find an element in a sorted matrix [duplicate]

Answer 1

Reading the previous comments I came up with this algorithm. It basically suppose that by starting in the upper-right corner, the matrix can be used as a BST with some "loops" (we don't care about this loops).

1 4 9
5 6 10

     9
    /  \
   4   10
  / \  /
 1   6
  \ /
   5

This algorithm is the same as searching in a BST and is really easy to understand. The worst case run time is O( n + m ).

public static boolean search( int[][] matrix, int value )
{   
    int rows = matrix.length;
    int columns = matrix[0].length;

    int i = 0;
    int j = columns - 1;

    while( i < rows
           && j >= 0 )
    {   
        if( matrix[i][j] == value )
        {
            System.out.println( "Found at " + i + " " + j );
            return true;
        }

        if( matrix[i][j] < value )
        {
            j--;
        }
        else
        {
            i++;
        }
    }

    return false;
}

Answer 2

Here's what I would try. Given an m by n matrix A, compare the value X with the entry A(m/2,n/2) (use floors if necessary).

If A(m/2,n/2) == X, done.

If A(m/2,n/2) < X, then there are 3 smaller matrices to check:

A(1:m/2, 1:n/2) 
A(1:m/2, n/2+1:n) 
A(m/2+1:m, 1:n/2) 

If A(m/2,n/2) > X, , then there are 3 smaller matrices to check:

A(m/2:m, n/2:n) 
A(1:m/2, n/2+1:n) 
A(m/2+1:m, 1:n/2) 

You can eliminate two of them (not always) by comparing the value to the smallest value in the corresponding matrix (the upper left value). Then you recursively try to find the value in each of the remaining matrices.

The complexity of this is approximately O((n*m)^0.8).

---

A row and column sorted matrix is a special case of a Young tableau. I did a google search for searching a Young tableau and found this article which gives 3 nice approaches (the first (and worst) of which is the one I described above).

Answer 3

JavaScript solution:

//start from the top right corner
//if value = el, element is found
//if value < el, move to the next row, element can't be in that row since row is sorted
//if value > el, move to the previous column, element can't be in that column since column is sorted

function find(matrix, el) {

  var row = 0; //first row
  var col = matrix[0].length - 1; //last column

  while (row < matrix.length && col >= 0) {
    if (matrix[row][col] === el) { //element is found
      return true;
    } else if (matrix[row][col] < el) {
      row++; //move to the next row
    } else {
      col--; //move to the previous column
    }
  }

  return false;

}