Suppose we have a 2D array\n\n "000000111000000"\n "000000101000000"\n "000000101100000"\n "000001100100000"\n\n\nImagine we project the 2D array to the bottom axis with the rule "if a column has any black pixel it's projection is black otherwise white". The projected 1D array is\n \n "000001111100000"\n\n> **Theorem**\n>\n> If there are only one black pixel region, then in a projected 1D array all the black pixels are connected.\n> \n> **Proof by contradiction**\n> \n> Assume to the contrary that there are disconnected black pixels at `i`\n> and `j` where `i < j` in the 1D projection array. Thus there exists one\n> column `k`, `k in (i, j)` and and the column `k` in the 2D array has no\n> black pixel. Therefore in the 2D array there exists at least 2 black\n> pixel regions separated by column `k` which contradicting the condition\n> of "only one black pixel region". \n> \n> Therefore we conclude that all the black pixels in the 1D projection\n> array is connected.\n\nThis means we can do a binary search in each half to find the boundaries, if we know one black pixel's position. And we do know that.\n\nTo find the left boundary, do the binary search in the `[0, y)` range and find the first column vector who has any black pixel. \n\nTo determine if a column vector has a black pixel is `O(m)` so the search in total is `O(m log n)`\n\nWe can do the same for the other boundaries. The area is then calculated by the boundaries.\nThus the algorithm runs in `O(m log n + n log m)`\n\n\n**Java**\n\n private char[][] image;\n public int minArea(char[][] iImage, int x, int y) {\n image = iImage;\n int m = image.length, n = image[0].length;\n int left = searchColumns(0, y, 0, m, true);\n int right = searchColumns(y + 1, n, 0, m, false);\n int top = searchRows(0, x, left, right, true);\n int bottom = searchRows(x + 1, m, left, right, false);\n return (right - left) * (bottom - top);\n }\n private int searchColumns(int i, int j, int top, int bottom, boolean opt) {\n while (i != j) {\n int k = top, mid = (i + j) / 2;\n while (k < bottom && image[k][mid] == '0') ++k;\n if (k < bottom == opt)\n j = mid;\n else\n i = mid + 1;\n }\n return i;\n }\n private int searchRows(int i, int j, int left, int right, boolean opt) {\n while (i != j) {\n int k = left, mid = (i + j) / 2;\n while (k < right && image[mid][k] == '0') ++k;\n if (k < right == opt)\n j = mid;\n else\n i = mid + 1;\n }\n return i;\n }\n // Runtime: 1 ms\n\n\n\n**C++**\n\n vector> *image;\n int minArea(vector> &iImage, int x, int y) {\n image = &iImage;\n int m = int(image->size()), n = int((*image)[0].size());\n int top = searchRows(0, x, 0, n, true);\n int bottom = searchRows(x + 1, m, 0, n, false);\n int left = searchColumns(0, y, top, bottom, true);\n int right = searchColumns(y + 1, n, top, bottom, false);\n return (right - left) * (bottom - top);\n }\n int searchRows(int i, int j, int low, int high, bool opt) {\n while (i != j) {\n int k = low, mid = (i + j) / 2;\n while (k < high && (*image)[mid][k] == '0') ++k;\n if (k < high == opt)\n j = mid;\n else\n i = mid + 1;\n }\n return i;\n }\n int searchColumns(int i, int j, int low, int high, bool opt) {\n while (i != j) {\n int k = low, mid = (i + j) / 2;\n while (k < high && (*image)[k][mid] == '0') ++k;\n if (k < high == opt)\n j = mid;\n else\n i = mid + 1;\n }\n return i;\n }\n // Runtime: 20 ms\n\n**Python**\n\n\n def minArea(self, image, x, y):\n top = self.searchRows(image, 0, x, True)\n bottom = self.searchRows(image, x + 1, len(image), False)\n left = self.searchColumns(image, 0, y, top, bottom, True)\n right = self.searchColumns(image, y + 1, len(image[0]), top, bottom, False)\n return (right - left) * (bottom - top)\n\n def searchRows(self, image, i, j, opt):\n while i != j:\n m = (i + j) / 2\n if ('1' in image[m]) == opt:\n j = m\n else:\n i = m + 1\n return i\n\n def searchColumns(self, image, i, j, top, bottom, opt):\n while i != j:\n m = (i + j) / 2\n if any(image[k][m] == '1' for k in xrange(top, bottom)) == opt:\n j = m\n else:\n i = m + 1\n return i\n # Runtime: 56 ms\n\n\n**Java (DRY)**\n\n\n private char[][] image;\n public int minArea(char[][] iImage, int x, int y) {\n image = iImage;\n int m = image.length, n = image[0].length;\n int top = search(0, x, 0, n, true, true);\n int bottom = search(x + 1, m, 0, n, false, true);\n int left = search(0, y, top, bottom, true, false);\n int right = search(y + 1, n, top, bottom, false, false);\n return (right - left) * (bottom - top);\n }\n private boolean isWhite(int mid, int k, boolean isRow) {\n return ((isRow) ? image[mid][k] : image[k][mid]) == '0';\n }\n private int search(int i, int j, int low, int high, boolean opt, boolean isRow) {\n while (i != j) {\n int k = low, mid = (i + j) / 2;\n while (k < high && isWhite(mid, k, isRow)) ++k;\n if (k < high == opt)\n j = mid;\n else\n i = mid + 1;\n }\n return i;\n }\n // Runtime: 2 ms\n\n**C++ (DRY)**\n\n vector> *image;\n int minArea(vector> &iImage, int x, int y) {\n image = &iImage;\n int m = int(image->size()), n = int((*image)[0].size());\n int top = search(0, x, 0, n, true, true);\n int bottom = search(x + 1, m, 0, n, false, true);\n int left = search(0, y, top, bottom, true, false);\n int right = search(y + 1, n, top, bottom, false, false);\n return (right - left) * (bottom - top);\n }\n bool isWhite(int mid, int k, bool isRow) {\n return ((isRow) ? (*image)[mid][k]:(*image)[k][mid]) == '0';\n }\n int search(int i, int j, int low, int high, bool opt, bool isRow) {\n while (i != j) {\n int k = low, mid = (i + j) / 2;\n while (k < high && isWhite(mid, k, isRow)) ++k;\n if (k < high == opt)\n j = mid;\n else\n i = mid + 1;\n }\n return i;\n }\n // Runtime: 24 ms\n\n\n\n**Python (DRY, from Stefan's cool [solution][1])**\n\n def minArea(self, image, x, y):\n top = self.search(0, x, lambda mid: '1' in image[mid])\n bottom = self.search(x + 1, len(image), lambda mid: '1' not in image[mid])\n left = self.search(0, y, lambda mid: any(image[k][mid] == '1' for k in xrange(top, bottom)))\n right = self.search(y + 1, len(image[0]), lambda mid: all(image[k][mid] == '0' for k in xrange(top, bottom)))\n return (right - left) * (bottom - top)\n\n def search(self, i, j, check):\n while i != j:\n mid = (i + j) / 2\n if check(mid):\n j = mid\n else:\n i = mid + 1\n return i\n # Runtime: 56 ms\n\n\n [1]: https://leetcode.com/discuss/68407/clear-binary-search-python

Smallest Rectangle Enclosing Black Pixels

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Problem Description

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