Search a 2D matrix

Time: O(LogM+LogN); Space: O(1); medium

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, 50]
],
target = 3

Output: True

Example 2:

Input: matrix =

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
],
target = 13

Output: False

1. Binary Search [O(LogM+LogN), O(1)]

[1]:

class Solution1(object):
    """
    Time: O(LogM+LogN)
    Space: O(1)
    """
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if not matrix:
            return False

        m, n = len(matrix), len(matrix[0])
        left, right = 0, m * n

        while left < right:
            mid = left + (right - left) // 2
            if matrix[mid // n][mid % n] >= target:
                right = mid
            else:
                left = mid + 1

        return left < m * n and matrix[left // n][left % n] == target

[2]:

s = Solution1()

matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
assert s.searchMatrix(matrix, target) == True

matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
assert s.searchMatrix(matrix, target) == False