A Boolean Matrix Question

Given a boolean matrix mat[M][N] of size M X N, modify it such that if a matrix cell mat[i][j] is 1 (or true) then make all the cells of ith row and jth column as 1. 

Examples:

Input: {{1, 0},
           {0, 0}}
Output: {{1, 1}
              {1, 0}}
Input: {{0, 0, 0},
            {0, 0, 1}}
Output: {{0, 0, 1},
               {1, 1, 1}}
 

Input: {{1, 0, 0, 1},
           {0, 0, 1, 0},
          {0, 0, 0, 0}}
Output: {{1, 1, 1, 1},
               {1, 1, 1, 1},
              {1, 0, 1, 1}}

A Boolean Matrix Question using Brute Force:

Approach: Using brute force

Assuming all the elements in the matrix are non-negative. Traverse through the matrix and if you find an element with value 1, then change all the elements in its row and column to -1, except when an element is 1. The reason for not changing other elements to 1, but -1, is because that might affect other columns and rows. Now traverse through the matrix again and if an element is -1 change it to 1, which will be the answer.

The Final Matrix is 
1 1 1 1 
1 4 5 1 
1 3 0 1 

Time Complexity:O((N*M)*(N + M)). O(N*M) for traversing through each element and (N+M)for traversing to row and column of elements having value 1.
Space Complexity:O(1)

Another Approach:

Follow the steps below to solve the problem

• Create two temporary arrays row[M] and col[N]. Initialize all values of row[] and col[] as 0. 
• Traverse the input matrix mat[M][N]. If you see an entry mat[i][j] as true, then mark row[i] and col[j] as true. 
• Traverse the input matrix mat[M][N] again. For each entry mat[i][j], check the values of row[i] and col[j]. If any of the two values (row[i] or col[j]) is true, then mark mat[i][j] as true.

Below is the implementation of the above approach:

Input Matrix 
1001
0010
0000
Matrix after modification 
1111
1111
1011

Time Complexity: O(M*N), Traversing over the matrix two times.
Auxiliary Space: O(M + N), Taking two arrays one of size M and another of size N.

Thanks to Dixit Sethi for suggesting this method. 

A Boolean Matrix Question using O(1) Space:

Intuition: Instead of taking two dummy arrays we can use the first row and column of the matrix for the same work. This will help to reduce the space complexity of the problem. While traversing for the second time the first row and column will be computed first, which will affect the values of further elements that’s why we traversing in the reverse direction.

Approach: Instead of taking two separate dummy arrays, take the first row and column of the matrix as the array for checking whether the particular column or row has the value 1 or not.Since matrix[0][0] are overlapping.Therefore take separate variable col0(say) to check if the 0th column has 1 or not and use matrix[0][0] to check if the 0th row has 1 or not. Now traverse from the last element to the first element and check if matrix[i][0]==1 || matrix[0][j]==1 and if true set matrix[i][j]=1, else continue.

Below is the implementation of above approach:

Input Matrix :
1 0 0 1 
0 0 1 0 
0 0 0 0 
Matrix After Modification :
1 1 1 1 
1 1 1 1 
1 0 1 1 

Time Complexity: O(M*N), Traversing over the matrix two times.
Auxiliary Space: O(1)

 Thanks to Sidh for suggesting this method.