How do you solve Sort Matrix by Diagonals on LeetCode?

Sort Matrix by Diagonals (LeetCode #3446) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n² log n) time complexity and O(n) space complexity. Key insight: Understanding how to identify and sort diagonals is crucial for solving matrix-related problems.

What is the time complexity of Sort Matrix by Diagonals?

The optimal solution for Sort Matrix by Diagonals runs in O(n² log n) time and uses O(n) space. The time complexity is O(n² log n) due to sorting the diagonal values. The space complexity remains O(n) for storing the diagonal values.

What is the brute force solution for Sort Matrix by Diagonals?

The brute force approach for Sort Matrix by Diagonals is Brute Force, with O(n²) time complexity and O(n) space complexity. We can extract each diagonal from the matrix, sort them according to the required order, and then place them back in their respective positions. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n² log n).

What algorithm or data structure is used to solve Sort Matrix by Diagonals?

Sort Matrix by Diagonals uses the following concepts: Array, Sorting, Matrix. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Sort Matrix by Diagonals?

Clarify the sorting order for each diagonal before starting your implementation. Consider edge cases, such as very small matrices or matrices with identical elements. Practice similar matrix manipulation problems to build confidence.

What are common mistakes when solving Sort Matrix by Diagonals?

Not correctly identifying which elements belong to which diagonal. Forgetting to sort the diagonals in the correct order before placing them back.

#3446

Sort Matrix by Diagonals

Medium

ArraySortingMatrixHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n² log n)
Space	O(n)	O(n)

💡

Intuition

Time O(n² log n)Space O(n)

We can optimize the process by directly sorting the diagonals while traversing the matrix, which reduces the number of passes over the data.

⚙️

Algorithm

3 steps

1Step 1: Create a map to hold the diagonal values as we traverse the matrix.
2Step 2: For each diagonal, sort the values in the required order during the traversal.
3Step 3: Place the sorted values back into the matrix in a single pass.

solution.py25 lines

1# Full working Python code
2from collections import defaultdict
3
4def diagonalSort(grid):
5    n = len(grid)
6    diagonals = defaultdict(list)
7
8    # Step 1: Fill diagonals
9    for i in range(n):
10        for j in range(n):
11            diagonals[i - j].append(grid[i][j])
12
13    # Step 2: Sort diagonals
14    for key in diagonals:
15        if key <= 0:
16            diagonals[key].sort(reverse=True)  # Bottom-left
17        else:
18            diagonals[key].sort()  # Top-right
19
20    # Step 3: Place sorted values back
21    for i in range(n):
22        for j in range(n):
23            grid[i][j] = diagonals[i - j].pop(0)
24
25    return grid

ℹ

Complexity note: The time complexity is O(n² log n) due to sorting the diagonal values. The space complexity remains O(n) for storing the diagonal values.

1Understanding how to identify and sort diagonals is crucial for solving matrix-related problems.
2Using a map to store diagonal values helps in organizing and sorting them efficiently.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.