Leetcode #37: Sudoku Solver
In this guide, we solve Leetcode #37 Sudoku Solver in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Write a program to solve a Sudoku puzzle by filling the empty cells. A sudoku solution must satisfy all of the following rules: Each of the digits 1-9 must occur exactly once in each row.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Hash Table, Backtracking, Matrix
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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:
Python Solution
class Solution:
def solveSudoku(self, board: List[List[str]]) -> None:
def dfs(k):
nonlocal ok
if k == len(t):
ok = True
return
i, j = t[k]
for v in range(9):
if row[i][v] == col[j][v] == block[i // 3][j // 3][v] == False:
row[i][v] = col[j][v] = block[i // 3][j // 3][v] = True
board[i][j] = str(v + 1)
dfs(k + 1)
row[i][v] = col[j][v] = block[i // 3][j // 3][v] = False
if ok:
return
row = [[False] * 9 for _ in range(9)]
col = [[False] * 9 for _ in range(9)]
block = [[[False] * 9 for _ in range(3)] for _ in range(3)]
t = []
ok = False
for i in range(9):
for j in range(9):
if board[i][j] == '.':
t.append((i, j))
else:
v = int(board[i][j]) - 1
row[i][v] = col[j][v] = block[i // 3][j // 3][v] = True
dfs(0)
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.