3461. Find The Minimum Area To Cover All Ones I¶

Difficulty: Medium

3461. Find the Minimum Area to Cover All Ones I

Medium

You are given a 2D binary array grid. Find a rectangle with horizontal and vertical sides with the smallest area, such that all the 1's in grid lie inside this rectangle.

Return the minimum possible area of the rectangle.

Example 1:

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

Output: 6

Explanation:

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

Explanation:

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¶

class Solution {
    public int minimumArea(int[][] grid) {
        int n = grid.length, m = grid[0].length;
        int minRow = Integer.MAX_VALUE, minCol = Integer.MAX_VALUE;
        int maxRow = Integer.MIN_VALUE, maxCol = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1) {
                    minRow = Math.min(minRow, i);
                    minCol = Math.min(minCol, j);
                    maxRow = Math.max(maxRow, i);
                    maxCol = Math.max(maxCol, j);
                }
            }
        }    
        return (maxRow - minRow + 1) * (maxCol - minCol + 1);
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here