10. Transpose of Matrix

The problem can be found at the following link: Question Link

My Approach

The problem of transposing a matrix can be solved by using two index variables, i and j. We can iterate over i from 0 to n - 1 and j from i to n - 1 to avoid unnecessary repetition.

• In each iteration, we swap the elements mat[i][j] with mat[j][i].

Time and Auxiliary Space Complexity

• Time Complexity: O(n^2) since a nested loops run for n iterations, where n is the size of the matrix.

• Auxiliary Space Complexity: O(1) since these operations were performed in-place, meaning it does not require any additional space.

Code (C++)

class Solution
{   
    public:
    void transpose(vector<vector<int>>& mat, int n)
    { 
        for(int i = 0; i<n;++i){
            for(int j = i; j<n;++j){
                swap(mat[i][j],mat[j][i]);
            }
        }
    }
};

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