Approach Summary
Treat the matrix as a flattened sorted array. Binary search on index 0..m*n-1, converting mid to (row, col) via division and modulo.
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(log(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 row=matrix.length;
int col=matrix[0].length;
int left=0;
int right=(row*col)-1;
while(left<=right){
int mid=left+(right-left)/2;
int i=mid/col;
int j=mid%col;
if(matrix[i][j]==target){
return true;
}
else if(matrix[i][j]<target){
left=mid+1;
}
else{
right=mid-1;
}
}
return false;
}
}class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int row=matrix.size();
int col=matrix[0].size();
int left=0;
int right=(row*col)-1;
while(left<=right){
int mid=left+(right-left)/2;
int i=mid/col;
int j=mid%col;
if(matrix[i][j]==target){
return true;
}
else if(matrix[i][j]<target){
left=mid+1;
}
else{
right=mid-1;
}
}
return false;
}
};class Solution:
def search_a_2d_matrix(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Search in a row-column sorted matrix such that the first integer of each row is greater than the last integer of the previous row.
*/
function searchA_2dMatrix(...args) {
// JavaScript implementation matching the optimal algorithm
} Step 2: Most Optimized
Approach 2: Optimal Interview Solution
Time: O(log(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 row=matrix.length;
int col=matrix[0].length;
int left=0;
int right=(row*col)-1;
while(left<=right){
int mid=left+(right-left)/2;
int i=mid/col;
int j=mid%col;
if(matrix[i][j]==target){
return true;
}
else if(matrix[i][j]<target){
left=mid+1;
}
else{
right=mid-1;
}
}
return false;
}
}class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int row=matrix.size();
int col=matrix[0].size();
int left=0;
int right=(row*col)-1;
while(left<=right){
int mid=left+(right-left)/2;
int i=mid/col;
int j=mid%col;
if(matrix[i][j]==target){
return true;
}
else if(matrix[i][j]<target){
left=mid+1;
}
else{
right=mid-1;
}
}
return false;
}
};class Solution:
def search_a_2d_matrix(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Search in a row-column sorted matrix such that the first integer of each row is greater than the last integer of the previous row.
*/
function searchA_2dMatrix(...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 Binary Search Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Binary Search pattern.
Framework: Binary Search • Interview Prep: Tested at FAANG •
Explore all Binary Search problems & invariants →
How to Recognize This Pattern
- Rows sorted, first element of each row > last of previous
- Binary search on virtual 1D array
Complexity Analysis
Time Complexity
O(log(m × n))
Space Complexity
O(1)
Tags
Array Binary Search Matrix
Frequently Asked Questions
What is the optimal time and space complexity for Search a 2D Matrix?
The optimal solution for Search a 2D Matrix achieves O(log(m × n)) runtime complexity and requires O(1) auxiliary space. Time Complexity is O(log(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 requires the Binary Search pattern?
In technical interviews, look for key problem signals: Rows sorted, first element of each row > last of previous; Binary search on virtual 1D array. Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Binary Search strategy.
What critical edge cases must be handled in Search a 2D Matrix?
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?
Search a 2D Matrix is a Medium-difficulty question tagged with high recurrence. It is actively asked in technical screening and onsite interview loops at Amazon, Microsoft, Google.