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.lengthn == matrix[i].length1 <= 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][0] == target:
return True
elif matrix[mid][0] > 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).