# C Program for Maximum size square sub-matrix with all 1s

Write a C program for a given binary matrix, the task is to find out the maximum size square sub-matrix with all 1s.

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Approach:

Let the given binary matrix be M[R][C]. The idea of the algorithm is to construct an auxiliary size matrix S[][] in which each entry S[i][j] represents the size of the square sub-matrix with all 1s including M[i][j] where M[i][j] is the rightmost and bottom-most entry in sub-matrix.

Step-by-step approach:

• Construct a sum matrix S[R][C] for the given M[R][C].
• Copy first row and first columns as it is from M[][] to S[][]
• For other entries, use the following expressions to construct S[][]
• 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 then
• S[i][j] = 0
• Find the maximum entry in S[R][C]
• Using the value and coordinates of maximum entry in S[i], print sub-matrix of M[][]

Below is the implementation of the above approach:

## C

 `// C code for Maximum size square``// sub-matrix with all 1s``#include ``#define bool int``#define R 6``#define C 5` `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"``);``    ``}``}` `/* 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;``}` `/* 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``();``}`

Output
```Maximum size sub-matrix is:
1 1 1
1 1 1
1 1 1
```

Time Complexity: O(m*n), where m is the number of rows and n is the number of columns in the given matrix.
Auxiliary Space: O(m*n), where m is the number of rows and n is the number of columns in the given matrix.

## C Program for Maximum size square sub-matrix with all 1s using Dynamic Programming:

In order to compute an entry at any position in the matrix we only need the current row and the previous row.

Below is the implementation of the above approach:

## C

 `#include ``#include ` `#define R 6``#define C 5` `void` `printMaxSubSquare(``bool` `M[R][C])``{``    ``int` `S[2][C];``    ``int` `Max = 0;` `    ``// Set all elements of S to 0 first``    ``for` `(``int` `i = 0; i < 2; i++)``        ``for` `(``int` `j = 0; j < C; j++)``            ``S[i][j] = 0;` `    ``// Construct the entries``    ``for` `(``int` `i = 0; i < R; i++)``        ``for` `(``int` `j = 0; j < C; j++)``        ``{``            ``// Compute the entry at the current position``            ``int` `Entrie = M[i][j];``            ``if` `(Entrie)``            ``{``                ``if` `(j)``                    ``Entrie = 1 + ((S[1][j - 1] < S[0][j - 1]) ? ((S[1][j - 1] < S[1][j]) ? S[1][j - 1] : S[1][j]) : ((S[0][j - 1] < S[1][j]) ? S[0][j - 1] : S[1][j]));``            ``}` `            ``// Save the last entry and add the new one``            ``S[0][j] = S[1][j];``            ``S[1][j] = Entrie;` `            ``// Keep track of the max square length``            ``Max = (Entrie > Max) ? Entrie : Max;``        ``}` `    ``// Print the square``    ``printf``(``"Maximum size sub-matrix is: \n"``);``    ``for` `(``int` `i = 0; i < Max; i++)``    ``{``        ``for` `(``int` `j = 0; j < Max; j++)``            ``printf``(``"1 "``);``        ``printf``(``"\n"``);``    ``}``}` `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);` `    ``return` `0;``}`

Output
```Maximum size sub-matrix is:
1 1 1
1 1 1
1 1 1
```

Time Complexity: O(m*n) where m is the number of rows and n is the number of columns in the given matrix.
Auxiliary space: O(n) where n is the number of columns in the given matrix.

Please refer complete article on Maximum size square sub-matrix with all 1s for more details!

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