[LeetCode][36. Valid Sudoku] 2 Approaches: HashSet and Array
By Long Luo
This article is the solution 2 Approaches: HashSet and Array of Problem 36. Valid Sudoku.
Here shows 2 Approaches to slove this problem: HashSet and Array.
HashSet
We can use a HashSet to record the number of occurrences of each number in each row, each column and each sub-box.
Traverse the Sudoku once, update the count in the HashMap during the traversal process, and determine whether the Sudoku board could be valid.1
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45public static boolean isValidSudoku(char[][] board) {
Map<Integer, Set<Integer>> rowMap = new HashMap<>();
Map<Integer, Set<Integer>> colMap = new HashMap<>();
for (int i = 0; i < 9; i++) {
rowMap.put(i, new HashSet<>());
colMap.put(i, new HashSet<>());
}
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
char ch = board[i][j];
if (ch != '.') {
int num = ch - '0';
Set<Integer> rowSet = rowMap.get(i);
if (!rowSet.add(num)) {
return false;
}
Set<Integer> colSet = colMap.get(j);
if (!colSet.add(num)) {
return false;
}
}
}
}
for (int subIdx = 0; subIdx < 9; subIdx++) {
int subRow = 3 * (subIdx / 3);
int subCol = 3 * (subIdx % 3);
Set<Integer> grpSet = new HashSet<>();
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
char ch = board[subRow + i][subCol + j];
if (ch != '.') {
int num = ch - '0';
if (!grpSet.add(num)) {
return false;
}
}
}
}
}
return true;
}
This is version 1.0 code.
In fact, we can only traversal once. The index of each sub-box is \(3 \times (i / 3) + j / 3\), so we can write better code.1
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37public static boolean isValidSudoku_better(char[][] board) {
Map<Integer, Set<Integer>> rowMap = new HashMap<>();
Map<Integer, Set<Integer>> colMap = new HashMap<>();
Map<Integer, Set<Integer>> subMap = new HashMap<>();
for (int i = 0; i < 9; i++) {
rowMap.put(i, new HashSet<>());
colMap.put(i, new HashSet<>());
subMap.put(i, new HashSet<>());
}
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
char ch = board[i][j];
if (ch != '.') {
int num = ch - '0';
Set<Integer> rowSet = rowMap.get(i);
if (!rowSet.add(num)) {
return false;
}
Set<Integer> colSet = colMap.get(j);
if (!colSet.add(num)) {
return false;
}
int subIdx = 3 * (i / 3) + j / 3;
Set<Integer> subSet = subMap.get(subIdx);
if (!subSet.add(num)) {
return false;
}
}
}
}
return true;
}
Analysis
- Time Complexity: \(O(1)\).
- Space Complexity: \(O(1)\).
Array
Since numbers in Sudoku range from \(1\) to \(9\), we can use array instead of the HashMap for counting.
We create a 2D Array, the rows and columns which record the number of occurrences of each number in each row and column of Sudoku, and create a 3D Array subboxes to record the number of occurrences of each number in each sub-box.
If the count is greater than \(1\), the Sudoku is not valid.1
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22public static boolean isValidSudoku_array(char[][] board) {
int[][] row = new int[9][9];
int[][] col = new int[9][9];
int[][][] subBox = new int[3][3][9];
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
char ch = board[i][j];
if (ch != '.') {
int num = ch - '0' - 1;
row[i][num]++;
col[j][num]++;
subBox[i / 3][j / 3][num]++;
if (row[i][num] > 1 || col[j][num] > 1 || subBox[i / 3][j / 3][num] > 1) {
return false;
}
}
}
}
return true;
}
Analysis
- Time Complexity: \(O(1)\).
- Space Complexity: \(O(1)\).
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