Approach Summary
Start at top-right. If target < current, move left. If target > current, move down. Each step eliminates a row or column.
Practice Mode
2 Progressive Approaches
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Solution Progression (Brute Force → Most Optimized)
Understand the architectural evolution from naive baseline to optimal complexity.
| Approach | Time | Space | Core Strategy / Trade-off |
|---|---|---|---|
| 1 Approach 1: Baseline Brute Force | O(2^n) | O(n) | Exhaustive check or standard baseline evaluating all candidates. |
| 2 Approach 2: Optimal Interview Solution | O(m+n) | O(1) | Optimal algorithmic solution utilizing direct state transitions and optimal data structures. |
Step 1: Baseline
Approach 1: Baseline Brute Force
Time: O(2^n)
Space: O(n)
Iterate or recurse over all possibilities to verify correctness before optimization.
Approach 1: Baseline Brute Force Implementation
Approach 1: Baseline Brute Force
class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
int i = 0;
int j = matrix[0].length-1;
int rowSize = matrix.length;
int colSize = matrix[0].length;
while(i >= 0 && j >= 0 && i < rowSize && j < colSize){
if(matrix[i][j] == target){
return true;
}
else if(matrix[i][j] < target){
i++;
}
else{
j--;
}
}
return false;
}
}class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int row=matrix.size();
int col=matrix[0].size();
int i=0;
int j=matrix[0].size()-1;
while(i>=0 && j>=0 && i<row && j<col){
if(matrix[i][j]==target){
return true;
}
else if(matrix[i][j]<target){
i++;
}
else{
j--;
}
}
return false;
}
};class Solution:
def search_a_2d_matrix_ii(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Search in a row-column sorted Matrix
*/
function searchA_2dMatrixIi(...args) {
// JavaScript implementation matching the optimal algorithm
} Step 2: Most Optimized
Approach 2: Optimal Interview Solution
Time: O(m+n)
Space: O(1)
Directly computes the result in optimal time without redundant passes.
Approach 2: Optimal Interview Solution Implementation
Approach 2: Optimal Interview Solution
class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
int i = 0;
int j = matrix[0].length-1;
int rowSize = matrix.length;
int colSize = matrix[0].length;
while(i >= 0 && j >= 0 && i < rowSize && j < colSize){
if(matrix[i][j] == target){
return true;
}
else if(matrix[i][j] < target){
i++;
}
else{
j--;
}
}
return false;
}
}class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int row=matrix.size();
int col=matrix[0].size();
int i=0;
int j=matrix[0].size()-1;
while(i>=0 && j>=0 && i<row && j<col){
if(matrix[i][j]==target){
return true;
}
else if(matrix[i][j]<target){
i++;
}
else{
j--;
}
}
return false;
}
};class Solution:
def search_a_2d_matrix_ii(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Search in a row-column sorted Matrix
*/
function searchA_2dMatrixIi(...args) {
// JavaScript implementation matching the optimal algorithm
}Edge Cases to Watch
- Empty or single-element input (n <= 1)
- Duplicates and boundary values near extrema
- Already sorted, reverse sorted, or uniform values
- Maximum constraint inputs to ensure no integer overflow or TLE
Algorithmic Pattern
View Pattern Guide →
Master the Matrix Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Matrix pattern.
How to Recognize This Pattern
- Rows and columns sorted independently — start at a corner
- Top-right corner: left decreases, down increases
Complexity Analysis
Time Complexity
O(m + n)
Space Complexity
O(1)
Tags
Array Binary Search Divide and Conquer Matrix
Frequently Asked Questions
What is the optimal time and space complexity for Search a 2D Matrix II?
The optimal solution for Search a 2D Matrix II achieves O(m + n) runtime complexity and requires O(1) auxiliary space. Time Complexity is O(m+n) because the algorithm processes elements efficiently in linear or logarithmic scans without unnecessary nested operations. Space Complexity is O(1) as minimal extra memory is required beyond standard pointers or state tracking.
How do I identify that Search a 2D Matrix II requires the Matrix pattern?
In technical interviews, look for key problem signals: Rows and columns sorted independently — start at a corner; Top-right corner: left decreases, down increases. Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Matrix strategy.
What critical edge cases must be handled in Search a 2D Matrix II?
Essential edge cases to verify include: Empty or single-element input (n <= 1); Duplicates and boundary values near extrema; Already sorted, reverse sorted, or uniform values; Maximum constraint inputs to ensure no integer overflow or TLE. Always walk through these scenarios with your interviewer before coding.
Which tech companies frequently ask Search a 2D Matrix II?
Search a 2D Matrix II is a Medium-difficulty question tagged with high recurrence. It is actively asked in technical screening and onsite interview loops at Amazon, Apple, Google, Microsoft.