,46,46,54,46,46,46,51},
{52,46,46,56,46,51,46,46,49},
{55,46,46,46,50,46,46,46,54},
{46,54,46,46,46,46,50,56,46},
{46,46,46,52,49,57,46,46,53},
{46,46,46,46,56,46,46,55,57}}
fmt.Println("Validitity of sudoku grid:", isValidSudoku(inputBoard))
}
// Output
// Validitity of sudoku grid: true
// Validitity of sudoku grid: false
// Validitity of sudoku grid: false
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Resources
How to Play Sudoku?
36. Valid Sudoku
Valid Sudoku - Amazon Interview Question - Leetcode 36 - Python