This problem asks you to find a target value in an m x n matrix where rows and columns are sorted. Binary search over a valid answer space is an optimal approach to solve it. Focus on applying binary search for each row or the entire matrix for efficient solution.
Problem Statement
Given a matrix of integers with m rows and n columns, where each row and each column is sorted in ascending order, search for a target value in this matrix. You need to determine if the target exists in the matrix and return true or false accordingly.
You must design an efficient algorithm to perform this search in a matrix that can be as large as 300x300, where the integers in the matrix are in the range from -10^9 to 10^9.
Examples
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 details omitted.
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
Example details omitted.
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
Solution Approach
Binary Search on Rows or Columns
You can apply binary search row by row or column by column. Start from the top-right or bottom-left corner of the matrix and eliminate a row or column based on the comparison with the target.
Divide and Conquer
You can treat the matrix as a set of sub-problems and solve them recursively. By dividing the matrix into smaller sections and applying binary search over valid answer spaces, you reduce the search space efficiently.
Time Complexity Optimization
With a binary search approach, the time complexity becomes O(m + n), where m is the number of rows and n is the number of columns. This is a significant improvement over brute force searching.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity of the solution depends on the approach chosen. A binary search per row or column leads to O(m + n) time complexity. Space complexity can be O(1) if done in-place without additional data structures.
What Interviewers Usually Probe
- The candidate applies binary search efficiently and explains the trade-off between row-by-row or column-by-column binary search.
- The candidate explains the divide and conquer approach in the context of the matrix.
- The candidate demonstrates an understanding of time and space complexity with regard to the constraints.
Common Pitfalls or Variants
Common pitfalls
- Not applying binary search correctly over the valid answer space, leading to inefficient solutions.
- Failing to reduce the problem size at each step, leading to a slower solution.
- Incorrectly handling edge cases such as empty rows, columns, or the target not existing in the matrix.
Follow-up variants
- Optimizing the solution for larger matrices by considering different search strategies.
- Handling special cases where the matrix contains duplicate values or non-standard sorts.
- Adjusting the approach to search for multiple targets in the same matrix.
How GhostInterview Helps
- Provides step-by-step guidance on applying binary search and matrix properties effectively.
- Helps in refining the interview-ready approach with precise time and space complexity analysis.
- Prepares candidates by suggesting interview-ready strategies and avoiding common pitfalls in matrix search problems.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
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.
Need direct help with Search a 2D Matrix II instead of spending more time grinding it?
Download GhostInterview when you want a LeetCode solver, not another long practice loop. Capture Search a 2D Matrix II from a screenshot, get the answer path and complexity, and use supported stealth workflows that stay outside captured layers.
Capture the prompt fast instead of rewriting the problem by hand.
Get the solution path, trade-offs, and complexity summary in one pass.
Stay outside captured layers on supported screen-share workflows.
Stay in the same pattern family
Solve the "Max Sum of Rectangle No Larger Than K" problem using binary search over the valid sum space to optimize space and time complexity.
Open problem page#378 Kth Smallest Element in a Sorted MatrixFind the kth smallest element in a sorted n x n matrix using efficient binary search or heap strategies for optimized memory usage.
Open problem page#74 Search a 2D MatrixSearch a 2D matrix efficiently using binary search over its linearized index, ensuring correctness in row-major sorted arrays.
Open problem page