Smallest Rectangle Enclosing Black Pixel

This page explains Java solution to problem Smallest Rectangle Enclosing Black Pixels using Depth First Search algorithm.

Problem Statement

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.

Input:
[
       "0010",
       "0110",
       "0100"
]
x = 0, y = 2
Output: 6

Solution

package com.vc.hard;

class SmallestRectangleEnclosingBlackPixels {

    private int top, bottom, left, right = 0;
    public int minArea(char[][] image, int x, int y) {
        if(image == null || image.length == 0) return 0;

        top = image.length - 1;
        bottom = 0;
        left = image[0].length - 1;
        right = 0;

        helper(image, x, y);

        return (right - left + 1) * (bottom - top + 1);
    }

    private int[][] dirs = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
    private void helper(char[][] image, int x, int y) {
        if(image[x][y] == 0) return;

        image[x][y] = 0;
        top = Math.min(x, top);
        bottom = Math.max(bottom, x);
        left = Math.min(left, y);
        right = Math.max(right, y);
        for(int[] dir: dirs) {
            int xNew = x + dir[0];
            int yNew = y + dir[1];
            if(xNew >= 0 && xNew < image.length && yNew >= 0 && yNew < image[0].length
                    && image[xNew][yNew] == '1') {
                helper(image, xNew, yNew);
            }
        }
    }
}

Time Complexity

O(M * N) Where
M is total number of rows in an input array
N is total number of cols in an inupt array

Space Complexity

O(1)