Problem: Given a binary matrix consisting only 0s and 1s, find the maximum size square sub-matrix with all 1s.
Example: Consider the below matrix.
0 1 1 0 1
1 1 0 1 0
0 1 1 1 0
1 1 1 1 0
1 1 1 1 1
0 0 0 0 0
The maximum square sub-matrix with all ’1′ bits is from (2,1) to (4,3)
1 1 1
1 1 1
1 1 1
This is a classic Dynamic Programming problem. This question was asked in Google interview.
Solution:
We will use a auxiliary matrix S[][] of same size for memoization. S[i][j] represents size of the square sub-matrix with all 1s including M[i][j]. ‘i’ and ‘j’ will be the last row and column respectively in square sub-matrix.
Steps to construct S[][]
(1) First copy the first row and first column as it is from M[][] to S[][]
(2) And for the remaining entries as mentioned do the following:
If M[i][j] is 1 then S[i][j] = min(S[i][j-1], S[i-1][j], S[i-1][j-1]) + 1 Else /*If M[i][j] is 0*/ S[i][j] = 0
(3) Find the maximum entry in S[][] and use it to construct the maximum size square sub-matrix.
For the given M[][] in above example, constructed S[][] would be:
0 1 1 0 1
1 1 0 1 0
0 1 1 1 0
1 1 2 2 0
1 2 2 3 1
0 0 0 0 0
The value of maximum entry in above matrix is 3 and coordinates of the entry are (4, 3). Using the maximum value and its coordinates, we can find out the required sub-matrix.
void printMaxSubSquare(bool M[R][C]) { int i,j; int S[R][C]; int maximum, i_index_max, j_index_max; maximum = S[0][0]; i_index_max = 0; j_index_max = 0; /* Set first column of S[][]*/ for(i = 0; i < R; i++) S[i][0] = M[i][0]; /* Set first row of S[][]*/ for(j = 0; j < C; j++) S[0][j] = M[0][j]; for(i = 1; i < R; i++) { for(j = 1; j < C; j++) { if(M[i][j] == 1) S[i][j] = min(S[i][j-1], S[i-1][j], S[i-1][j-1]) + 1; else S[i][j] = 0; } } for(i = 0; i < R; i++) { for(j = 0; j < C; j++) { if(maximum < S[i][j]) { maximum = S[i][j]; i_index_max = i; j_index_max = j; } } } printf("Maximum size sub-matrix is: \n"); for(i = i_index_max; i > i_index_max - maximum; i--) { for(j = j_index_max; j > j_index_max - maximum; j--) { printf("%d ", M[i][j]); } printf("\n"); } }
Complexity:
Time Complexity: O(m*n) where m is number of rows and n is number of columns in the given matrix.
Space Complexity: O(m*n) where m is number of rows and n is number of columns in the given matrix.
can i have recursive solution….of this question..