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# 240\_Search a 2D Matrix II

\[Medium]

Write an efficient algorithm that searches for a value in an mxn matrix. This matrix has the following properties:

* Integers in each row are sorted in ascending from left to right.
* Integers in each column are sorted in ascending from top to bottom.

**Example:**

Consider the following matrix:

```
[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]
```

Given target = `5`, return `true`.

Given target = `20`, return `false`.

## Solution:

**Idea**:

* Start from top-right, check if the element equals to target.
  * If equal, return True.
  * If larger than target, move to the left.
  * If smaller than target, move to below.

**Time Complexity**: $$O(m + n)$$ where m is the number of row, n is the number of column.

**Space Complexity**: $$O(1)$$

```python
def searchMatrix(matrix, target):
    """
    :type matrix: List[List[int]]
    :type target: int
    :rtype: bool
    """
    # corner case when matrix is empty
    if not matrix: return False

    # start from top-right corner
    i, j = 0, len(matrix[0]) - 1
    # keep searching while there are unclassified rows and columns
    while i < len(matrix) and j >= 0:
        if matrix[i][j] == target:
            return True
        elif matrix[i][j] > target:
            j -= 1
        else:  
            # matrix[i][j] < target
            i += 1                
    return False
```