Sudoku using Backtracking

Like all other Backtracking problems, Sudoku can be solved by assigning numbers one by one to empty cells. Before assigning a number, check whether it is safe to assign. 

Check that the same number is not present in the current row, current column and current 3X3 subgrid. After checking for safety, assign the number, and recursively check whether this assignment leads to a solution or not. If the assignment doesn’t lead to a solution, then try the next number for the current empty cell. And if none of the number (1 to 9) leads to a solution, return false and print no solution exists.

Follow the steps below to solve the problem:

• Create a function that checks after assigning the current index the grid becomes unsafe or not. Keep Hashmap for a row, column and boxes. If any number has a frequency greater than 1 in the hashMap return false else return true; hashMap can be avoided by using loops.
• Create a recursive function that takes a grid.
• Check for any unassigned location. 
  • If present then assigns a number from 1 to 9.
  • Check if assigning the number to current index makes the grid unsafe or not. 
  • If safe then recursively call the function for all safe cases from 0 to 9.
  • If any recursive call returns true, end the loop and return true. If no recursive call returns true then return false.
• If there is no unassigned location then return true.

Below is the implementation of the above approach:

Time complexity: O(9^(N*N)), For every unassigned index, there are 9 possible options so the time complexity is O(9^(n*n)). The time complexity remains the same but there will be some early pruning so the time taken will be much less than the naive algorithm but the upper bound time complexity remains the same.
Space Complexity: O(N*N), To store the output array a matrix is needed.

Given a partially filled 9×9 2D array ‘grid[9][9]’, the goal is to assign digits (from 1 to 9) to the empty cells so that every row, column, and subgrid of size 3×3 contains exactly one instance of the digits from 1 to 9. 

Examples: 

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

Input: grid
{ { 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, 0, 1, 5, 9, 8, 7 },
{ 9, 7, 4, 8, 6, 0, 1, 2, 5 },
{ 8, 5, 1, 7, 9, 2, 6, 4, 3 },
{ 1, 3, 8, 0, 4, 7, 2, 0, 6 },
{ 6, 9, 2, 3, 5, 1, 8, 7, 4 },
{ 7, 4, 5, 0, 8, 6, 3, 1, 0 } };
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 
Explanation: Each row, column and 3*3 box of the output matrix contains unique numbers.

Naive Approach:

The naive approach is to generate all possible configurations of numbers from 1 to 9 to fill the empty cells. Try every configuration one by one until the correct configuration is found, i.e. for every unassigned position fill the position with a number from 1 to 9. After filling all the unassigned positions check if the matrix is safe or not. If safe print else recurs for other cases.

Follow the steps below to solve the problem:

• Create a function that checks if the given matrix is valid sudoku or not. Keep Hashmap for the row, column and boxes. If any number has a frequency greater than 1 in the hashMap return false else return true;
• Create a recursive function that takes a grid and the current row and column index.
• Check some base cases. 
  • If the index is at the end of the matrix, i.e. i=N-1 and j=N then check if the grid is safe or not, if safe print the grid and return true else return false. 
  • The other base case is when the value of column is N, i.e j = N, then move to next row, i.e. i++ and j = 0.
• If the current index is not assigned then fill the element from 1 to 9 and recur for all 9 cases with the index of next element, i.e. i, j+1. if the recursive call returns true then break the loop and return true.
• If the current index is assigned then call the recursive function with the index of the next element, i.e. i, j+1

Below is the implementation of the above approach:

Time complexity: O(9^(N*N)), For every unassigned index, there are 9 possible options so the time complexity is O(9^(n*n)).
Space Complexity: O(N*N), To store the output array a matrix is needed.