notes/OJ notes/pages/Leetcode Search-a-2D-Matrix.md

# Leetcode Search-a-2D-Matrix

#### 2022-06-13 15:33

---

##### Algorithms:

#algorithm #binary_search

##### Data structures:

#DS #vector #vector_2d

##### Difficulty:

#leetcode #coding_problem #difficulty-easy

##### Related topics:

##### Links:

- [Link to problem](https://leetcode.com/problems/search-a-2d-matrix/submissions/)

---

### Problem

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 from left to right.
- The first integer of each row is greater than the last integer of the previous row.

#### Examples

**Example 1:**

![exapmle1](https://assets.leetcode.com/uploads/2020/10/05/mat.jpg)

```markdown
**Input:** matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
**Output:** true
```

**Example 2:**

![exapmle2](https://assets.leetcode.com/uploads/2020/10/05/mat2.jpg)

```matrix
**Input:** matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
**Output:** false
```

#### Constraints

- m == matrix.length
- n == matrix[i].length
- 1 <= m, n <= 100
- -104 <= matrix[i][j], target <= 104

### Thoughts

Binary search algorithm, with simple 2-d to 1-d conversion

> [!tip] Binary search while loops
> In binary search, remember to check for:
>
> - l <= r
>   And use:
> - l = mid + 1
> - r = mid - 1
>   Init r as m \* n - 1 to avoid mid being OOB.

### Solution

==note how I calculated the mid, i, j, and how I changed r and l==

```cpp
class Solution {
public:
  bool searchMatrix(vector<vector<int>> &matrix, int target) {
    int m = matrix.size();
    int n = matrix[0].size();
    int l = 0;
    // set r to len - 1 to avoid mid being bigger than len.
    int r = m * n - 1;
    int i; // the key to speed is to precaculate i and j, to save time.
    int j;
    int mid;

    do {
      mid = l + (r - l) / 2;
      i = mid / n;
      j = mid % n;
      if (target == matrix[i][j]) {
        return true;
      } else if (target < matrix[i][j]) {
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    } while (l <= r);

    return false;
  }
};

```