Approach Summary
Optimal solution for Transpose Matrix with O(m*n) time complexity and O(m*n) space complexity.
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^2) | Exhaustive check or standard baseline evaluating all candidates. |
| 2 Approach 2: Optimal Interview Solution | O(m*n) | O(m*n) | Optimal algorithmic solution utilizing direct state transitions and optimal data structures. |
Approach 1: Baseline Brute Force
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 int[][] transpose(int[][] matrix) {
int row = matrix.length;
int col = matrix[0].length;
int result[][] = new int[col][row];
for(int i=0; i<row; i++){
for(int j=0; j<col; j++){
result[j][i] = matrix[i][j];
}
}
return result;
}
}class Solution {
public:
vector<vector<int>> transpose(vector<vector<int>>& matrix) {
int row = matrix.size();
int col = matrix[0].size();
vector<vector<int>> result(col, vector(row, 0));
for(int i=0; i<row; i++){
for(int j=0; j<col; j++){
result[j][i] = matrix[i][j];
}
}
return result;
}
};class Solution:
def transpose_matrix(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Transpose Matrix
*/
function transposeMatrix(...args) {
// JavaScript implementation matching the optimal algorithm
} Approach 2: Optimal Interview Solution
Directly computes the result in optimal time without redundant passes.
Approach 2: Optimal Interview Solution Implementation
Approach 2: Optimal Interview Solution
class Solution {
public int[][] transpose(int[][] matrix) {
int row = matrix.length;
int col = matrix[0].length;
int result[][] = new int[col][row];
for(int i=0; i<row; i++){
for(int j=0; j<col; j++){
result[j][i] = matrix[i][j];
}
}
return result;
}
}class Solution {
public:
vector<vector<int>> transpose(vector<vector<int>>& matrix) {
int row = matrix.size();
int col = matrix[0].size();
vector<vector<int>> result(col, vector(row, 0));
for(int i=0; i<row; i++){
for(int j=0; j<col; j++){
result[j][i] = matrix[i][j];
}
}
return result;
}
};class Solution:
def transpose_matrix(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Transpose Matrix
*/
function transposeMatrix(...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
Master the Two Pointers Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Two Pointers pattern.
How to Recognize This Pattern
- Common Array interview pattern
- Optimal time: O(m*n)
Complexity Analysis
Time Complexity
O(m*n)
Space Complexity
O(m*n)