867. Transpose Matrix

📝 Problem Details

• Title: 867. Transpose Matrix
• Link: https://leetcode.com/problems/transpose-matrix/
• Difficulty: Easy
• Tags/Categories: Matrix Arrays

💭What Were My Initial Thoughts?

- the initial matrix is of size (n*m)
- create a new matrix with the inverse dimensions of (m*n)
- iterate through the matrix, reversing the location of the element in the new matrix 
- return the new matrix

🤔What Did I Struggle With?

~

💡 Explanation of Solution

same as initial thoughts

⌛ Complexity Analysis

Time Complexity: O(n*m) as we are iterating over every element in the matrix
Space Compleixty: O(m*n) for the construction of the transposed matrix 

💻 Implementation of Solution

class Solution {
public:
    vector<vector<int>> transpose(vector<vector<int>>& matrix) {
        int n = matrix.size();    // Number of rows in the input matrix
        int m = matrix[0].size(); // Number of columns in the input matrix
        vector<vector<int>> newMatrix(m, vector<int>(n)); // Transposed dimensions
 
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                newMatrix[j][i] = matrix[i][j]; // Transpose logic
            }
        }
        return newMatrix;
    }
};