#3195

Medium

LeetCode Problem Workspace

Find the Minimum Area to Cover All Ones I

Find the smallest rectangle to cover all 1's in a 2D binary matrix, focusing on array and matrix handling.

Problem Statement

You are given a 2D binary array grid. The goal is to find a rectangle, with horizontal and vertical sides, that has the smallest area, and contains all 1's in the grid.

Return the area of the smallest rectangle that can cover all the 1's. For example, given the grid [[0,1,0],[1,0,1]], the smallest rectangle has an area of 6, formed by covering the entire section from the top-left to bottom-right 1's.

Examples

Example 1

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

Output: 6

The smallest rectangle has a height of 2 and a width of 3, so it has an area of 2 * 3 = 6 .

Example 2

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

Output: 1

The smallest rectangle has both height and width 1, so its area is 1 * 1 = 1 .

Constraints

1 <= grid.length, grid[i].length <= 1000
grid[i][j] is either 0 or 1.
The input is generated such that there is at least one 1 in grid.

Solution Approach

Identify Boundaries of the Rectangle

The solution requires identifying the minimum and maximum row and column indices that contain 1's. This can be done by iterating through the grid once and noting the farthest left, right, top, and bottom positions of the 1's.

Calculate Rectangle Dimensions

Once you have the boundaries, compute the rectangle's height and width. The height is the difference between the maximum and minimum row indices, and the width is the difference between the maximum and minimum column indices.

Return the Area

After calculating the height and width, multiply them to get the area of the rectangle that covers all 1's.

Complexity Analysis

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

The time complexity of this approach is O(n * m) where n and m are the number of rows and columns in the grid, respectively. This is due to the need to traverse the grid to find the boundary indices. The space complexity is O(1) as only a few variables are used to track the minimum and maximum indices.

What Interviewers Usually Probe

The candidate should efficiently identify the boundaries of the rectangle.
They should avoid unnecessary iterations or reprocessing of the grid.
A good solution will have O(n * m) time complexity with constant space usage.

Common Pitfalls or Variants

Common pitfalls

Failing to correctly identify the boundaries of the rectangle.
Overcomplicating the problem by using extra data structures.
Incorrectly calculating the area after finding the bounds.

Follow-up variants

Consider the case when the grid has only one 1.
What if the grid has a large number of 1's, and performance is a concern?
How would you handle grids where the 1's are distributed sparsely?

FAQ

What is the main approach to solving the 'Find the Minimum Area to Cover All Ones I' problem?

The main approach is to find the minimum and maximum row and column indices of cells containing 1's, then calculate the area of the rectangle formed by these coordinates.

How do I handle grids where there is only one 1 in the grid?

If there's only one 1 in the grid, the smallest rectangle is just a 1x1 area, with an area of 1.

What should I avoid when solving this problem?

Avoid unnecessary iterations and extra space usage; only focus on finding the boundaries of the rectangle.

How does GhostInterview help me with this problem?

GhostInterview guides you through efficiently identifying the boundaries of the rectangle and calculating the area with minimal complexity.

What is the time complexity of this problem?

The time complexity is O(n * m), where n is the number of rows and m is the number of columns in the grid.

terminal

Solution

Solution 1: Find Minimum and Maximum Boundaries

We can traverse `grid`, finding the minimum boundary of all `1`s, denoted as $(x_1, y_1)$, and the maximum boundary, denoted as $(x_2, y_2)$. Then, the area of the minimum rectangle is $(x_2 - x_1 + 1) \times (y_2 - y_1 + 1)$.

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class Solution:
    def minimumArea(self, grid: List[List[int]]) -> int:
        x1 = y1 = inf
        x2 = y2 = -inf
        for i, row in enumerate(grid):
            for j, x in enumerate(row):
                if x == 1:
                    x1 = min(x1, i)
                    y1 = min(y1, j)
                    x2 = max(x2, i)
                    y2 = max(y2, j)
        return (x2 - x1 + 1) * (y2 - y1 + 1)

Continue Practicing

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