Leetcode: Smallest Rectangle Enclosing Black Pixels Posted on May 26, 2018September 23, 2019 by braindenny Smallest Rectangle Enclosing Black Pixels Similar Problems: Series: Island & Follow-up Number of Distinct Islands CheatSheet: Leetcode For Code Interview Tag: #island, #dfs 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 Leave me comments, if you have better ways to solve. Solution: dfs island issues Similar Problems: Number of Distinct Islands CheatSheet: Leetcode For Code Interview Tag: #island, #dfs // 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) } Post Views: 7