Sort Matrix by Diagonals

Medium

Asked by:

10 views

Topics:

Arrays

You are given an n x n square matrix of integers grid. Return the matrix such that:

The diagonals in the bottom-left triangle (including the middle diagonal) are sorted in non-increasing order.
The diagonals in the top-right triangle are sorted in non-decreasing order.

Example 1:

Input: grid = [[1,7,3],[9,8,2],[4,5,6]]

Output: [[8,2,3],[9,6,7],[4,5,1]]

Explanation:

The diagonals with a black arrow (bottom-left triangle) should be sorted in non-increasing order:

[1, 8, 6] becomes [8, 6, 1].
[9, 5] and [4] remain unchanged.

The diagonals with a blue arrow (top-right triangle) should be sorted in non-decreasing order:

[7, 2] becomes [2, 7].
[3] remains unchanged.

Example 2:

Input: grid = [[0,1],[1,2]]

Output: [[2,1],[1,0]]

Explanation:

The diagonals with a black arrow must be non-increasing, so [0, 2] is changed to [2, 0]. The other diagonals are already in the correct order.

Example 3:

Input: grid = [[1]]

Output: [[1]]

Explanation:

Diagonals with exactly one element are already in order, so no changes are needed.

Constraints:

grid.length == grid[i].length == n
1 <= n <= 10
-10⁵ <= grid[i][j] <= 10⁵

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the dimensions of the matrix (number of rows and columns)?
Are the values in the matrix guaranteed to be non-negative integers, or can they include negative numbers or zero?
Can the matrix be empty, or can individual rows be empty?
Are there any constraints on the range of integer values within the matrix?
Is it possible for diagonals to contain duplicate values, and if so, should they maintain their relative order after sorting?

Brute Force Solution

Approach

To sort the matrix by diagonals using a brute force approach, we'll identify all the distinct diagonals and then sort the numbers within each diagonal independently. Finally, we'll place these sorted numbers back into their original diagonal positions.

Here's how the algorithm would work step-by-step:

First, identify all the diagonals that run from the top-left to the bottom-right of the grid.
For each identified diagonal, gather all the numbers that belong to it.
Sort the collected numbers for each diagonal from smallest to largest.
After sorting all the numbers within each diagonal, carefully put them back into the grid along their respective diagonals, in the new sorted order.
This process ensures that all numbers on every diagonal are now in ascending order.

Code Implementation

def sort_matrix_by_diagonals_brute_force(matrix):
    rows = len(matrix)
    columns = len(matrix[0])
    diagonals = {}

    # Group elements by their diagonal identifier (row - column)
    for rowIndex in range(rows):
        for columnIndex in range(columns):
            diagonalIdentifier = rowIndex - columnIndex
            if diagonalIdentifier not in diagonals:
                diagonals[diagonalIdentifier] = []
            diagonals[diagonalIdentifier].append(matrix[rowIndex][columnIndex])

    # Sort each diagonal independently
    for diagonalIdentifier in diagonals:
        # Sorting ensures elements are in ascending order for each diagonal
        diagonals[diagonalIdentifier].sort()

    # Place sorted elements back into the matrix
    for rowIndex in range(rows):
        for columnIndex in range(columns):
            diagonalIdentifier = rowIndex - columnIndex
            # Pop from the front of the sorted list to maintain order
            matrix[rowIndex][columnIndex] = diagonals[diagonalIdentifier].pop(0)

    return matrix

Big(O) Analysis

Time Complexity

O(m*n*log(min(m,n))) – Let m be the number of rows and n be the number of columns in the matrix. The brute force approach involves identifying each diagonal, collecting its elements, sorting them, and then placing them back. There are m + n - 1 diagonals in total. The length of the longest diagonal is min(m,n). Collecting elements for each diagonal takes time proportional to the number of elements on that diagonal. Sorting these elements for the longest diagonal will take O(min(m,n) * log(min(m,n))) time. Since we do this for all diagonals, and the total number of elements is m*n, the dominant cost comes from sorting, leading to an overall time complexity of O(m*n*log(min(m,n))). This is because while we process all m*n elements, the sorting operation within each diagonal dominates the complexity.

Space Complexity

O(N*M) – The brute force approach requires storing all elements of the matrix temporarily to sort each diagonal independently. This is typically done by creating a list or array for each diagonal. In the worst case, where we store all elements of the matrix to sort them, the auxiliary space complexity will be proportional to the total number of elements in the matrix, which is N rows multiplied by M columns. Therefore, the space complexity is O(N*M).

Optimal Solution

Approach

The most efficient way to sort a matrix by its diagonals is to treat each diagonal as a separate collection of numbers. We can gather all the numbers belonging to a single diagonal, sort that collection, and then place them back onto their original diagonal in the sorted order. Repeating this for every diagonal ensures the entire matrix is sorted correctly.

Here's how the algorithm would work step-by-step:

Identify all the diagonals that run from the top-left to the bottom-right.
For each diagonal, gather all the numbers that belong to it.
Sort the collected numbers for that specific diagonal in ascending order.
Place the sorted numbers back onto their original diagonal in the matrix.
Repeat this process for every diagonal in the matrix.
Once all diagonals have been processed, the entire matrix will be sorted by its diagonals.

Code Implementation

def sort_matrix_by_diagonals(matrix):
    number_of_rows = len(matrix)
    number_of_columns = len(matrix[0])

    # Group elements by diagonal index.
    diagonals_by_identifier = {}

    for row_index in range(number_of_rows):
        for column_index in range(number_of_columns):
            diagonal_identifier = row_index - column_index
            if diagonal_identifier not in diagonals_by_identifier:
                diagonals_by_identifier[diagonal_identifier] = []
            diagonals_by_identifier[diagonal_identifier].append(matrix[row_index][column_index])

    # Sort each collected diagonal.
    for diagonal_identifier in diagonals_by_identifier:
        diagonals_by_identifier[diagonal_identifier].sort()

    # Place sorted elements back into the matrix.
    for row_index in range(number_of_rows):
        for column_index in range(number_of_columns):
            diagonal_identifier = row_index - column_index
            # Ensure correct placement by popping from the sorted list.
            matrix[row_index][column_index] = diagonals_by_identifier[diagonal_identifier].pop(0)

    return matrix

Big(O) Analysis

Time Complexity

O(M*N log K) – Let M be the number of rows and N be the number of columns in the matrix. The key insight is that the sum of row and column indices (r + c) is constant for each diagonal. There are M + N - 1 such diagonals. The length of each diagonal, K, is at most min(M, N). For each of the M + N - 1 diagonals, we extract its elements (O(K) time), sort them (O(K log K) time), and then place them back (O(K) time). Therefore, the total time complexity is the sum over all diagonals: sum(O(K log K)) for K up to min(M, N). In the worst case, this approximates to (M+N) * O(min(M,N) log min(M,N)). A simpler bound can be derived if we consider that each element is visited twice (once to extract and once to place back) and inserted into a temporary list and sorted once, making it O(M*N log K) where K is the maximum diagonal length.

Space Complexity

O(min(M, N)) – The primary auxiliary space usage comes from temporarily storing the elements of each diagonal to sort them. For a matrix of dimensions M rows and N columns, the longest diagonal can have at most min(M, N) elements. Therefore, we need an auxiliary data structure, such as a list or array, to hold these elements, leading to space complexity proportional to the length of the longest diagonal. The total auxiliary space complexity is thus O(min(M, N)), where M is the number of rows and N is the number of columns.

Edge Cases

Case	How to Handle
Empty matrix or matrix with zero rows/columns	The solution should gracefully handle an empty matrix by returning it as is, as no operations are possible.
Single-row or single-column matrix	Such matrices effectively have only one diagonal each, which the solution should sort correctly.
1x1 matrix	A 1x1 matrix has a single diagonal with one element, which is already sorted and should be returned unchanged.
Matrix with all identical values	The sorting logic should still work, placing identical values adjacent to each other within their respective diagonals.
Matrix with negative numbers, zeros, or large boundary values	Standard integer sorting algorithms handle these values correctly, so no special handling is needed.
Matrix where diagonals have varying lengths	The approach of collecting elements per diagonal and then sorting them handles varying lengths automatically.
Very large matrices exceeding memory limits	For extremely large matrices, an in-place sorting approach for each diagonal might be necessary to conserve memory.
Rectangular matrices (non-square)	The diagonal definition and sorting logic should apply universally to both square and rectangular matrices.

Empty matrix or matrix with zero rows/columns

How to Handle:

The solution should gracefully handle an empty matrix by returning it as is, as no operations are possible.

Single-row or single-column matrix

How to Handle:

Such matrices effectively have only one diagonal each, which the solution should sort correctly.

1x1 matrix

How to Handle:

A 1x1 matrix has a single diagonal with one element, which is already sorted and should be returned unchanged.

Matrix with all identical values

How to Handle:

The sorting logic should still work, placing identical values adjacent to each other within their respective diagonals.

Matrix with negative numbers, zeros, or large boundary values

How to Handle:

Standard integer sorting algorithms handle these values correctly, so no special handling is needed.

Matrix where diagonals have varying lengths

How to Handle:

The approach of collecting elements per diagonal and then sorting them handles varying lengths automatically.

Very large matrices exceeding memory limits

How to Handle:

For extremely large matrices, an in-place sorting approach for each diagonal might be necessary to conserve memory.

Rectangular matrices (non-square)

How to Handle:

The diagonal definition and sorting logic should apply universally to both square and rectangular matrices.