Smallest Rectangle Enclosing Black Pixels
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An image is represented by a binary matrix with 0 as a white pixel and 1 as a black pixel. The black pixels are connected, i.e., there is only one black region. Pixels are connected horizontally and vertically. Given the location (x, y) of one of the black pixels, return the area of the smallest (axis-aligned) rectangle that encloses all black pixels.
Example:
Input: [ "0010", "0110", "0100" ] and x = 0, y = 2 Output: 6
Github: code.dennyzhang.com
Credits To: leetcode.com
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- Solution: dfs island issues
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// Blog link: https://code.dennyzhang.com/smallest-rectangle-enclosing-black-pixels // Basic Ideas: dfs // region: (max_x-min_x+1)*(max_y-min_y+1) // Complexity: Time O(n*m), Space O(1) var min_x, max_x, min_y, max_y int func dfs(image [][]byte, x int, y int) { if x<0 || x>=len(image) || y<0 || y>=len(image[0]) { return } if image[x][y] != '1' { return } image[x][y] = '2' if x<min_x { min_x = x } if x>max_x { max_x = x } if y<min_y { min_y = y } if y>max_y { max_y = y } dfs(image, x+1, y) dfs(image, x-1, y) dfs(image, x, y+1) dfs(image, x, y-1) } func minArea(image [][]byte, x int, y int) int { if len(image) == 0 { return 0 } min_x, max_x = x, x min_y, max_y = y, y dfs(image, x, y) return (max_x-min_x+1)*(max_y-min_y+1) }
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