# 240. Search a 2D Matrix II

#### Medium

***

Write an efficient algorithm that searches for a value `target` in an `m x n` integer matrix `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 1:**

![](https://assets.leetcode.com/uploads/2020/11/24/searchgrid2.jpg)

```
Input: 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]], target = 5
Output: true
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/11/24/searchgrid.jpg)

```
Input: 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]], target = 20
Output: false
```

 

**Constraints:**

* `m == matrix.length`
* `n == matrix[i].length`
* `1 <= n, m <= 300`
* `-109 <= matrix[i][j] <= 109`
* All the integers in each row are **sorted** in ascending order.
* All the integers in each column are **sorted** in ascending order.
* `-109 <= target <= 109`

`Time Complexity : O(300.m.logn)`

```python
class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        index = 0
        rows = len(matrix)
        cols = len(matrix[0])
        rowes = []
        for i in range(rows):
            if matrix[i][0] <= target <= matrix[i][cols-1]:
                rowes.append(i)
            if matrix[i][0] > target:
                break
        for row in rowes:
            column = bisect.bisect_left(matrix[row], target)
            if matrix[row][column] == target or matrix[row][column-1] == target:
                return True
        return False
```