Transpose a 2D vector matrix

Question

Is there a way to find transpose a 2D vector matrix without allocating another 2D vector??

Sample Testcase

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[1,4,7],[2,5,8],[3,6,9]]

Code that I tried

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int i,j, n=matrix.size(),temp;
        for(i=0; i<n; i++) {
            for(j=0; j<n; j++) {
                temp=matrix[i][j];
                matrix[i][j]=matrix[j][i];
                matrix[j][i]=temp;
            }
        }
        for(i=0; i<n; i++) {
            for(j=0; j<n; j++) {
                cout<<matrix[i][j]<<" ";
            }
            cout<<endl;
        }
    }
};

Printed Output:

1 2 3 
4 5 6 
7 8 9

Expected Output:

1 4 7
2 5 9
3 6 9

Edited:

New Code:

class Solution {
public:
    void transpose(vector<vector<int>>& mat, int n) {
        int i,j,temp;
        for(i=0; i<n; i++) {
            for(j=0; j<n; j++) {
                temp=mat[i][j];
                mat[i][j]=mat[j][i];
                mat[j][i]=temp;
            }
        }
    }
    void rotate(vector<vector<int>>& matrix) {
        int n=matrix.size(),i,j;
        transpose(matrix, n);
        for(i=0; i<n; i++) {
            for(j=0; j<n; j++) {
                cout<<matrix[i][j]<<" ";
            }
            cout<<endl;
        }
    }
};

Thanks in advance

No @paddy is right, swap the elements you need to move around e.g. std::swap(a[0][2],a[2][0]). — Pepijn Kramer
– Pepijn Kramer, Commented Jan 5, 2022 at 8:34
No it will be done in place. By passing the matrix by reference to rotate you don't make a copy — Pepijn Kramer
– Pepijn Kramer, Commented Jan 5, 2022 at 8:36
You're swapping every possible combination. You should only traverse a single triangle either side of the diagonal. e.g. for (j = 0; j < i; j++) <-- note j < i not j < n. It would be very straight forward for you to work through this by hand on paper and work out how your swaps should go. Also, use std::swap instead of the 3-line version you're doing. — paddy
– paddy, Commented Jan 5, 2022 at 8:47

Amaarockz · Accepted Answer · 2022-01-05 09:44:30Z

1

Below is the code for inplace(Fixed space) || n*n matrix

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int i,j,n=matrix.size();
        for(i=0; i<n; i++) {
            // Instead of j starting with 0 every time, its needs to start from i+1
            for(j=i+1; j<n; j++) {
                swap(matrix[i][j], matrix[j][i]);
            }
        }
        for(int i=0; i<n;i++) {
            reverse(matrix[i].begin(), matrix[i].end());
        }
    }
};

edited Jan 5, 2022 at 9:44

answered Jan 5, 2022 at 8:58

Amaarockz

4,6942 gold badges12 silver badges31 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

paddy Over a year ago

Note that there's no point in starting the j-loop at i because that is on the diagonal. Either use i+1 or use the loop I suggested in the question comments.

Amaarockz Over a year ago

that's a valid point

Collectives™ on Stack Overflow

Transpose a 2D vector matrix

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related