What is the primary approach for solving the Search a 2D Matrix II problem?

The primary approach is to apply binary search over the valid answer space, either row by row or column by column, and reduce the search space at each step.

What is the time complexity of the binary search approach in this problem?

The time complexity is O(m + n), where m is the number of rows and n is the number of columns in the matrix.

How can GhostInterview assist with this problem?

GhostInterview provides insights into the problem's patterns, ensures candidates apply binary search correctly, and helps optimize the solution based on time and space complexity.

What are common mistakes when solving Search a 2D Matrix II?

Common mistakes include not applying binary search correctly, not reducing the search space effectively, and mishandling edge cases like missing targets.

How can I optimize my solution for larger matrices?

You can optimize your solution by carefully applying binary search per row or column and understanding the trade-offs between row-by-row versus column-by-column searches.

#240

Medium

auto_awesome二分·搜索·答案·空间

LeetCode 题解工作台

搜索二维矩阵 II

编写一个高效的算法来搜索 m x n 矩阵 matrix 中的一个目标值 target 。该矩阵具有以下特性：每行的元素从左到右升序排列。每列的元素从上到下升序排列。示例 1：输入： matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[1…

数组二分查找分治矩阵

题目描述

编写一个高效的算法来搜索 m x n 矩阵 matrix 中的一个目标值 target 。该矩阵具有以下特性：

每行的元素从左到右升序排列。
每列的元素从上到下升序排列。

示例 1：

输入：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
输出：true

示例 2：

输入：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
输出：false

提示：

m == matrix.length
n == matrix[i].length
1 <= n, m <= 300
-10⁹ <= matrix[i][j] <= 10⁹
每行的所有元素从左到右升序排列
每列的所有元素从上到下升序排列
-10⁹ <= target <= 10⁹

lightbulb

解题思路

方法一：二分查找

由于每一行的所有元素升序排列，因此，对于每一行，我们可以使用二分查找找到第一个大于等于 $\textit{target}$ 的元素，然后判断该元素是否等于 $\textit{target}$ 。如果等于 $\textit{target}$ ，说明找到了目标值，直接返回 $\text{true}$ 。如果不等于 $\textit{target}$ ，说明这一行的所有元素都小于 $\textit{target}$ ，应该继续搜索下一行。

如果所有行都搜索完了，都没有找到目标值，说明目标值不存在，返回 $\text{false}$ 。

时间复杂度 $O(m \times \log n)$ ，其中 $m$ 和 $n$ 分别为矩阵的行数和列数。空间复杂度 $O(1)$ 。

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class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        for row in matrix:
            j = bisect_left(row, target)
            if j < len(matrix[0]) and row[j] == target:
                return True
        return False

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
The candidate applies binary search efficiently and explains the trade-off between row-by-row or column-by-column binary search.
question_mark
The candidate explains the divide and conquer approach in the context of the matrix.
question_mark
The candidate demonstrates an understanding of time and space complexity with regard to the constraints.

warning

常见陷阱

外企场景

error
Not applying binary search correctly over the valid answer space, leading to inefficient solutions.
error
Failing to reduce the problem size at each step, leading to a slower solution.
error
Incorrectly handling edge cases such as empty rows, columns, or the target not existing in the matrix.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Optimizing the solution for larger matrices by considering different search strategies.
arrow_right_alt
Handling special cases where the matrix contains duplicate values or non-standard sorts.
arrow_right_alt
Adjusting the approach to search for multiple targets in the same matrix.

help

常见问题

外企场景

继续练习

#363 矩形区域不超过 K 的最大数值和 #378 有序矩阵中第 K 小的元素 #74 搜索二维矩阵