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Algorithm to Solve Sudoku | Sudoku Solver

Last Updated : 31 Jan, 2025
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Given an incomplete Sudoku in the form of matrix mat[][] of order 9*9, the task is to complete the Sudoku.
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.

Note: Zeros in the mat[][] indicate blanks, which are to be filled with some number between 1 to 9. You can not replace the element in the cell which is not blank.

Examples:

Input:

Suduko-example-question


Output:

Suduko-example-answer

Explanation: Each row, column and 3*3 box of the output matrix contains unique numbers.

[Naive Approach] Using Backtracking

The idea is to use backtracking and recursively generate all possible configurations of numbers from 1 to 9 to fill the empty cells of matrix mat[][]. To do so, for every unassigned cell, fill the cell with a number from 1 to 9 one by one. After filling the unassigned cell check if the matrix is safe or not. If safe, move to the next cell else backtrack for other cases.

To check if it is safe to place value num in the cell mat[i][j], iterate through all the columns of row i, rows of column j and the 3*3 matrix containing cell (i, j) and check if they already has value num, if so return false, else return true.

C++ 
// C++ Program to solve Sudoku problem #include <iostream> #include <vector> using namespace std;  // Function to check if it is safe to place num at mat[row][col] bool isSafe(vector<vector<int>> &mat, int row, int col, int num) {      // Check if num exist in the row     for (int x = 0; x <= 8; x++)         if (mat[row][x] == num)             return false;      // Check if num exist in the col     for (int x = 0; x <= 8; x++)         if (mat[x][col] == num)             return false;      // Check if num exist in the 3x3 sub-matrix     int startRow = row - (row % 3), startCol = col - (col % 3);      for (int i = 0; i < 3; i++)         for (int j = 0; j < 3; j++)             if (mat[i + startRow][j + startCol] == num)                 return false;      return true; }  // Function to solve the Sudoku problem bool solveSudokuRec(vector<vector<int>> &mat, int row, int col) {     int n = mat.size();      // base case: Reached nth column of last row     if (row == n - 1 && col == n)         return true;      // If last column of the row go to next row     if (col == n) {         row++;         col = 0;     }      // If cell is already occupied then move forward     if (mat[row][col] != 0)         return solveSudokuRec(mat, row, col + 1);      for (int num = 1; num <= n; num++) {          // If it is safe to place num at current position         if (isSafe(mat, row, col, num)) {             mat[row][col] = num;             if (solveSudokuRec(mat, row, col + 1))                 return true;             mat[row][col] = 0;         }     }      	return false; }  void solveSudoku(vector<vector<int>> &mat) {   	solveSudokuRec(mat, 0, 0); }  int main() {     vector<vector<int>> mat = {         {3, 0, 6, 5, 0, 8, 4, 0, 0},        	{5, 2, 0, 0, 0, 0, 0, 0, 0},        	{0, 8, 7, 0, 0, 0, 0, 3, 1},         {0, 0, 3, 0, 1, 0, 0, 8, 0},        	{9, 0, 0, 8, 6, 3, 0, 0, 5},        	{0, 5, 0, 0, 9, 0, 6, 0, 0},         {1, 3, 0, 0, 0, 0, 2, 5, 0},        	{0, 0, 0, 0, 0, 0, 0, 7, 4},        	{0, 0, 5, 2, 0, 6, 3, 0, 0}};  	solveSudoku(mat);        	for (int i = 0; i < mat.size(); i++) {         for (int j = 0; j < mat.size(); j++)             cout << mat[i][j] << " ";         cout << endl;     }      return 0; }
Java 
// Java Program to solve Sudoku problem import java.util.Arrays;  class GfG {      // Function to check if it is safe to place num at mat[row][col]     static boolean isSafe(int[][] mat, int row, int col, int num) {         // Check if num exists in the row         for (int x = 0; x < 9; x++)             if (mat[row][x] == num)                 return false;          // Check if num exists in the col         for (int x = 0; x < 9; x++)             if (mat[x][col] == num)                 return false;          // Check if num exists in the 3x3 sub-matrix         int startRow = row - (row % 3), startCol = col - (col % 3);          for (int i = 0; i < 3; i++)             for (int j = 0; j < 3; j++)                 if (mat[i + startRow][j + startCol] == num)                     return false;          return true;     }      // Function to solve the Sudoku problem     static boolean solveSudokuRec(int[][] mat, int row, int col) {                // base case: Reached nth column of the last row         if (row == 8 && col == 9)             return true;          // If last column of the row go to the next row         if (col == 9) {             row++;             col = 0;         }          // If cell is already occupied then move forward         if (mat[row][col] != 0)             return solveSudokuRec(mat, row, col + 1);          for (int num = 1; num <= 9; num++) {                        // If it is safe to place num at current position             if (isSafe(mat, row, col, num)) {                 mat[row][col] = num;                 if (solveSudokuRec(mat, row, col + 1))                     return true;                 mat[row][col] = 0;             }         }          return false;     }      static void solveSudoku(int[][] mat) {         solveSudokuRec(mat, 0, 0);     }      public static void main(String[] args) {         int[][] mat = {             {3, 0, 6, 5, 0, 8, 4, 0, 0},             {5, 2, 0, 0, 0, 0, 0, 0, 0},             {0, 8, 7, 0, 0, 0, 0, 3, 1},             {0, 0, 3, 0, 1, 0, 0, 8, 0},             {9, 0, 0, 8, 6, 3, 0, 0, 5},             {0, 5, 0, 0, 9, 0, 6, 0, 0},             {1, 3, 0, 0, 0, 0, 2, 5, 0},             {0, 0, 0, 0, 0, 0, 0, 7, 4},             {0, 0, 5, 2, 0, 6, 3, 0, 0}         };          solveSudoku(mat);          for (int i = 0; i < mat.length; i++) {             for (int j = 0; j < mat[i].length; j++)                 System.out.print(mat[i][j] + " ");             System.out.println();         }     } }
Python 
# Python Program to solve Sudoku problem  # Function to check if it is safe to place num at mat[row][col] def isSafe(mat, row, col, num):     # Check if num exists in the row     for x in range(9):         if mat[row][x] == num:             return False      # Check if num exists in the col     for x in range(9):         if mat[x][col] == num:             return False      # Check if num exists in the 3x3 sub-matrix     startRow = row - (row % 3)     startCol = col - (col % 3)      for i in range(3):         for j in range(3):             if mat[i + startRow][j + startCol] == num:                 return False      return True  # Function to solve the Sudoku problem def solveSudokuRec(mat, row, col):     # base case: Reached nth column of the last row     if row == 8 and col == 9:         return True      # If last column of the row go to the next row     if col == 9:         row += 1         col = 0      # If cell is already occupied then move forward     if mat[row][col] != 0:         return solveSudokuRec(mat, row, col + 1)      for num in range(1, 10):         # If it is safe to place num at current position         if isSafe(mat, row, col, num):             mat[row][col] = num             if solveSudokuRec(mat, row, col + 1):                 return True             mat[row][col] = 0      return False  def solveSudoku(mat):     solveSudokuRec(mat, 0, 0)  if __name__ == "__main__":     mat = [         [3, 0, 6, 5, 0, 8, 4, 0, 0],         [5, 2, 0, 0, 0, 0, 0, 0, 0],         [0, 8, 7, 0, 0, 0, 0, 3, 1],         [0, 0, 3, 0, 1, 0, 0, 8, 0],         [9, 0, 0, 8, 6, 3, 0, 0, 5],         [0, 5, 0, 0, 9, 0, 6, 0, 0],         [1, 3, 0, 0, 0, 0, 2, 5, 0],         [0, 0, 0, 0, 0, 0, 0, 7, 4],         [0, 0, 5, 2, 0, 6, 3, 0, 0]     ]      solveSudoku(mat)      for row in mat:         print(" ".join(map(str, row)))
C# 
// C# Program to solve Sudoku problem  using System;  class GfG {      // Function to check if it is safe to place num at mat[row][col]     static bool isSafe(int[,] mat, int row, int col, int num) {         // Check if num exists in the row         for (int x = 0; x < 9; x++)             if (mat[row, x] == num)                 return false;          // Check if num exists in the col         for (int x = 0; x < 9; x++)             if (mat[x, col] == num)                 return false;          // Check if num exists in the 3x3 sub-matrix         int startRow = row - (row % 3), startCol = col - (col % 3);          for (int i = 0; i < 3; i++)             for (int j = 0; j < 3; j++)                 if (mat[i + startRow, j + startCol] == num)                     return false;          return true;     }      // Function to solve the Sudoku problem     static bool solveSudokuRec(int[,] mat, int row, int col) {                // base case: Reached nth column of the last row         if (row == 8 && col == 9)             return true;          // If last column of the row go to the next row         if (col == 9) {             row++;             col = 0;         }          // If cell is already occupied then move forward         if (mat[row, col] != 0)             return solveSudokuRec(mat, row, col + 1);          for (int num = 1; num <= 9; num++) {             // If it is safe to place num at current position             if (isSafe(mat, row, col, num)) {                 mat[row, col] = num;                 if (solveSudokuRec(mat, row, col + 1))                     return true;                 mat[row, col] = 0;             }         }          return false;     }      static void solveSudoku(int[,] mat) {         solveSudokuRec(mat, 0, 0);     }      public static void Main() {         int[,] mat = {             {3, 0, 6, 5, 0, 8, 4, 0, 0},             {5, 2, 0, 0, 0, 0, 0, 0, 0},             {0, 8, 7, 0, 0, 0, 0, 3, 1},             {0, 0, 3, 0, 1, 0, 0, 8, 0},             {9, 0, 0, 8, 6, 3, 0, 0, 5},             {0, 5, 0, 0, 9, 0, 6, 0, 0},             {1, 3, 0, 0, 0, 0, 2, 5, 0},             {0, 0, 0, 0, 0, 0, 0, 7, 4},             {0, 0, 5, 2, 0, 6, 3, 0, 0}         };          solveSudoku(mat);          for (int i = 0; i < 9; i++) {             for (int j = 0; j < 9; j++)                 Console.Write(mat[i, j] + " ");             Console.WriteLine();         }     } }
JavaScript 
// JavaScript Program to solve Sudoku problem  // Function to check if it is safe to place num at mat[row][col] function isSafe(mat, row, col, num) {     // Check if num exists in the row     for (let x = 0; x < 9; x++)         if (mat[row][x] === num)             return false;      // Check if num exists in the col     for (let x = 0; x < 9; x++)         if (mat[x][col] === num)             return false;      // Check if num exists in the 3x3 sub-matrix     const startRow = row - (row % 3),           startCol = col - (col % 3);      for (let i = 0; i < 3; i++)         for (let j = 0; j < 3; j++)             if (mat[i + startRow][j + startCol] === num)                 return false;      return true; }  // Function to solve the Sudoku problem function solveSudokuRec(mat, row, col) {      // base case: Reached nth column of the last row     if (row === 8 && col === 9)         return true;      // If last column of the row go to the next row     if (col === 9) {         row++;         col = 0;     }      // If cell is already occupied then move forward     if (mat[row][col] !== 0)         return solveSudokuRec(mat, row, col + 1);      for (let num = 1; num <= 9; num++) {         // If it is safe to place num at current position         if (isSafe(mat, row, col, num)) {             mat[row][col] = num;             if (solveSudokuRec(mat, row, col + 1))                 return true;             mat[row][col] = 0;         }     }      return false; }  function solveSudoku(mat) {     solveSudokuRec(mat, 0, 0); }  // Driver Code const mat = [     [3, 0, 6, 5, 0, 8, 4, 0, 0],     [5, 2, 0, 0, 0, 0, 0, 0, 0],     [0, 8, 7, 0, 0, 0, 0, 3, 1],     [0, 0, 3, 0, 1, 0, 0, 8, 0],     [9, 0, 0, 8, 6, 3, 0, 0, 5],     [0, 5, 0, 0, 9, 0, 6, 0, 0],     [1, 3, 0, 0, 0, 0, 2, 5, 0],     [0, 0, 0, 0, 0, 0, 0, 7, 4],     [0, 0, 5, 2, 0, 6, 3, 0, 0] ];  solveSudoku(mat);  mat.forEach(row => console.log(row.join(" ")));

Output
3 1 6 5 7 8 4 9 2  5 2 9 1 3 4 7 6 8  4 8 7 6 2 9 5 3 1  2 6 3 4 1 5 9 8 7  9 7 4 8 6 3 1 2 5  8 5 1 7 9 2 6 4 3  1 3 8 9 4 7 2 5 6  6 9 2 3 5 1 8 7 4  7 4 5 2 8 6 3 1 9

Time complexity: O(n*9(n*n)), For every unassigned index, there are 9 possible options and for each index, we are checking other columns, rows and boxes.
Auxiliary Space: O(1)

[Expected Approach] Using Bit Masking with Backtracking - O(9(n*n)) Time and O(n) Space

In the above approach, isSafe() function which is used to check if it is safe to place number num in cell (i, j) searches for num in each row, col and box. The idea is to optimize this using Bit Masking. To do so, create three arrays rows[], cols[], boxs[] of size n to mark the used value in row, column and box respectively. The element row[i] marks the number already been used in row i, and so do cols[] and boxs[] for columns and boxes. To mark the number num of row i, set the bit num from left of row[i] and operate similarly for cols[] and boxs[]. Similarly, to unmark the value num, unset the bits set in current step.

C++ 
// C++ Program to solve Sudoku problem #include <iostream> #include <vector> using namespace std;  // Function to heck if it is safe to place num at mat[row][col] bool isSafe(vector<vector<int>> &mat, int i, int j, int num,          vector<int> &row, vector<int> &col, vector<int> &box) {        if( (row[i] & (1 << num)) || (col[j] & (1 << num)) ||        					(box[i / 3 * 3 + j / 3] & (1 << num)) )         return false;          return true; }  bool sudokuSolverRec(vector<vector<int>> &mat, int i, int j,      		vector<int> &row, vector<int> &col, vector<int> &box) {     int n = mat.size();  	// base case: Reached nth column of last row     if (i == n - 1 && j == n)         return true;      // If reached last column of the row go to next row     if (j == n) {         i++;         j = 0;     }        // If cell is already occupied then move forward     if (mat[i][j] != 0)         return sudokuSolverRec(mat, i, j + 1, row, col, box);      for (int num = 1; num <= n; num++) {                  // If it is safe to place num at current position         if (isSafe(mat, i, j, num, row, col, box)) {             mat[i][j] = num;                      	// Update masks for the corresponding row, column and box             row[i] |= (1 << num);             col[j] |= (1 << num);             box[i / 3 * 3 + j / 3] |= (1 << num);                        if (sudokuSolverRec(mat, i, j + 1, row, col, box))                 return true;           	           	// Unmask the number num in the corresponding row, column and box masks             mat[i][j] = 0;             row[i] &= ~(1 << num);             col[j] &= ~(1 << num);             box[i / 3 * 3 + j / 3] &= ~(1 << num);         }     }        return false; }  void solveSudoku(vector<vector<int>> &mat) {   	int n = mat.size();     vector<int> row(n, 0), col(n, 0), box(n, 0);      // Set the bits in bitmasks for values that are initital present        for (int i = 0; i < n; i++) {         for (int j = 0; j < n; j++) {             if (mat[i][j] != 0) {                 row[i] |= (1 << mat[i][j]);                 col[j] |= (1 << mat[i][j]);                 box[ (i / 3) * 3 + j / 3] |= (1 << mat[i][j]);             }         }     }      sudokuSolverRec(mat, 0, 0, row, col, box); }  int main() {     vector<vector<int>> mat = {         {3, 0, 6, 5, 0, 8, 4, 0, 0},        	{5, 2, 0, 0, 0, 0, 0, 0, 0},        	{0, 8, 7, 0, 0, 0, 0, 3, 1},         {0, 0, 3, 0, 1, 0, 0, 8, 0},        	{9, 0, 0, 8, 6, 3, 0, 0, 5},        	{0, 5, 0, 0, 9, 0, 6, 0, 0},         {1, 3, 0, 0, 0, 0, 2, 5, 0},        	{0, 0, 0, 0, 0, 0, 0, 7, 4},        	{0, 0, 5, 2, 0, 6, 3, 0, 0}};  	solveSudoku(mat);        	for (int i = 0; i < mat.size(); i++) {         for (int j = 0; j < mat.size(); j++)             cout << mat[i][j] << " ";         cout << endl;     }      return 0; }
Java 
// Java Program to solve Sudoku problem import java.util.Arrays;  class GfG {      // Function to check if it is safe to place num at mat[row][col]     static boolean isSafe(int[][] mat, int i, int j, int num,                            int[] row, int[] col, int[] box) {         if ((row[i] & (1 << num)) != 0 || (col[j] & (1 << num)) != 0 ||              (box[i / 3 * 3 + j / 3] & (1 << num)) != 0)             return false;                  return true;     }      static boolean sudokuSolverRec(int[][] mat, int i, int j,                                     int[] row, int[] col, int[] box) {         int n = mat.length;          // base case: Reached nth column of last row         if (i == n - 1 && j == n)             return true;          // If reached last column of the row go to next row         if (j == n) {             i++;             j = 0;         }          // If cell is already occupied then move forward         if (mat[i][j] != 0)             return sudokuSolverRec(mat, i, j + 1, row, col, box);          for (int num = 1; num <= n; num++) {             // If it is safe to place num at current position             if (isSafe(mat, i, j, num, row, col, box)) {                 mat[i][j] = num;                  // Update masks for the corresponding row, column and box                 row[i] |= (1 << num);                 col[j] |= (1 << num);                 box[i / 3 * 3 + j / 3] |= (1 << num);                  if (sudokuSolverRec(mat, i, j + 1, row, col, box))                     return true;                  // Unmask the number num in the corresponding row, column and box masks                 mat[i][j] = 0;                 row[i] &= ~(1 << num);                 col[j] &= ~(1 << num);                 box[i / 3 * 3 + j / 3] &= ~(1 << num);             }         }          return false;     }      static void solveSudoku(int[][] mat) {         int n = mat.length;         int[] row = new int[n];         int[] col = new int[n];         int[] box = new int[n];          // Set the bits in bitmasks for values that are initially present         for (int i = 0; i < n; i++) {             for (int j = 0; j < n; j++) {                 if (mat[i][j] != 0) {                     row[i] |= (1 << mat[i][j]);                     col[j] |= (1 << mat[i][j]);                     box[(i / 3) * 3 + j / 3] |= (1 << mat[i][j]);                 }             }         }          sudokuSolverRec(mat, 0, 0, row, col, box);     }      public static void main(String[] args) {         int[][] mat = {             {3, 0, 6, 5, 0, 8, 4, 0, 0},             {5, 2, 0, 0, 0, 0, 0, 0, 0},             {0, 8, 7, 0, 0, 0, 0, 3, 1},             {0, 0, 3, 0, 1, 0, 0, 8, 0},             {9, 0, 0, 8, 6, 3, 0, 0, 5},             {0, 5, 0, 0, 9, 0, 6, 0, 0},             {1, 3, 0, 0, 0, 0, 2, 5, 0},             {0, 0, 0, 0, 0, 0, 0, 7, 4},             {0, 0, 5, 2, 0, 6, 3, 0, 0}         };          solveSudoku(mat);          for (int i = 0; i < mat.length; i++) {             for (int j = 0; j < mat[i].length; j++)                 System.out.print(mat[i][j] + " ");             System.out.println();         }     } }
Python 
# Python Program to solve Sudoku problem  def isSafe(mat, i, j, num, row, col, box):     if (row[i] & (1 << num)) or (col[j] & (1 << num)) or (box[i // 3 * 3 + j // 3] & (1 << num)):         return False     return True  def sudokuSolverRec(mat, i, j, row, col, box):     n = len(mat)      # base case: Reached nth column of last row     if i == n - 1 and j == n:         return True      # If reached last column of the row go to next row     if j == n:         i += 1         j = 0      # If cell is already occupied then move forward     if mat[i][j] != 0:         return sudokuSolverRec(mat, i, j + 1, row, col, box)      for num in range(1, n + 1):         # If it is safe to place num at current position         if isSafe(mat, i, j, num, row, col, box):             mat[i][j] = num              # Update masks for the corresponding row, column and box             row[i] |= (1 << num)             col[j] |= (1 << num)             box[i // 3 * 3 + j // 3] |= (1 << num)              if sudokuSolverRec(mat, i, j + 1, row, col, box):                 return True              # Unmask the number num in the corresponding row, column and box masks             mat[i][j] = 0             row[i] &= ~(1 << num)             col[j] &= ~(1 << num)             box[i // 3 * 3 + j // 3] &= ~(1 << num)      return False  def solveSudoku(mat):     n = len(mat)     row = [0] * n     col = [0] * n     box = [0] * n      # Set the bits in bitmasks for values that are initially present     for i in range(n):         for j in range(n):             if mat[i][j] != 0:                 row[i] |= (1 << mat[i][j])                 col[j] |= (1 << mat[i][j])                 box[(i // 3) * 3 + j // 3] |= (1 << mat[i][j])      sudokuSolverRec(mat, 0, 0, row, col, box)  if __name__ == "__main__":     mat = [         [3, 0, 6, 5, 0, 8, 4, 0, 0],         [5, 2, 0, 0, 0, 0, 0, 0, 0],         [0, 8, 7, 0, 0, 0, 0, 3, 1],         [0, 0, 3, 0, 1, 0, 0, 8, 0],         [9, 0, 0, 8, 6, 3, 0, 0, 5],         [0, 5, 0, 0, 9, 0, 6, 0, 0],         [1, 3, 0, 0, 0, 0, 2, 5, 0],         [0, 0, 0, 0, 0, 0, 0, 7, 4],         [0, 0, 5, 2, 0, 6, 3, 0, 0]     ]      solveSudoku(mat)      for row in mat:         print(" ".join(map(str, row)))
C# 
// C# Program to solve Sudoku problem using bitmasks using System;  class GfG {      // Function to check if it is safe to place num at mat[row, col]     static bool isSafe(int[,] mat, int i, int j, int num,                         	int[] row, int[] col, int[] box) {                if ((row[i] & (1 << num)) != 0 || (col[j] & (1 << num)) != 0 ||             (box[i / 3 * 3 + j / 3] & (1 << num)) != 0)             return false;          return true;     }      static bool sudokuSolverRec(int[,] mat, int i, int j,                                  int[] row, int[] col, int[] box) {         int n = mat.GetLength(0);          // base case: Reached nth column of last row         if (i == n - 1 && j == n)             return true;          // If reached last column of the row, go to next row         if (j == n) {             i++;             j = 0;         }          // If cell is already occupied, then move forward         if (mat[i, j] != 0)             return sudokuSolverRec(mat, i, j + 1, row, col, box);          for (int num = 1; num <= n; num++) {              // If it is safe to place num at current position             if (isSafe(mat, i, j, num, row, col, box)) {                 mat[i, j] = num;                  // Update masks for the corresponding row, column, and box                 row[i] |= (1 << num);                 col[j] |= (1 << num);                 box[i / 3 * 3 + j / 3] |= (1 << num);                  if (sudokuSolverRec(mat, i, j + 1, row, col, box))                     return true;                  // Unmask the number num in the corresponding row, column and box masks                 mat[i, j] = 0;                 row[i] &= ~(1 << num);                 col[j] &= ~(1 << num);                 box[i / 3 * 3 + j / 3] &= ~(1 << num);             }         }          return false;     }      static void solveSudoku(int[,] mat) {         int n = mat.GetLength(0);         int[] row = new int[n];         int[] col = new int[n];         int[] box = new int[n];          // Set the bits in bitmasks for values that are initially present         for (int i = 0; i < n; i++) {             for (int j = 0; j < n; j++) {                 if (mat[i, j] != 0) {                     row[i] |= (1 << mat[i, j]);                     col[j] |= (1 << mat[i, j]);                     box[(i / 3) * 3 + j / 3] |= (1 << mat[i, j]);                 }             }         }          sudokuSolverRec(mat, 0, 0, row, col, box);     }      public static void Main(string[] args) {         int[,] mat = {             {3, 0, 6, 5, 0, 8, 4, 0, 0},             {5, 2, 0, 0, 0, 0, 0, 0, 0},             {0, 8, 7, 0, 0, 0, 0, 3, 1},             {0, 0, 3, 0, 1, 0, 0, 8, 0},             {9, 0, 0, 8, 6, 3, 0, 0, 5},             {0, 5, 0, 0, 9, 0, 6, 0, 0},             {1, 3, 0, 0, 0, 0, 2, 5, 0},             {0, 0, 0, 0, 0, 0, 0, 7, 4},             {0, 0, 5, 2, 0, 6, 3, 0, 0}         };          solveSudoku(mat);          for (int i = 0; i < mat.GetLength(0); i++) {             for (int j = 0; j < mat.GetLength(1); j++)                 Console.Write(mat[i, j] + " ");             Console.WriteLine();         }     } }
JavaScript 
// JavaScript Program to solve Sudoku problem using bitmasks  // Function to check if it is safe to place num at mat[row][col] function isSafe(mat, i, j, num, row, col, box) {     if ((row[i] & (1 << num)) !== 0 || (col[j] & (1 << num)) !== 0 ||         (box[Math.floor(i / 3) * 3 + Math.floor(j / 3)] & (1 << num)) !== 0)         return false;      return true; }  function sudokuSolverRec(mat, i, j, row, col, box) {     const n = mat.length;      // base case: Reached nth column of last row     if (i === n - 1 && j === n)         return true;      // If reached last column of the row, go to next row     if (j === n) {         i++;         j = 0;     }      // If cell is already occupied, then move forward     if (mat[i][j] !== 0)         return sudokuSolverRec(mat, i, j + 1, row, col, box);      for (let num = 1; num <= n; num++) {          // If it is safe to place num at current position         if (isSafe(mat, i, j, num, row, col, box)) {             mat[i][j] = num;              // Update masks for the corresponding row, column, and box             row[i] |= (1 << num);             col[j] |= (1 << num);             box[Math.floor(i / 3) * 3 + Math.floor(j / 3)] |= (1 << num);              if (sudokuSolverRec(mat, i, j + 1, row, col, box))                 return true;              // Unmask the number num in the corresponding row, column and box masks             mat[i][j] = 0;             row[i] &= ~(1 << num);             col[j] &= ~(1 << num);             box[Math.floor(i / 3) * 3 + Math.floor(j / 3)] &= ~(1 << num);         }     }      return false; }  function solveSudoku(mat) {     const n = mat.length;     const row = new Array(n).fill(0);     const col = new Array(n).fill(0);     const box = new Array(n).fill(0);      // Set the bits in bitmasks for values that are initially present     for (let i = 0; i < n; i++) {         for (let j = 0; j < n; j++) {             if (mat[i][j] !== 0) {                 row[i] |= (1 << mat[i][j]);                 col[j] |= (1 << mat[i][j]);                 box[Math.floor(i / 3) * 3 + Math.floor(j / 3)] |= (1 << mat[i][j]);             }         }     }      sudokuSolverRec(mat, 0, 0, row, col, box); }  // Driver Code const mat = [     [3, 0, 6, 5, 0, 8, 4, 0, 0],     [5, 2, 0, 0, 0, 0, 0, 0, 0],     [0, 8, 7, 0, 0, 0, 0, 3, 1],     [0, 0, 3, 0, 1, 0, 0, 8, 0],     [9, 0, 0, 8, 6, 3, 0, 0, 5],     [0, 5, 0, 0, 9, 0, 6, 0, 0],     [1, 3, 0, 0, 0, 0, 2, 5, 0],     [0, 0, 0, 0, 0, 0, 0, 7, 4],     [0, 0, 5, 2, 0, 6, 3, 0, 0] ];  solveSudoku(mat);  for (let i = 0; i < mat.length; i++) {     console.log(mat[i].join(" ")); }

Output
3 1 6 5 7 8 4 9 2  5 2 9 1 3 4 7 6 8  4 8 7 6 2 9 5 3 1  2 6 3 4 1 5 9 8 7  9 7 4 8 6 3 1 2 5  8 5 1 7 9 2 6 4 3  1 3 8 9 4 7 2 5 6  6 9 2 3 5 1 8 7 4  7 4 5 2 8 6 3 1 9

Time complexity: O(9(n*n))
Auxiliary Space: O(n)



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