How do you solve Rank Transform of a Matrix on LeetCode?

Rank Transform of a Matrix (LeetCode #1632) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n log(m * n)) time complexity and O(m * n) space complexity. Key insight: Sorting elements allows for efficient rank assignment.

What is the time complexity of Rank Transform of a Matrix?

The optimal solution for Rank Transform of a Matrix runs in O(m * n log(m * n)) time and uses O(m * n) space. The complexity is dominated by the sorting step, which is O(m * n log(m * n)), where m * n is the total number of elements in the matrix.

What is the brute force solution for Rank Transform of a Matrix?

The brute force approach for Rank Transform of a Matrix is Brute Force, with O(m * n * m * n) time complexity and O(1) space complexity. In the brute force approach, we will iterate through each element in the matrix and compare it with every other element in its row and column to determine its rank. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(m * n log(m * n)).

What are the interview tips for Rank Transform of a Matrix?

Always clarify the problem constraints and requirements before diving into coding. Think about edge cases, such as matrices with all identical elements. Discuss your thought process and approach with the interviewer to demonstrate your understanding.

What are common mistakes when solving Rank Transform of a Matrix?

Not considering the same value elements correctly when assigning ranks. Overlooking the need to track maximum ranks in rows and columns.

#1632

Rank Transform of a Matrix

Hard

ArrayUnion-FindGraph TheoryTopological SortSortingMatrixHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m * n * m * n)	O(m * n log(m * n))
Space	O(1)	O(m * n)

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Intuition

Time O(m * n log(m * n))Space O(m * n)

In the optimal solution, we will sort the elements and assign ranks based on their positions in the sorted order, while keeping track of the maximum rank in their respective rows and columns. This is more efficient than the brute force approach.

⚙️

Algorithm

3 steps

1Step 1: Create a list of tuples containing each element's value and its coordinates (row, col).
2Step 2: Sort this list based on the values of the elements.
3Step 3: Iterate through the sorted list and for each element, calculate its rank based on the maximum rank in its row and column, updating the result matrix accordingly.

solution.py16 lines

1# Full working Python code
2
3def rankTransform(matrix):
4    m, n = len(matrix), len(matrix[0])
5    result = [[0] * n for _ in range(m)]
6    elements = [(matrix[i][j], i, j) for i in range(m) for j in range(n)]
7    elements.sort()
8    row_rank = [0] * m
9    col_rank = [0] * n
10    
11    for value, i, j in elements:
12        rank = max(row_rank[i], col_rank[j]) + 1
13        result[i][j] = rank
14        row_rank[i] = rank
15        col_rank[j] = rank
16    return result

ℹ

Complexity note: The complexity is dominated by the sorting step, which is O(m * n log(m * n)), where m * n is the total number of elements in the matrix.

1Sorting elements allows for efficient rank assignment.
2Maintaining row and column ranks helps to derive the correct rank without redundant comparisons.

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