Problem Description

The problem provides us with an m x n matrix, named mat, and asks us to return an array of all the elements in the matrix arranged in a specific order. This order follows the diagonal patterns of the matrix. We start from the top-left corner, move diagonally up and to the right, and upon reaching the top or right boundary, we move to the next diagonal, which starts from the leftmost or bottommost element of the current boundary. The process continues until all elements are traversed. It is important to note that when moving along the diagonals, the order of traversal alternates. Diagonals that have a bottom-left to top-right orientation are traversed from top to bottom, while diagonals with a top-right to bottom-left orientation are traversed from bottom to top.

Intuition

The solution uses a single-pass algorithm that iterates over the potential starting points for each diagonal in the matrix. A crucial observation is that if you list the starting points, they follow the border of the matrix, first going downward along the left side and then going rightward along the bottom side. The total number of such diagonals would correspond to m + n - 1, considering every element of the first and last row and column except the bottom-right corner element, which is accounted for twice if we simply sum m and n.

The solution assigns two pointers (i and j) that represent the row and column positions, respectively. These pointers scan the matrix along the current diagonal. The direction of scanning alternates with each diagonal: if we're on an even-numbered diagonal (using 0-based indexing), the elements need to be appended in reverse, and for odd-numbered diagonals, the order is maintained as is. The result of each diagonal scan is then concatenated to build the final solution.

Solution Approach

The solution approach involves iterating through all the possible diagonals in the matrix. These diagonals are indexed by a variable called k which runs from 0 to m + n - 2 inclusive.

Here are the steps taken by the algorithm:

1. A loop begins with k starting from 0 and running up to m + n - 1. Each iteration of this loop corresponds to a diagonal in the matrix.

2. For each k, we determine the starting position (i, j) of the current diagonal. If k is less than n (number of columns), then i starts at 0 and j starts at k. Otherwise, when k is at least n, i starts at k - n + 1 and j starts at n - 1.

3. A temporary list t is created to store the elements of the current diagonal. We use while loop to navigate through the diagonal and collect elements from mat[i][j] as long as i < m and j >= 0, incrementing i and decrementing j.

4. The direction of iteration along the diagonal is important. After collecting elements in the temporary list, if k is even, the order of the elements is reversed (using t[::-1]), to adhere to the proper diagonal traversal sequence as specified by the problem.

5. The elements in t (in the correct order) are then added to ans, which is the list that will ultimately be returned.

6. After all iterations, ans contains all matrix elements in the desired diagonal order, and the list is returned.

This approach uses straightforward indexing to access the matrix elements and employs a list to collect the output. The conditional reversal of the temporary list accounts for the zigzag pattern of the traversal, alternating between diagonals. Such an approach is space-efficient since it does not require additional space proportional to the size of the matrix, besides the output list. It also traverses each element exactly once, making it time-efficient.

Example Walkthrough

Let's walk through an example to illustrate the solution approach with a 3 x 4 matrix.

Given matrix mat:

1  2  3  4
5  6  7  8
9 10 11 12

We need to collect the elements in the matrix following a specific diagonal pattern:

1. Start with k = 0, which corresponds to the first diagonal, which is just the element 1. Since k is even, we collect elements from top to bottom. Our answer list ans begins with [1].

2. Move to k = 1, representing the second diagonal [2, 5]. k is odd, so we reverse the order when appending to ans. The answer list becomes [1, 5, 2].

3. For k = 2, the diagonal is [3, 6, 9]. k is even, so append in original order: ans = [1, 5, 2, 3, 6, 9].

4. k = 3 gives diagonal [4, 7, 10]. k is odd, so reverse it: ans = [1, 5, 2, 3, 6, 9, 10, 7, 4].

5. For k = 4, the diagonal is [8, 11]. k is even: ans = [1, 5, 2, 3, 6, 9, 10, 7, 4, 8, 11].

6. Finally, k = 5 corresponds to the last diagonal which is simply [12]. Since k is odd, it remains as is: ans = [1, 5, 2, 3, 6, 9, 10, 7, 4, 8, 11, 12].

The final output is [1, 5, 2, 3, 6, 9, 10, 7, 4, 8, 11, 12], the matrix elements arranged in the desired diagonal order.

Through these steps, the solution iterates over each possible starting point for the diagonals, collects the elements in either forward or reverse order depending on the index of the diagonal, and constructs the output list efficiently. The k loop guarantees that we cover all possible diagonals in the matrix. The given solution is a simple yet effective means to convert 2-dimensional matrix traversal into a 1-dimensional list that preserves the diagonal ordering constraint.

Solution Implementation

Time and Space Complexity

The given code snippet traverses through all the elements of a two-dimensional matrix mat of size m x n diagonally, collecting the elements of each diagonal and appending them to the final result in a specified order.

Time Complexity: O(m * n)

The time complexity of the code is determined by the number of iterations needed to traverse all the elements of the matrix. Since the matrix has m rows and n columns, there are m * n elements in total. The outer loop runs for m + n - 1 iterations, where each diagonal is processed. In each iteration of this loop, a while-loop runs, collecting elements along the current diagonal. Each element in the matrix is processed exactly once inside the nested while-loops. Thus, the total number of operations is proportional to the number of elements in the matrix, leading to a time complexity of O(m * n).

Space Complexity: O(min(m, n))

The space complexity of the code is primarily due to the auxiliary data structure t used for storing each diagonal before appending it to the result list ans. The length of a diagonal is at most min(m, n), because the diagonals are limited either by the number of rows or the number of columns. The reversing operation, t[::-1], creates a new list of the same size as t, and thus, the temporary space required is also in the order of min(m, n). Since the additional space needed is not dependent on the total number of elements but rather on the longer dimension of the input matrix, the space complexity is O(min(m, n)).

The result list ans grows to hold all m * n elements, but this does not contribute to space complexity in the analysis, as it is required to hold the output of the function. Hence, we only consider the additional space used by the algorithm beyond the output space.

Learn more about how to find time and space complexity quickly using problem constraints.