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# C++ Program to Find all rectangles filled with 0

• Last Updated : 02 Aug, 2022

We have one 2D array, filled with zeros and ones. We have to find the starting point and ending point of all rectangles filled with 0. It is given that rectangles are separated and do not touch each other however they can touch the boundary of the array.A rectangle might contain only one element.

Examples:

```input = [
[1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 0, 0, 0, 1],
[1, 0, 1, 0, 0, 0, 1],
[1, 0, 1, 1, 1, 1, 1],
[1, 0, 1, 0, 0, 0, 0],
[1, 1, 1, 0, 0, 0, 1],
[1, 1, 1, 1, 1, 1, 1]
]

Output:
[
[2, 3, 3, 5], [3, 1, 5, 1], [5, 3, 6, 5]
]

Explanation:
We have three rectangles here, starting from
(2, 3), (3, 1), (5, 3)```
```Input = [
[1, 0, 1, 1, 1, 1, 1],
[1, 1, 0, 1, 1, 1, 1],
[1, 1, 1, 0, 0, 0, 1],
[1, 0, 1, 0, 0, 0, 1],
[1, 0, 1, 1, 1, 1, 1],
[1, 1, 1, 0, 0, 0, 0],
[1, 1, 1, 1, 1, 1, 1],
[1, 1, 0, 1, 1, 1, 0]
]

Output:
[
[0, 1, 0, 1], [1, 2, 1, 2], [2, 3, 3, 5],
[3, 1, 4, 1], [5, 3, 5, 6], [7, 2, 7, 2],
[7, 6, 7, 6]
]```

Step 1: Look for the 0 row-wise and column-wise
Step 2: When you encounter any 0, save its position in the output array, and using loop change all related 0 with this position in any common number so that we can exclude it from processing next time.
Step 3: When you change all related 0 in Step 2, store the last processed 0’s location in the output array in the same index.
Step 4: Take Special care when you touch the edge, by not subtracting -1 because the loop has broken on the exact location.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include `   `using` `namespace` `std;`   `void` `findend(``int` `i, ``int` `j, vector >& a,` `             ``vector >& output, ``int` `index)` `{` `    ``int` `x = a.size();` `    ``int` `y = a[0].size();`   `    ``// flag to check column edge case,` `    ``// initializing with 0` `    ``int` `flagc = 0;`   `    ``// flag to check row edge case,` `    ``// initializing with 0` `    ``int` `flagr = 0;` `    ``int` `n, m;`   `    ``for` `(m = i; m < x; m++) {`   `        ``// loop breaks where first 1 encounters` `        ``if` `(a[m][j] == 1) {` `            ``flagr = 1; ``// set the flag` `            ``break``;` `        ``}`   `        ``// pass because already processed` `        ``if` `(a[m][j] == 5)` `            ``continue``;`   `        ``for` `(n = j; n < y; n++) {` `            ``// loop breaks where first 1 encounters` `            ``if` `(a[m][n] == 1) {` `                ``flagc = 1; ``// set the flag` `                ``break``;` `            ``}`   `            ``// fill rectangle elements with any` `            ``// number so that we can exclude` `            ``// next time` `            ``a[m][n] = 5;` `        ``}` `    ``}`   `    ``if` `(flagr == 1)` `        ``output[index].push_back(m - 1);` `    ``else` `        ``// when end point touch the boundary` `        ``output[index].push_back(m);`   `    ``if` `(flagc == 1)` `        ``output[index].push_back(n - 1);` `    ``else` `        ``// when end point touch the boundary` `        ``output[index].push_back(n);` `}`   `void` `get_rectangle_coordinates(vector > a)` `{`   `    ``// retrieving the column size of array` `    ``int` `size_of_array = a.size();`   `    ``// output array where we are going` `    ``// to store our output` `    ``vector > output;`   `    ``// It will be used for storing start` `    ``// and end location in the same index` `    ``int` `index = -1;`   `    ``for` `(``int` `i = 0; i < size_of_array; i++) {` `        ``for` `(``int` `j = 0; j < a[0].size(); j++) {` `            ``if` `(a[i][j] == 0) {`   `                ``// storing initial position` `                ``// of rectangle` `                ``output.push_back({ i, j });`   `                ``// will be used for the` `                ``// last position` `                ``index = index + 1;` `                ``findend(i, j, a, output, index);` `            ``}` `        ``}` `    ``}`   `    ``cout << ``"["``;` `    ``int` `aa = 2, bb = 0;`   `    ``for` `(``auto` `i : output) {` `        ``bb = 3;` `        ``cout << ``"["``;` `        ``for` `(``int` `j : i) {` `            ``if` `(bb)` `                ``cout << j << ``", "``;` `            ``else` `                ``cout << j;` `            ``bb--;` `        ``}` `        ``cout << ``"]"``;` `        ``if` `(aa)` `            ``cout << ``", "``;` `        ``aa--;` `    ``}` `    ``cout << ``"]"``;` `}`   `// Driver code` `int` `main()` `{` `    ``vector > tests = {` `        ``{ 1, 1, 1, 1, 1, 1, 1 }, { 1, 1, 1, 1, 1, 1, 1 },` `        ``{ 1, 1, 1, 0, 0, 0, 1 }, { 1, 0, 1, 0, 0, 0, 1 },` `        ``{ 1, 0, 1, 1, 1, 1, 1 }, { 1, 0, 1, 0, 0, 0, 0 },` `        ``{ 1, 1, 1, 0, 0, 0, 1 }, { 1, 1, 1, 1, 1, 1, 1 }` `    ``};`   `    ``get_rectangle_coordinates(tests);`   `    ``return` `0;` `}`   `// This code is contributed by mohit kumar 29.`

Output:

`[[2, 3, 3, 5], [3, 1, 5, 1], [5, 3, 6, 5]]`

Time Complexity:O(n^2)

Space Complexity: O(1)

Please refer complete article on Find all rectangles filled with 0 for more details!

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