How do you solve Largest Submatrix With Rearrangements on LeetCode?

Largest Submatrix With Rearrangements (LeetCode #1727) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n log n) time complexity and O(n) space complexity. Key insight: Understanding how to manipulate matrix dimensions can lead to significant optimizations.

What is the time complexity of Largest Submatrix With Rearrangements?

The optimal solution for Largest Submatrix With Rearrangements runs in O(m * n log n) time and uses O(n) space. The complexity arises from iterating through each row and sorting the heights, which is efficient compared to the brute-force method.

What is the brute force solution for Largest Submatrix With Rearrangements?

The brute force approach for Largest Submatrix With Rearrangements is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible submatrix within the binary matrix to find the largest area of 1s. This is simple but inefficient, as it requires examining all combinations of rows and columns. The optimal Optimal Solution approach improves this to O(m * n log n).

What algorithm or data structure is used to solve Largest Submatrix With Rearrangements?

Largest Submatrix With Rearrangements uses the following concepts: Array, Greedy, Sorting, Matrix. The recommended patterns to study are: Dynamic Programming, Sorting.

What are the interview tips for Largest Submatrix With Rearrangements?

Always clarify the problem constraints and requirements before jumping into coding. Think about how you can optimize your brute-force solution before implementing it. Practice explaining your thought process clearly as you work through the problem.

What are common mistakes when solving Largest Submatrix With Rearrangements?

Failing to account for the need to rearrange columns when calculating areas. Not properly updating heights for consecutive 1s.

#1727

Largest Submatrix With Rearrangements

Medium

ArrayGreedySortingMatrixDynamic ProgrammingSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m * n log n)
Space	O(1)	O(n)

💡

Intuition

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

The optimal approach focuses on calculating the height of consecutive 1s for each column and then sorting these heights for each row. This allows us to efficiently compute the maximum area of 1s by rearranging columns.

⚙️

Algorithm

4 steps

1Step 1: Create an array to store the heights of consecutive 1s for each column.
2Step 2: For each row, update the heights based on the current row's values.
3Step 3: Sort the heights in non-increasing order and calculate the area for each height considering its position.
4Step 4: Keep track of the maximum area found.

solution.py16 lines

1# Full working Python code
2
3def largestSubmatrix(matrix):
4    if not matrix:
5        return 0
6    m, n = len(matrix), len(matrix[0])
7    heights = [0] * n
8    max_area = 0
9    for row in matrix:
10        for j in range(n):
11            heights[j] = heights[j] + 1 if row[j] == 1 else 0
12        sorted_heights = sorted(heights, reverse=True)
13        for i in range(n):
14            max_area = max(max_area, sorted_heights[i] * (i + 1))
15    return max_area
16

ℹ

Complexity note: The complexity arises from iterating through each row and sorting the heights, which is efficient compared to the brute-force method.

1Understanding how to manipulate matrix dimensions can lead to significant optimizations.
2Sorting can help in rearranging columns effectively to maximize areas.

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