LeetCode 3195: Find the Minimum Area to Cover All Ones I Solution
Master LeetCode problem 3195 (Find the Minimum Area to Cover All Ones I), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
Problem Explanation
Explanation:
To find the minimum area to cover all ones in the given 2D binary array, we need to determine the minimum bounding rectangle that contains all the ones. We can achieve this by calculating the minimum and maximum row and column indices where ones are present. The area of the rectangle can then be determined by the difference between the maximum and minimum row indices multiplied by the difference between the maximum and minimum column indices.
- Initialize variables to store the minimum row, maximum row, minimum column, and maximum column indices where ones are present.
- Iterate through the grid to find the minimum and maximum row and column indices where ones are present.
- Calculate the area of the minimum bounding rectangle using the formula:
(maxRow - minRow + 1) * (maxCol - minCol + 1). - Return the calculated area as the minimum possible area to cover all ones.
Time Complexity: O(m*n) where m and n are the dimensions of the grid. Space Complexity: O(1)
:
Solution Code
class Solution {
public int minArea(char[][] grid) {
int minRow = grid.length, maxRow = 0;
int minCol = grid[0].length, maxCol = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == '1') {
minRow = Math.min(minRow, i);
maxRow = Math.max(maxRow, i);
minCol = Math.min(minCol, j);
maxCol = Math.max(maxCol, j);
}
}
}
return (maxRow - minRow + 1) * (maxCol - minCol + 1);
}
}Try It Yourself
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