# Question

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 in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

Example:

Consider the following 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]
]


Given target = 5, return true.

Given target = 20, return false.

# Solution

class Solution:
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
if len(matrix) == 0 or len(matrix[0]) == 0:
return False
if target < matrix[0][0] or target > matrix[-1][-1]:
return False
elif target == matrix[0][0] or target == matrix[-1][-1]:
return True

Y, X = len(matrix), len(matrix[0])
x, y = X, Y
top, bottom, left, right = 0, y - 1, 0, x - 1

while bottom - top > 1 or right - left > 1:
x_mid = (left + right) // 2
y_mid = (top + bottom) // 2
mid_val = matrix[y_mid][x_mid]
if target > mid_val:
left = x_mid
top = y_mid
elif target < mid_val:
right = x_mid
bottom = y_mid
else:
return True

x, y = left, top

for i in range(y + 1, Y):
if target == matrix[i][0] or target == matrix[i][-1]:
return True
elif matrix[i][0] < target < matrix[i][-1]:
left, right = 0, X - 1
while right - left > 1:
mid = (left + right) // 2
mid_val = matrix[i][mid]
if target > mid_val:
left = mid
elif target < mid_val:
right = mid
else:
return True

for j in range(x + 1, X):
if target == matrix[0][j] or target == matrix[y][j]:
return True
elif matrix[0][j] < target < matrix[y][j]:
top, bottom = 0, y
while bottom - top > 1:
mid = (top + bottom) // 2
mid_val = matrix[mid][j]
if target > mid_val:
top = mid
elif target < mid_val:
bottom = mid
else:
return True

return False