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# Sudoku Solver LeetCode Solution

## Problem – Sudoku Solver

Write a program to solve a Sudoku puzzle by filling the empty cells.

A sudoku solution must satisfy all of the following rules:

1. Each of the digits `1-9` must occur exactly once in each row.
2. Each of the digits `1-9` must occur exactly once in each column.
3. Each of the digits `1-9` must occur exactly once in each of the 9 `3x3` sub-boxes of the grid.

The `'.'` character indicates empty cells.

Example 1:

``````Input: board = [["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"]]
Output: [["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
Explanation: The input board is shown above and the only valid solution is shown below:
``````

Constraints:

• `board.length == 9`
• `board[i].length == 9`
• `board[i][j]` is a digit or `'.'`.
• It is guaranteed that the input board has only one solution.

### Sudoku Solver LeetCode Solution in Java

``````public class Solution {
public void solveSudoku(char[][] board) {
if(board == null || board.length == 0)
return;
solve(board);
}

public boolean solve(char[][] board){
for(int i = 0; i < board.length; i++){
for(int j = 0; j < board[0].length; j++){
if(board[i][j] == '.'){
for(char c = '1'; c <= '9'; c++){//trial. Try 1 through 9
if(isValid(board, i, j, c)){
board[i][j] = c; //Put c for this cell

if(solve(board))
return true; //If it's the solution return true
else
board[i][j] = '.'; //Otherwise go back
}
}

return false;
}
}
}
return true;
}

private boolean isValid(char[][] board, int row, int col, char c){
for(int i = 0; i < 9; i++) {
if(board[i][col] != '.' && board[i][col] == c) return false; //check row
if(board[row][i] != '.' && board[row][i] == c) return false; //check column
if(board[3 * (row / 3) + i / 3][ 3 * (col / 3) + i % 3] != '.' &&
board[3 * (row / 3) + i / 3][3 * (col / 3) + i % 3] == c) return false; //check 3*3 block
}
return true;
}
}
``````

### Sudoku Solver LeetCode Solution in Python

``````class Solution:
# @param board, a 9x9 2D array
# Solve the Sudoku by modifying the input board in-place.
# Do not return any value.
def solveSudoku(self, board):
self.board = board
self.solve()

def findUnassigned(self):
for row in range(9):
for col in range(9):
if self.board[row][col] == ".":
return row, col
return -1, -1

def solve(self):
row, col = self.findUnassigned()
#no unassigned position is found, puzzle solved
if row == -1 and col == -1:
return True
for num in ["1","2","3","4","5","6","7","8","9"]:
if self.isSafe(row, col, num):
self.board[row][col] = num
if self.solve():
return True
self.board[row][col] = "."
return False

def isSafe(self, row, col, ch):
boxrow = row - row%3
boxcol = col - col%3
if self.checkrow(row,ch) and self.checkcol(col,ch) and self.checksquare(boxrow, boxcol, ch):
return True
return False

def checkrow(self, row, ch):
for col in range(9):
if self.board[row][col] == ch:
return False
return True

def checkcol(self, col, ch):
for row in range(9):
if self.board[row][col] == ch:
return False
return True

def checksquare(self, row, col, ch):
for r in range(row, row+3):
for c in range(col, col+3):
if self.board[r][c] == ch:
return False
return True
``````

### Sudoku Solver LeetCode Solution in C++

``````class Solution {
struct cell // encapsulates a single cell on a Sudoku board
{
uint8_t value; // cell value 1..9 or 0 if unset
// number of possible (unconstrained) values for the cell
uint8_t numPossibilities;
// if bitset[v] is 1 then value can't be v
bitset<10> constraints;
cell() : value(0), numPossibilities(9),constraints() {};
};
array<array<cell,9>,9> cells;

// sets the value of the cell to [v]
// the function also propagates constraints to other cells and deduce new values where possible
bool set(int i, int j, int v)
{
// updating state of the cell
cell& c = cells[i][j];
if (c.value == v)
return true;
if (c.constraints[v])
return false;
c.constraints = bitset<10>(0x3FE); // all 1s
c.constraints.reset(v);
c.numPossibilities = 1;
c.value = v;

// propagating constraints
for (int k = 0; k<9; k++) {
// to the row:
if (i != k && !updateConstraints(k, j, v))
return false;
// to the column:
if (j != k && !updateConstraints(i, k, v))
return false;
// to the 3x3 square:
int ix = (i / 3) * 3 + k / 3;
int jx = (j / 3) * 3 + k % 3;
if (ix != i && jx != j && !updateConstraints(ix, jx, v))
return false;
}
return true;
}
// update constraints of the cell i,j by excluding possibility of 'excludedValue'
// once there's one possibility left the function recurses back into set()
bool updateConstraints(int i, int j, int excludedValue)
{
cell& c = cells[i][j];
if (c.constraints[excludedValue]) {
return true;
}
if (c.value == excludedValue) {
return false;
}
c.constraints.set(excludedValue);
if (--c.numPossibilities > 1)
return true;
for (int v = 1; v <= 9; v++) {
if (!c.constraints[v]) {
return set(i, j, v);
}
}
assert(false);
}

// backtracking state - list of empty cells
vector<pair<int, int>> bt;

// find values for empty cells
bool findValuesForEmptyCells()
{
// collecting all empty cells
bt.clear();
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
if (!cells[i][j].value)
bt.push_back(make_pair(i, j));
}
}
// making backtracking efficient by pre-sorting empty cells by numPossibilities
sort(bt.begin(), bt.end(), [this](const pair<int, int>&a, const pair<int, int>&b) {
return cells[a.first][a.second].numPossibilities < cells[b.first][b.second].numPossibilities; });
return backtrack(0);
}

// Finds value for all empty cells with index >=k
bool backtrack(int k)
{
if (k >= bt.size())
return true;
int i = bt[k].first;
int j = bt[k].second;
// fast path - only 1 possibility
if (cells[i][j].value)
return backtrack(k + 1);
auto constraints = cells[i][j].constraints;
// slow path >1 possibility.
// making snapshot of the state
array<array<cell,9>,9> snapshot(cells);
for (int v = 1; v <= 9; v++) {
if (!constraints[v]) {
if (set(i, j, v)) {
if (backtrack(k + 1))
return true;
}
// restoring from snapshot,
// note: computationally this is cheaper
// than alternative implementation with undoing the changes
cells = snapshot;
}
}
return false;
}
public:
void solveSudoku(vector<vector<char>> &board) {
cells = array<array<cell,9>,9>(); // clear array
// Decoding input board into the internal cell matrix.
// As we do it - constraints are propagated and even additional values are set as we go
// (in the case if it is possible to unambiguously deduce them).
for (int i = 0; i < 9; i++)
{
for (int j = 0; j < 9; j++) {
if (board[i][j] != '.' && !set(i, j, board[i][j] - '0'))
return; // sudoku is either incorrect or unsolvable
}
}
// if we're lucky we've already got a solution,
// however, if we have empty cells we need to use backtracking to fill them
if (!findValuesForEmptyCells())
return; // sudoku is unsolvable

// copying the solution back to the board
for (int i = 0; i < 9; i++)
{
for (int j = 0; j < 9; j++) {
if (cells[i][j].value)
board[i][j] = cells[i][j].value + '0';
}
}
}
};
``````
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