#3197

Hard

LeetCode Problem Workspace

Find the Minimum Area to Cover All Ones II

Find the minimum area to cover all 1's in a 2D binary grid using three non-overlapping rectangles.

Problem Statement

You are given a 2D binary array grid. Your task is to find three non-overlapping rectangles, each having non-zero areas, with horizontal and vertical sides. These rectangles should be positioned in such a way that they cover all the 1's in the grid.

Return the minimum possible sum of the areas of these rectangles. The rectangles may touch but cannot overlap. Keep in mind that the grid contains at least three 1's, and the size constraints are such that grid.length and grid[i].length are both at most 30.

Examples

Example 1

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

Output: 5

Example 2

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

Output: 5

Constraints

1 <= grid.length, grid[i].length <= 30
grid[i][j] is either 0 or 1.
The input is generated such that there are at least three 1's in grid.

Solution Approach

Rectangle Enumeration and Minimization

Enumerate all possible rectangle placements in the grid that cover the 1's. This step involves finding the top-left and bottom-right corners of potential rectangles. Use dynamic programming or backtracking to minimize the sum of the areas of the rectangles while ensuring no overlap between them.

Divide the Grid into Smaller Sections

Divide the grid into smaller sections by selecting possible cuts, either horizontally or vertically. Each cut will divide the grid into two regions, and you can then calculate the area of the rectangles covering the 1's in each region. This approach helps reduce the complexity by narrowing down the search space for possible rectangle placements.

Optimize the Rectangle Selection

Once the possible rectangles are identified, the next step is to select the ones that minimize the total area. You can use a greedy approach or dynamic programming to find the optimal combination of rectangles that cover all 1's while minimizing the area.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depend on the final approach chosen. If dynamic programming or backtracking is used to enumerate rectangle placements, the complexity will be related to the number of possible rectangle configurations. Given the grid's constraints, an efficient solution will focus on minimizing redundant calculations and optimizing the search space.

What Interviewers Usually Probe

Strong understanding of dynamic programming or backtracking.
Ability to optimize grid-based problems with enumeration.
Clear approach to solving problems involving minimizing area and avoiding overlap.

Common Pitfalls or Variants

Common pitfalls

Failing to handle cases where rectangles must touch but not overlap.
Overcomplicating the enumeration of potential rectangles.
Missing optimization steps to minimize the area of rectangles.

Follow-up variants

Extending the problem to cover more than three rectangles.
Considering different shapes of rectangles (not just aligned with axes).
Handling grids with larger sizes (e.g., up to 100x100).

FAQ

What is the minimum area to cover all ones in a grid?

The problem asks to find the minimum sum of areas of three non-overlapping rectangles that cover all 1's in a 2D grid.

How do you minimize the area of rectangles in this problem?

By using enumeration and optimization techniques like dynamic programming, you can select rectangles that minimize the total area while ensuring they do not overlap.

What is the primary pattern in 'Find the Minimum Area to Cover All Ones II'?

The primary pattern is 'Array plus Matrix', as it involves working with a 2D grid and enumerating possible rectangle placements to minimize the area.

How can I handle large grids in this problem?

You can optimize your solution by dividing the grid into smaller sections and using dynamic programming to handle large grids more efficiently.

What are common pitfalls in this problem?

Common pitfalls include failing to optimize rectangle enumeration, missing overlap constraints, and overcomplicating the approach to minimizing area.

terminal

Solution

Solution 1: Enumeration

According to the problem description, we can use two dividing lines to split the rectangle into three parts. We calculate the minimum rectangular area containing all $1$s for each part and then take the minimum sum of the areas of the three parts.

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class Solution:
    def minimumSum(self, grid: List[List[int]]) -> int:
        def f(i1: int, j1: int, i2: int, j2: int) -> int:
            x1 = y1 = inf
            x2 = y2 = -inf
            for i in range(i1, i2 + 1):
                for j in range(j1, j2 + 1):
                    if grid[i][j] == 1:
                        x1 = min(x1, i)
                        y1 = min(y1, j)
                        x2 = max(x2, i)
                        y2 = max(y2, j)
            return (x2 - x1 + 1) * (y2 - y1 + 1)

        m, n = len(grid), len(grid[0])
        ans = m * n
        for i1 in range(m - 1):
            for i2 in range(i1 + 1, m - 1):
                ans = min(
                    ans,
                    f(0, 0, i1, n - 1)
                    + f(i1 + 1, 0, i2, n - 1)
                    + f(i2 + 1, 0, m - 1, n - 1),
                )
        for j1 in range(n - 1):
            for j2 in range(j1 + 1, n - 1):
                ans = min(
                    ans,
                    f(0, 0, m - 1, j1)
                    + f(0, j1 + 1, m - 1, j2)
                    + f(0, j2 + 1, m - 1, n - 1),
                )
        for i in range(m - 1):
            for j in range(n - 1):
                ans = min(
                    ans,
                    f(0, 0, i, j) + f(0, j + 1, i, n - 1) + f(i + 1, 0, m - 1, n - 1),
                )
                ans = min(
                    ans,
                    f(0, 0, i, n - 1)
                    + f(i + 1, 0, m - 1, j)
                    + f(i + 1, j + 1, m - 1, n - 1),
                )

                ans = min(
                    ans,
                    f(0, 0, i, j) + f(i + 1, 0, m - 1, j) + f(0, j + 1, m - 1, n - 1),
                )
                ans = min(
                    ans,
                    f(0, 0, m - 1, j)
                    + f(0, j + 1, i, n - 1)
                    + f(i + 1, j + 1, m - 1, n - 1),
                )
        return ans

Continue Practicing

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