LeetCode – Sudoku Solver (Java)

Write a program to solve a Sudoku puzzle by filling the empty cells.

Java Solution

public void solveSudoku(char[][] board) {
    helper(board);
}
 
private boolean helper(char[][] board){
    for(int i=0; i<9; i++){
        for(int j=0; j<9; j++){
            if(board[i][j]!='.'){
                continue;
            }
 
            for(char k='1'; k<='9'; k++){
                board[i][j]=k;
                if(isValid(board, i, j) && helper(board)){
                    return true;
                }
                board[i][j]='.';
            }
 
            return false;
        }
    }
 
    return true; //return true if all cells are checked
}
 
private boolean isValid(char[][] board, int i, int j){
    HashSet<Character> set = new HashSet<>();
 
    for(int k=0; k<9; k++){
        if(set.contains(board[i][k])){
            return false;
        }
        if(board[i][k]!='.'){
            set.add(board[i][k]);
        }
 
    }
 
    set.clear();
 
    for(int k=0; k<9; k++){
        if(set.contains(board[k][j])){
            return false;
        }
        if(board[k][j]!='.'){
            set.add(board[k][j]);
        }
    }
 
    set.clear();
 
    int x=i/3 * 3;
    int y=j/3 * 3;
    for(int m=x; m<x+3; m++){
        for(int n=y; n<y+3; n++){
            if(set.contains(board[m][n])){
                return false;
            }
            if(board[m][n]!='.'){
                set.add(board[m][n]);
            }
        }
    }
 
    set.clear();
 
    return true;
}

Improved Version

public void solveSudoku(char[][] board) {
    helper(board);
}
 
private boolean helper(char[][] board) {
    for (int i = 0; i < 9; i++) {
        for (int j = 0; j < 9; j++) {
            if (board[i][j] != '.') {
                continue;
            }
 
            for (char k = '1'; k <= '9'; k++) {
                if (isValid(board, i, j, k)) {
                    board[i][j] = k;
                    if (helper(board)) {
                        return true;
                    }
                    board[i][j] = '.';
                }
            }
            return false;
        }
    }
 
    return true; //return true if all cells are checked
}
 
private boolean isValid(char[][] board, int row, int col, char c) {
    for (int i = 0; i < 9; i++) {
        if (board[i][col] != '.' && board[i][col] == c) {
            return false;
        }
 
        if (board[row][i] != '.' && board[row][i] == c) {
            return false;
        }
 
        if (board[3 * (row / 3) + i / 3][3 * (col / 3) + i % 3] != '.'
                &&
                board[3 * (row / 3) + i / 3][3 * (col / 3) + i % 3] == c) {
            return false;
        }
    }
    return true;
}

The time complexity is O(9^m) where m represents the number of blanks to be filled. No extra space is needed.

Category >> Algorithms  
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  • ryanlr

    true.

  • Winfred Jj

    ‘return true’ does matter