# LeetCode 74. Search a 2D Matrix

## Description

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

Write an efficient algorithm that searches for a value in an `m x n` 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.

Example 1:

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

Example 2:

```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`

## Explanation

Use binary search to find which row the target potentially locates. Then also use binary search to search those potential rows.

## Python Solution

``````class Solution:
def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
start = 0
end = len(matrix) - 1

while start + 1 < end:
mid = start + (end - start) // 2

if matrix[mid] == target:
return True
elif matrix[mid] > target:
end = mid
else:
if self.binary_search(matrix[mid], target):
return True

start = mid

if self.binary_search(matrix[start], target) or self.binary_search(matrix[end], target):
return True

return False

def binary_search(self, nums, target):
start = 0
end = len(nums) - 1

while start + 1 < end:
mid = start + (end - start) // 2

if nums[mid] == target:
return True
elif nums[mid] > target:
end = mid
else:
start = mid

if nums[start] == target or nums[end] == target:
return True

return False``````
• Time Complexity: O(log(N)).
• Space Complexity: O(1).