# 0240. Search a 2D Matrix II

https://leetcode.com/problems/search-a-2d-matrix-ii

## Description

Write an efficient algorithm that searches for a `target` value in an `m x n` integer `matrix`. The `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`

## ac

```java
// it's not binary search. strange... binary search would be even more ugly.
// information from description must be useful somehow, try to utilize it.
class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int row = matrix.length;
        // edge case
        if (row == 0) return false;
        int col = matrix[0].length;
        if (col == 0 || matrix[0][0] > target || matrix[row-1][col-1] < target) return false;

        // loop
        int r = 0, c = col - 1;
        while (r < row && c >= 0) {
            if (matrix[r][c] == target) {
                return true;
            } else if (matrix[r][c] < target) {
                r++;
            } else {
                c--;
            }
        }

        // can't find
        return false;
    }
}

// first row, last col. while row < m && col >= 0
// binRow -> binCol
```
