Reshape the Matrix
“Reshape the Matrix” can be approximately mapped to “Matrix Diagonal Sum”
Reasoning:
Both are related to 2D matrices and require knowledge of matrix manipulation and traversal. While not exactly the same, they share a similar concept of iterating through the elements of a matrix.
“Reshape the Matrix” involves reshaping the given matrix into a different dimension, which requires a clear understanding of matrix representation and manipulation. This problem is more complex as it involves more significant matrix alterations.
“Matrix Diagonal Sum”, on the other hand, is a simpler problem and involves calculating the sum of the elements on the diagonals of a square matrix. Although the operation performed on the matrix is different, it still requires an understanding of how to access and manipulate matrix elements, which is a key aspect in both problems.
Thus, they can be seen as related in terms of the skills and concepts required to solve them, making them approximately isomorphic.
In Python, you can approach this problem by first checking if the total number of elements in the original matrix (found by multiplying the number of rows and columns) is equal to the total number of elements in the reshaped matrix (found similarly).
If these totals are not equal, we cannot perform the reshaping operation and must return the original matrix. Otherwise, we can create a new list with the reshaped dimensions and fill it with the elements from the original matrix in rowmajor order.
Python solution:


This solution works by “flattening” the original matrix into a single list, then placing each element into the reshaped matrix in the correct location. The division and modulus operations calculate the correct row and column indices for each element.
The line reshaped[i//c][i%c] = mat[i//n][i%n]
is responsible for populating the reshaped matrix from the original matrix in rowmajor order.
The expression reshaped[i//c][i%c]
calculates the row and column indices in the reshaped matrix for the ith element (in rowmajor order).
Here’s how it works:
i//c
performs integer division by the number of columns in the reshaped matrix (c), giving the row index. Essentially, for every increment of ‘c’ elements, we move down to the next row.i%c
gives the remainder when i is divided by the number of columns in the reshaped matrix (c), giving the column index. This cycles from 0 to c1 as i increments, moving rightward across each row.
On the right side of the assignment, mat[i//n][i%n]
is the ith element of the original matrix in rowmajor order.
Here’s the breakdown:
i//n
calculates the row index in the original matrix. Similar to before, for every ’n’ elements, we move down to the next row.i%n
calculates the column index in the original matrix. This moves rightward across each row from 0 to n1 as i increments.
In this way, we can iterate over each element in the original matrix (in rowmajor order) and assign it to the corresponding position in the reshaped matrix.


Problem Classification
This problem falls under the domain of “Array Manipulation” and “Matrix Operations”. The reshaping of an array or matrix is a common operation in linear algebra and data analysis tasks.
What components:
Input: The inputs to the problem are:
 An m x n matrix
mat
.  Two integers
r
andc
which represent the number of rows and columns of the reshaped matrix, respectively.
 An m x n matrix
Output: The output is a reshaped matrix. If the reshaping operation with the given parameters is not possible or illegal, the original matrix should be returned as is.
Condition: The reshaped matrix should be filled with all the elements of the original matrix in the same rowtraversing order as they were in the original matrix.
The problem can be classified as a “Transformation” problem. It requires transforming an existing data structure (matrix) into a new shape while preserving the original data and the order in which it’s presented. A key aspect of this problem involves understanding how elements in a matrix can be reindexed based on the reshaping requirements. It also requires knowledge about matrix dimensions, and an understanding of when a matrix can and cannot be reshaped (e.g., total number of elements in the original and reshaped matrices must be the same).
Language Agnostic Coding Drills
 Dissect the code and identify each distinct concept it contains
The given code can be divided into several distinct coding concepts:
a. Variable and list initialization: This is used to create empty lists to hold data.
b. Nested looping through multidimensional lists (or matrices): This is done to traverse all elements of the matrix.
c. Appending elements to a list: This is used to build up the “flattened” list, which is a 1D representation of the matrix.
d. Conditional statements: The if
condition is used to check if the reshaping is possible or not. If it’s not possible, the original matrix is returned.
e. Slicing lists: This is used to grab chunks of the flattened list and form the new reshaped matrix.
f. Multidimensional list creation: This is used to build the final reshaped matrix.
 Coding concepts or drills in order of increasing difficulty
a. Variable and list initialization: Beginner level. It involves simply declaring variables and initializing them.
b. Appending elements to a list: Beginner level. It requires understanding the basic list operations.
c. Conditional statements: Beginner level. The understanding of if
conditionals and comparison operators is essential here.
d. Nested looping through multidimensional lists (or matrices): Intermediate level. It requires understanding of how loops work and how nested loops can be used to traverse multidimensional data structures.
e. Slicing lists: Intermediate level. It requires understanding of indexing and the slicing syntax.
f. Multidimensional list creation: Advanced level. It involves using list slicing in combination with a loop to dynamically create a list of lists, which represents the reshaped matrix.
 Problemsolving approach
a. The problem starts by flattening the input matrix, i.e., converting it to a onedimensional list. This is done by looping over the matrix and appending each element to the flatten
list.
b. Then, it checks if the reshaping is possible. This is done by checking if the total number of elements in the flattened list is equal to the total number of elements in the reshaped matrix (r * c
). If the reshaping is not possible, it returns the original matrix.
c. If the reshaping is possible, it proceeds by dividing the flattened list into chunks of size c
using list slicing, and appends each chunk as a row in the reshaped matrix.
d. Finally, it returns the reshaped matrix.
Targeted Drills in Python
 Pythonbased coding drills for each concept
a. Variable and list initialization:


b. Appending elements to a list:


c. Conditional statements:


d. Nested looping through multidimensional lists (or matrices):


e. Slicing lists:


f. Multidimensional list creation:


 Problemspecific concepts
The problemspecific concept in this problem is reshaping a matrix. The reshaping is done by first flattening the matrix and then slicing the flattened list into chunks and adding them as rows in the new matrix.
 Integrating the drills
The drills can be integrated together to solve the problem as follows:
a. Start by initializing an empty list to hold the flattened matrix.
b. Use a nested loop to traverse the input matrix and append each element to the flattened list.
c. Check if reshaping is possible. If not, return the original matrix. The condition for reshaping to be possible is that the total number of elements in the reshaped matrix should be equal to the total number of elements in the original matrix.
d. If reshaping is possible, initialize an empty list to hold the reshaped matrix.
e. Use a loop and list slicing to create chunks from the flattened list, and append each chunk as a row in the reshaped matrix.
f. Finally, return the reshaped matrix.