25(June) Left Rotate Matrix K times

25. Left Rotate Matrix K times

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

Problem Description

You are given an integer k and a matrix mat. Return a matrix where it is rotated left k times.

Example:

Input:

k = 1
mat = [
  [1, 2, 3],
  [4, 5, 6],
  [7, 8, 9]
]

Output:

2 3 1
5 6 4
8 9 7

Explanation: After rotating the matrix by one position to the left, it becomes:

2 3 1
5 6 4
8 9 7

My Approach

1. Initialization:

• Determine the number of rows (n) and columns (m) in the matrix mat.

• Create a new matrix ans of the same dimensions to store the result after rotation.

1. Rotation Calculation:

• Calculate the effective number of rotations needed using k % m to handle cases where k is larger than m.

• Iterate over each element in the original matrix mat, and determine its new position in the matrix ans after rotation.

1. Return:

• Return the matrix ans containing the elements after k left rotations.

Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n * m), as we iterate over all elements in the matrix once.

• Expected Auxiliary Space Complexity: O(n * m), as we use an additional matrix of the same size to store the rotated elements.

Code (C++)

class Solution {
public:
    vector<vector<int>> rotateMatrix(int k, vector<vector<int>> mat) {
        int n = mat.size();
        int m = mat[0].size();
        vector<vector<int>> ans(n, vector<int>(m));

        k %= m;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int newCol = (j - k + m) % m;
                ans[i][newCol] = mat[i][j];
            }
        }
        return ans;
    }
};

Code (Java)

class Solution {
    int[][] rotateMatrix(int k, int mat[][]) {
        int n = mat.length;
        int m = mat[0].length;
        int[][] ans = new int[n][m];

        k %= m;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int newCol = (j - k + m) % m;
                ans[i][newCol] = mat[i][j];
            }
        }
        return ans;
    }
}

Code (Python)

class Solution:
    def rotateMatrix(self, k, mat):
        n = len(mat)
        m = len(mat[0])
        ans = [[0] * m for _ in range(n)]

        k %= m

        for i in range(n):
            for j in range(m):
                new_col = (j - k + m) % m
                ans[i][new_col] = mat[i][j]

        return ans

Contribution and Support

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