# Dynamic Programming – Maximum size square sub-matrix with all 1s

Objective: Given a matrix of 0’s and 1’s (binary matrix). Find out Maximum size square sub-matrix with all 1’s.

Example:

Maximum size square sub-matrix with all 1s Example

Approach:

Base Cases:

• If only one row is given then cells with 1’s will be the Maximum size square sub-matrix with size = 1.
• If only one column is given then cells with 1’s will be the Maximum size square sub-matrix with size = 1.

Dynamic Programming (Bottom-up).

• Create an auxiliary array of the same size as given input array. We will fill the auxiliary array with Maximum size square sub-matrix with all 1’s possible with respect to the particular cell.
• Once the auxiliary is fill, scan it and find out the maximum entry in it, this will the size of Maximum size square sub-matrix with all 1’s in the given matrix.

How to fill the auxiliary matrix??

• Copy the first row and first column from the given array to auxiliary array. (Read the base cases described earlier).
• For filling rest of cells, check if particular cell’s value is 0 in given array, if yes then put 0 against to that cell in auxiliary array.
• check if particular cell’s value is 1 in given array, if yes then put Minimum (value in the left cell, top cell and left-top diagonal cell) + 1 in auxiliary cell.

Recursive Equation:

```For Given arrA[][], create auxiliary array sub[][].
```

#### Base Cases: sub[i][0] = arrA[i][0] i = 0 to row Count // copy the first column sub[0][i] = arrA[0][i] i = 0 to column Count // copy the first row

for rest of the cells
sub[i][j] = 0 if arrA[i][j]=0               = Min(arrA[i-1][j], arrA[i][j-1], arrA[i-1][j-1] )
At the End, scan the sub[][] and find out the maximum entry in it.

Example:

Maximum size square sub-matrix – Auxiliary Array

Complete Code:

```Output:
Maximum size square sub-matrix with all 1s: 3

```

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If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
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### 2 Responses

1. Mithun Mathew says:

Why is it required to go through the matrix sub again to find the maxSize?

Why not keep track of maxSize when sub is being populated?
int maxSize = 0;
for (int i = 1; i < row; i++) {
for (int j = 1; j < cols; j++) {
sub[i][j] = (arrA[i][j] == 0)? 0: Math.min(sub[i – 1][j – 1], Math.min(sub[i][j – 1], sub[i – 1][j])) + 1;
maxSize = Math.max(maxSize, sub[i][j]);
}
}
return maxSize;

• tutorialhorizon says:

Yes we can do that as well. nice approach.

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