How to efficiently search in an ordered matrix? [duplicate]

Answer 1

Split the Matrix in 4 submatrices. If bottom right of a sub-matrix is less than key, discard it. If the top left of a sub-matrix is bigger than the key, discard it. Repeat the splitting procedure for the remaining sub-matrices.

[Update] For some pseudo code (and a discussion of complexity) see Jeffrey L Whitledge's answer of this question.

Answer 2

// the matrix is like this, from left to right is ascending, and
// from down to up is ascending, but the second row'start is not always bigger than the first row's end, which is diff from [leetcode]https://oj.leetcode.com/problems/search-a-2d-matrix/
// 1   5   7    9
// 4   6   10   15
// 8   11  12   19
// 14  16  18   21
// time complexity is O(x+y), x is the count of row, and y is the count of column

public boolean searchMatrix2(int[][] matrix, int target) {
    int rowCount = matrix.length;
    if(rowCount == 0) return false;

    int colCount = matrix[0].length;
    if(colCount == 0) return false;

    //first find the target row, needs O(x)
    int targetRow = 0;
    while(targetRow < rowCount-1 && matrix[targetRow+1][0] <= target) {
        targetRow++;
    }
    //than find the target in the target row, needs O(y), so the total is O(x)+O(y)
    boolean result = false;
    for(int i = 0; i < colCount; i ++) {
        if(matrix[targetRow][i] == target) {
            result = true;
            break;
        }
    }
    return result;
}

Actually, we can use binary search twice, find the target row by binary search first, then find the target in the row by binary search, so the time complexity is O(lgx) + O(lgy), is O(lgx + lgy), better the O(x+y).

Answer 3

Start at the last element of the first row(top-right corner).
Compare it with the key. We have 3 cases:

• If they are equal we are done.

• If key is greater than that element then it means key cannot be present in that row so move the search to the element below it.

• If key is less than that element then it means key could be present in that row towards left and cannot be present in the column further down, so move the search to the element left of it.

Keep doing it till you find the element or you cannot further move(key does not exist).

Pseudo code:

Let R be number of rows
Let C be number of columns

Let i = 0
Let j = C-1

found = false
while( i>=0 && i<R) && (j>=0 && j<C) )
   if (matrix[i][j] == key )
      found = true
      break
   else if( matrix[i][j] > key )
       j--
   else if( matrix[i][j] < key )
       i++
end-while