Interchanging the rows of a Matrix

Problem Statement - link #

Given a matrix of dimensions n1 x m1. Interchange its rows in-place such that the first row will become the last row and so on.

Your Task: You dont need to read input or print anything. Complete the function interchangeRows() that takes matrix as input parameter and modifies the matrix in-place such that the first row becomes the last row and so on.

Expected Time Complexity: O(n1*m1) Expected Auxiliary Space: O(1)

Examples #

Example 1:

Input:
Input:
n1 = 4, m1 = 4
matrix[][] = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15,16]]
Output: 
13 14 15 16 9 10 11 12 5 6 7 8 1 2 3 4
Explanation:
Matrix after exchanging rows:
13 14 15 16
 9 10 11 12
 5  6  7  8
 1  2  3  4
Note: Output is printed row-wise linearly. 

Example 2:

Input:
n1 = 5, m1 = 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
matrix[][] = [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12], [13, 14, 15]]
Output: 
13 14 15 10 11 12 7 8 9 4 5 6 1 2 3
Explanation:
After interchanging rows:
13 14 15
10 11 12
 7  8  9
 4  5  6
 1  2  3

Constraints #

• 1 <= n1, m1 <= 100
• 1 <= matrixij <= 1000

Solutions #

class Solution{
    public:
    //Function to interchange the rows of a matrix.
    void interchangeRows(vector<vector<int> > &matrix)
    {
        // code here 
        int r = matrix.size(), c = matrix[0].size();
        for(int i=0; i<r/2; i++)
            for(int j=0; j<c; j++)
                swap(matrix[i][j], matrix[r-1-i][j]);
    }

};