240. Search a 2D Matrix II

Medium

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Write an efficient algorithm that searches for a value target in an m x n integer matrix 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 1:

Input: 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]], target = 5
Output: true

Example 2:

Input: 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]], target = 20
Output: false

Constraints:

• m == matrix.length

• n == matrix[i].length

• 1 <= n, m <= 300

• -109 <= matrix[i][j] <= 109

• All the integers in each row are sorted in ascending order.

• All the integers in each column are sorted in ascending order.

• -109 <= target <= 109

Time Complexity : O(300.m.logn)

class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        index = 0
        rows = len(matrix)
        cols = len(matrix[0])
        rowes = []
        for i in range(rows):
            if matrix[i][0] <= target <= matrix[i][cols-1]:
                rowes.append(i)
            if matrix[i][0] > target:
                break
        for row in rowes:
            column = bisect.bisect_left(matrix[row], target)
            if matrix[row][column] == target or matrix[row][column-1] == target:
                return True
        return False