Commonly Asked Data Structure Interview Questions on Matrix

Last Updated : 19 Jun, 2025

A Matrix can be considered as array of arrays or 2D array. We use matrix data structure to store two dimensional data.

Theoretical Questions for Interviews on Matrix

1. What is the difference between row-major and column-major order in matrix representation?

In row-major order, elements in the same row are stored consecutively in memory, so accessing them is typically faster. In contrast, column-major order is more efficient when performing column operations because the elements of the same column are stored together.

2. How would you check if a given matrix is symmetric?

A matrix is symmetric if it is equal to its transpose. Check if each element at position (i, j) is equal to the element at position (j, i) for all valid indices.

Read more about Transpose of a matrix

3. Explain how to rotate a matrix by 90 degrees.

To rotate a matrix by 90 degrees clockwise, you can transpose the matrix and then reverse each row. First, for an n x n matrix, you swap the element at position (i, j) with the element at (j, i) to perform the transpose. Afterwards, you reverse each row in the transposed matrix. This algorithm runs in O(n^2) time.

4. What are some efficient methods for performing matrix multiplication using arrays?

Matrix multiplication is the process of multiplying two matrices. For two matrices, A (m x n) and B (n x p), the resulting matrix C will have dimensions (m x p). A straightforward approach involves three nested loops: one for the rows of A, one for the columns of B, and one for the summing of the products of corresponding elements. This gives a time complexity of O(m * n * p).

Strassen's Algorithm: This is an optimized algorithm for matrix multiplication that reduces the time complexity from O(n³) to O(n^2.81) using divide-and-conquer techniques.

5. How would you find the kth smallest element in a 2D matrix sorted both row-wise and column-wise?

Use a min-heap. Insert the first element of each row into the heap, then pop the smallest element, and push the next element from the same row into the heap. Repeat this process k times to get the kth smallest element.

6. Given a 2D matrix, write an algorithm to traverse the matrix in a spiral order.

To traverse a matrix in spiral order, you need to systematically move through the matrix starting from the top-left corner. The order of traversal should follow the pattern: left to right, top to bottom, right to left, and bottom to top.

After completing one cycle (left-to-right, top-to-bottom, right-to-left, and bottom-to-top), you shrink the boundaries and repeat the process until all elements are covered. This approach is usually achieved by maintaining four boundaries (top, bottom, left, and right) and adjusting them after each traversal step.

Read more about code implementation for traversing the matrix in a spiral order

7. What are the applications of sparse matrices?

Sparse matrices are matrices in which most of the elements are zero. They are used in various applications such as:

Representing graphs (adjacency matrices), optimization problems, and scientific computing.
The primary advantage of sparse matrices is that they save memory and computation time by only storing the non-zero elements.
Common techniques for sparse matrix storage include compressed sparse row (CSR) or compressed sparse column (CSC) formats, which allow for efficient storage and access.

8. How can you flatten a 2D matrix into a 1D array without using extra space?

To flatten a 2D matrix into a 1D array without using extra space, you can convert the 2D index into a 1D index by using the formula:

index_1D = (i * num_columns) + j

Where i is the row index and j is the column index, and num_columns is the number of columns in the 2D matrix. This operation allows you to access each element of the matrix in a single dimension without using additional memory.

9. How would you perform a binary search on a 2D matrix that is sorted row-wise and column-wise?

To perform binary search in such a matrix, treat it as a 1D sorted array. You can do this by calculating the row and column indices from a single index in the flattened matrix.
Start with the entire matrix range and calculate the middle index.
Compare the element at the middle with the target.
If the middle element is smaller, search the right part; if it’s larger, search the left part.
This approach leverages the sorted structure of both rows and columns.

Time Complexity: O(log(n * m))

Please refer Search element in a sorted matrix for details