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 <bits/stdc++.h> using namespace std; void findend(int i, int j, vector<vector<int> >& a, vector<vector<int> >& 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<vector<int> > a) { // retrieving the column size of array int size_of_array = a.size(); // output array where we are going // to store our output vector<vector<int> > 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<vector<int> > 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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