Find if there is a rectangle in binary matrix with corners as 1 in C++

Suppose we have a binary matrix. We have to find if there is any rectangle or sequence in the given matrix whose all four corners are equal to 1. The matrix is like

1	0	0	1	0
0	0	1	0	1
0	0	0	1	0
1	0	1	0	1

The result will be yes. Here one rectangle is present, whose corners are with 1s.

1	0	1
0	1	0
1	0	1

To solve this we will use one efficient approach. We will follow these steps −

• Scan the matrix from top to bottom line by line

• For each line remember each combination of two 1’s and push that into a hash-set.

• If we ever find that combination again in the later line, we will get our rectangle.

Example

Live Demo

#include<iostream>
#include<unordered_set>
#include<unordered_map>
#include<vector>
using namespace std;
bool isRectanglePresent(const vector<vector<int> >& matrix) {
   int rows = matrix.size();
   if (rows == 0)
   return false;
   int columns = matrix[0].size();
   unordered_map<int, unordered_set<int> > table;
   for (int i = 0; i < rows; ++i) {
      for (int j = 0; j < columns - 1; ++j) {
         for (int k = j + 1; k < columns; ++k) {
            if (matrix[i][j] == 1 && matrix[i][k] == 1) {
               if (table.find(j) != table.end() && table[j].find(k) != table[j].end())
                  return true;
               if (table.find(k) != table.end() && table[k].find(j) != table[k].end())
                  return true;
               table[j].insert(k);
               table[k].insert(j);
            }
         }
      }
   }
   return false;
}
int main() {
   vector<vector<int> > matrix = {
      { 1, 0, 0, 1, 0 },
      { 0, 0, 1, 0, 1 },
      { 0, 0, 0, 1, 0 },
      { 1, 0, 1, 0, 1 }
   };
   if (isRectanglePresent(matrix))
      cout << "Rectangle is present";
   else
      cout << "Rectangle is not present";
}

Output

Rectangle is present