π‘ Problem Formulation: We are tasked to identify the maximum sized rectangle consisting of 1’s in a binary matrix, with the allowance of swapping columns to maximize the rectangle’s area. Given a 2D binary matrix, such as [[0, 1, 0, 1, 0], [1, 1, 0, 1, 1], [0, 1, 1, 0, 0]], the output should be the dimensions of the largest rectangle of 1’s obtainable through column swaps, for instance (2, 3) indicating a rectangle of 2 by 3.
Method 1: Dynamic Programming with Column Reordering
This method involves an intelligent processing of the binary matrix by extending the concept of maximum histogram area. First, we calculate the “height” of consecutive 1’s for each column. Next, we reorder these heights in non-descending order for each row and apply the largest rectangle in histogram approach to find the maximum area at that row considering column swaps.
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Here’s an example:
def maximalRectangle(matrix):
if not matrix:
return 0
max_area = 0
dp = [[0] * len(matrix[0]) for _ in range(len(matrix))]
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if matrix[i][j] == '1':
dp[i][j] = dp[i - 1][j] + 1 if i != 0 else 1
sorted_row = sorted(dp[i], reverse=True)
for k in range(len(sorted_row)):
max_area = max(max_area, sorted_row[k] * (k + 1))
return max_area
# Example matrix
matrix = [["0", "1", "0", "1", "0"],
["1", "1", "0", "1", "1"],
["0", "1", "1", "0", "0"]]
print(maximalRectangle(matrix))
Output:
6
The code defines a function maximalRectangle() which utilizes dynamic programming to calculate the maximum area of a rectangle of 1’s. It iterates over each row, computing an array of “heights” representing the number of consecutive 1’s up to that row, then sorts it to simulate column swapping. The maximum area is updated as it goes along, and the final result reflects the largest area that can be achieved.
Summary/Discussion
- Method 1: Dynamic Programming with Column Reordering. Strengths: Methodically assesses each possible rectangle and guarantees the largest area. Weaknesses: Could be inefficient for large matrices due to repeated sorting and lack of early exit conditions.
