Tags: "leetcode", "matrix", "binary-search", access_time 2-min read

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Search a 2D Matrix

Created: October 30, 2018 by [lek-tin]

Last updated: October 30, 2018

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

Solution

# Treat the 2-d matrix as a 1-d list of length rows * cols.
# Binary search indices from 0 to rows * cols - 1.
# Time - O(log m + log n)
# Space - O(1)
class Solution(object):
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if not matrix or not matrix[0]:
            return False

        rows, cols = len(matrix), len(matrix[0])
        low, high = 0, rows * cols - 1

        while high >= low:

            mid = (high + low) // 2
            # convert mid to a row and column
            value = matrix[mid // cols][mid % cols]

            if target == value:
                return True
            if target > value:
                low = mid + 1
            else:
                high = mid - 1

        return False