We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsDisclaimersPrivacyTermsContactAbout Us

Reshape the Matrix

Difficulty: Easy

Problem Description

You are given an m x n matrix mat and two integers r and c representing the number of rows and the number of columns of the wanted reshaped matrix. The reshaped matrix should be filled with all the elements of the original matrix in the same row-traversing order as they were. If the reshape operation with the given parameters is possible and legal, output the new reshaped matrix; otherwise, output the original matrix.

Key Insights

  • The total number of elements in the original matrix must equal the total number of elements in the reshaped matrix (i.e., m * n == r * c).
  • If the reshape operation is not possible, we should return the original matrix.
  • The elements should be filled in a row-wise manner in the new matrix.

Space and Time Complexity

Time Complexity: O(m * n) - We need to traverse all elements of the original matrix once. Space Complexity: O(r * c) - We create a new matrix of size r x c.

Solution

To solve the problem, we will use the following steps:

  1. Calculate the total number of elements in the original matrix.
  2. Check if the reshape operation is possible by comparing the total number of elements in the original matrix with the desired dimensions of the new matrix.
  3. If reshaping is possible, create a new matrix of dimensions r x c and fill it with elements from the original matrix in a row-wise manner.
  4. If reshaping is not possible, return the original matrix.

The main data structures used here are two-dimensional arrays (for both the original and reshaped matrices) and simple integer variables to track indices.

Code Solutions

def reshape(mat, r, c):
    m, n = len(mat), len(mat[0])
    if m * n != r * c:
        return mat
    
    # Create the reshaped matrix
    reshaped = [[0] * c for _ in range(r)]
    
    for i in range(m):
        for j in range(n):
            index = i * n + j  # Calculate the index in the 1D representation
            reshaped[index // c][index % c] = mat[i][j]  # Fill the reshaped matrix
            
    return reshaped
← Back to All Questions