leetcode: 74. 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.
# For example,
# 
# Consider the following matrix:
# 
# [
#   [1,   3,  5,  7],
#   [10, 11, 16, 20],
#   [23, 30, 34, 50]
# ]
# Given target = 3, return true.

class Solution():
    def searchMatrix(self, x, target):
        if not x:
            return False
        m, n = len(x), len(x[0])
        left, right = 0, m*n-1
        while left <= right:
            mid = (left + right) // 2
            if x[mid // n][mid % n] == target:
                return True
            elif x[mid // n][mid % n] > target:
                right = mid - 1
            else:
                left = mid + 1
        return 0 <= left < m*n and x[left // n][left % n] == target


if __name__ == "__main__":
    x = [[1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50]]
    assert Solution().searchMatrix(x, 3) == True