Given a binary matrix, find out the maximum size square sub-matrix with all 1s.
For example, consider the below binary matrix.
Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.
C/C++
// C/C++ code for Maximum size square // sub-matrix with all 1s #include <stdio.h> #define bool int #define R 6 #define C 5 /* UTILITY FUNCTIONS *//* Function to get minimum of three values */int min(int a, int b, int c) { int m = a; if (m > b) m = b; if (m > c) m = c; return m; } void printMaxSubSquare(bool M[R][C]) { int i, j; int S[R][C]; int max_of_s, max_i, max_j; /* 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]; /* Construct other entries of S[][]*/ 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; } } /* Find the maximum entry, and indexes of maximum entry in S[][] */ max_of_s = S[0][0]; max_i = 0; max_j = 0; for (i = 0; i < R; i++) { for (j = 0; j < C; j++) { if (max_of_s < S[i][j]) { max_of_s = S[i][j]; max_i = i; max_j = j; } } } printf("Maximum size sub-matrix is: \n"); for (i = max_i; i > max_i - max_of_s; i--) { for (j = max_j; j > max_j - max_of_s; j--) { printf("%d ", M[i][j]); } printf("\n"); } } /* Driver function to test above functions */int main() { bool M[R][C] = { { 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 } }; printMaxSubSquare(M); getchar(); } |
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Output:
Maximum size sub-matrix is: 1 1 1 1 1 1 1 1 1
Please refer complete article on Maximum size square sub-matrix with all 1s for more details!
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