Sudoku Validator - Amazon Top Interview Questions
Problem Statement :
Sudoku is a puzzle where you're given a 9 by 9 grid with digits. The objective is to fill the grid with the constraint that every row, column, and box (3 by 3 subgrid) must contain all of the digits from 1 to 9, and numbers shouldn't repeat within a row, column, or box.
Given a filled out sudoku board, return whether it's valid.
Constraints
n = 9 where n is number of rows and columns in matrix
Example 1
Input
matrix = [
[4, 2, 6, 5, 7, 1, 3, 9, 8],
[8, 5, 7, 2, 9, 3, 1, 4, 6],
[1, 3, 9, 4, 6, 8, 2, 7, 5],
[9, 7, 1, 3, 8, 5, 6, 2, 4],
[5, 4, 3, 7, 2, 6, 8, 1, 9],
[6, 8, 2, 1, 4, 9, 7, 5, 3],
[7, 9, 4, 6, 3, 2, 5, 8, 1],
[2, 6, 5, 8, 1, 4, 9, 3, 7],
[3, 1, 8, 9, 5, 7, 4, 6, 2]
]
Output
True
Solution :
Solution in C++ :
bool solve(vector<vector<int>>& matrix) {
vector<map<int, int>> rows(9);
vector<map<int, int>> columns(9);
vector<map<int, int>> squares(9);
for (int i = 0; i < matrix.size(); i++) {
for (int j = 0; j < matrix[i].size(); j++) {
if (!(0 < matrix[i][j] && matrix[i][j] < 10)) {
return false;
}
if (rows[i][matrix[i][j]]) {
cout << "repeated in row, number: " << matrix[i][j];
return false;
}
rows[i][matrix[i][j]]++;
if (columns[j][matrix[i][j]]) {
cout << "repeated in column, number: " << matrix[i][j];
return false;
}
columns[j][matrix[i][j]]++;
int square = (i / 3 * 3) + (j / 3);
if (squares[square][matrix[i][j]]) {
cout << "repeated in square, number: " << matrix[i][j];
return false;
}
squares[square][matrix[i][j]]++;
}
}
return true;
}
Solution in Java :
import java.util.*;
class Solution {
public boolean solve(int[][] matrix) {
// Build a set of nums 1-9
Set<Integer> allNums = new HashSet<>();
for (int i = 1; i < 10; i++) {
allNums.add(i);
}
for (int i = 0; i < matrix.length; i++) {
Set<Integer> row = new HashSet<>(allNums);
Set<Integer> col = new HashSet<>(allNums);
for (int j = 0; j < matrix.length; j++) {
// Check row
if (!row.remove(matrix[i][j]))
return false;
// Check column
if (!col.remove(matrix[j][i]))
return false;
}
}
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
Set<Integer> box = new HashSet<>(allNums);
for (int k = 0; k < 3; k++) {
for (int l = 0; l < 3; l++) {
// Check box
if (!box.remove(matrix[i * 3 + k][j * 3 + l]))
return false;
}
}
}
}
// We validated all numbers 1-9 from all 9*9*9 cases
return true;
}
}
Solution in Python :
class Solution:
def solve(self, matrix):
row = defaultdict(set)
col = defaultdict(set)
box = defaultdict(set)
# box = r // 3 * 3 + c // 3
for i in range(9):
for j in range(9):
if matrix[i][j] < 1 or matrix[i][j] > 9:
return False
row[i].add(matrix[i][j])
col[j].add(matrix[i][j])
box[i // 3 * 3 + j // 3].add(matrix[i][j])
for i in range(9):
if len(row[i]) != 9 or len(col[i]) != 9 or len(box[i]) != 9:
return False
return True
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