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{ Sudoku Validator }

The Question

Write a function that determines if a Sudoku solution is valid. Your input will be a 2-D array that represents a 9x9 matrix. Sudoku has three rules:

• Every row must contain the numbers from 1-9 (no repeats)
• Every column must also contain every number from 1-9 
• Every 3x3 subgroup/square must contain each number from 1-9

Representing the data

Your input is a 2-D array that represents a 9x9 matrix. For example: 

[
  [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]
]

One Possible Solution

function sudokuValidator(solution) {
    for (var i=0; i < 9; i++) { //check the rows
        var curRow = [ ];
        for (var j =0; j < 9; j++)  {
            if (curRow.indexOf(solution[i][j]) > -1)
                return false;
            curRow.push(solution[i][j])
        }
    }
    for (var k=0; k < 9; k++) { //check the columns
        var curCol = [ ];
        for (var m =0; m < 9; m++)  {
            if (curCol.indexOf(solution[m][k]) > -1)
                return false;
            curCol.push(solution[m][k])
        }
    }
    for(var p=0; p<9; p+=3){ //check the squares
        for(var q=0; q<9; q+=3){
            var curSquare = [ ];
                for(var l=p; l<p+3; l++){
                    for(var n=q; n<q+3; n++){
                        if (curSquare.indexOf(solution[l][n]) > -1)
                            return false;
                        curSquare.push(solution[l][n]);
                    } 
                }
        }
    }
    return true;
}

Another Solution

'use strict'
function validSolution(solution){
    function check(arr){
        return arr.sort()
               .filter(function(val, index){
                    return val===index+1;
                })
                .length===9;
    }

    for(let i=0;i<9;i++){
        var col=[ ];
        var row=[ ];
        var square=[ ];
        for(let j=0;j<9;j++){
            col.push(solution[j][i]);
            row.push(solution[i][j]);
            square.push(solution[Math.floor(j/3)+(i%3)*3][j%3+Math.floor(i/3)*3]);
        }

    if(!check(col) || !check(row) || !check(square)) return false;
    }

    return true;

}

Efficiency  

• O(n²) at best since you need to check every space of the board in the worst case
• Even with multiple for loops (the first solution), you check the board 3 times which is O(3n²) which is still O(n²) 

Conclusion  

REPL.IT (solution 1)

REPL.IT (solution 2)

- Be creative and show you can think of a problem in different ways

-Be able to discuss efficiency in terms of Big-O notation

- Know how to represent matrices in 2D arrays