Valid Sudoku

Medium

Python

def is_valid_sudoku(board):
    rows=[set() for _ in range(9)]; cols=[set() for _ in range(9)]; boxes=[set() for _ in range(9)]
    for r in range(9):
        for c in range(9):
            v=board[r][c]
            if v=='.': continue
            b=(r//3)*3+c//3
            if v in rows[r] or v in cols[c] or v in boxes[b]: return False
            rows[r].add(v); cols[c].add(v); boxes[b].add(v)
    return True

Java

public boolean isValidSudoku(char[][] board){
    Set[] rows=new HashSet[9],cols=new HashSet[9],boxes=new HashSet[9];
    for(int i=0;i<9;i++){rows[i]=new HashSet<>();cols[i]=new HashSet<>();boxes[i]=new HashSet<>();}
    for(int r=0;r<9;r++) for(int c=0;c<9;c++){
        char v=board[r][c]; if(v=='.') continue;
        int b=(r/3)*3+c/3;
        if(!rows[r].add(v)||!cols[c].add(v)||!boxes[b].add(v)) return false;
    } return true;
}

Key Insight

Java Set.add() returns false on dup. Python 'v in set' checks existence.

Python → Java Differences

List comprehension init vs explicit loop
'v in set' vs add() returning false
Array of Sets needs init Java

Python

def is_valid_sudoku(board):
    rows=[set() for _ in range(9)]; cols=[set() for _ in range(9)]; boxes=[set() for _ in range(9)]
    for r in range(9):
        for c in range(9):
            v=board[r][c]
            if v=='.': continue
            b=(r//3)*3+c//3
            if v in rows[r] or v in cols[c] or v in boxes[b]: return False
            rows[r].add(v); cols[c].add(v); boxes[b].add(v)
    return True

Java

public boolean isValidSudoku(char[][] board){
    Set[] rows=new HashSet[9],cols=new HashSet[9],boxes=new HashSet[9];
    for(int i=0;i<9;i++){rows[i]=new HashSet<>();cols[i]=new HashSet<>();boxes[i]=new HashSet<>();}
    for(int r=0;r<9;r++) for(int c=0;c<9;c++){
        char v=board[r][c]; if(v=='.') continue;
        int b=(r/3)*3+c/3;
        if(!rows[r].add(v)||!cols[c].add(v)||!boxes[b].add(v)) return false;
    } return true;
}

Algorithm Steps

1. Track per row col box
2. Any dup: false

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