# Find length of the largest region in Boolean Matrix

Consider a matrix with rows and columns, where each cell contains either a ‘0’ or a ‘1’ and any cell containing a 1 is called a filled cell. Two cells are said to be connected if they are adjacent to each other horizontally, vertically, or diagonally .If one or more filled cells are also connected, they form a region. find the length of the largest region.

Examples:

```Input : M[][5] = { 0 0 1 1 0
1 0 1 1 0
0 1 0 0 0
0 0 0 0 1 }
Output : 6
Ex: in the following example, there are 2 regions one with length 1 and the other as 6.
so largest region : 6
```

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

Idea is based on the problem or finding number of islands in Boolean 2D-matrix
A cell in 2D matrix can be connected to at most 8 neighbors. So in DFS, we make recursive calls for 8 neighbors. We keep track of the visited 1’s in every DFS and update maximum length region.

Below is C++ implementation of above idea.

```// Program to find the length of the largest
// region in boolean 2D-matrix
#include<bits/stdc++.h>
using namespace std;
#define ROW 4
#define COL 5

// A function to check if a given cell (row, col)
// can be included in DFS
int isSafe(int M[][COL], int row, int col,
bool visited[][COL])
{
// row number is in range, column number is in
// range and value is 1 and not yet visited
return (row >= 0) && (row < ROW) &&
(col >= 0) && (col < COL) &&
(M[row][col] && !visited[row][col]);
}

// A utility function to do DFS for a 2D boolean
// matrix. It only considers the 8 neighbours as
void DFS(int M[][COL], int row, int col,
bool visited[][COL], int &count)
{
// These arrays are used to get row and column
// numbers of 8 neighbours of a given cell
static int rowNbr[] = {-1, -1, -1, 0, 0, 1, 1, 1};
static int colNbr[] = {-1, 0, 1, -1, 1, -1, 0, 1};

// Mark this cell as visited
visited[row][col] = true;

// Recur for all connected neighbours
for (int k = 0; k < 8; ++k)
{
if (isSafe(M, row + rowNbr[k], col + colNbr[k],
visited))
{
// increment region length by one
count++;
DFS(M, row + rowNbr[k], col + colNbr[k],
visited, count);
}
}
}

// The main function that returns largest  length region
// of a given boolean 2D matrix
int  largest(int M[][COL])
{
// Make a bool array to mark visited cells.
// Initially all cells are unvisited
bool visited[ROW][COL];
memset(visited, 0, sizeof(visited));

// Initialize result as 0 and travesle through the
// all cells of given matrix
int result  = INT_MIN;
for (int i = 0; i < ROW; ++i)
{
for (int j = 0; j < COL; ++j)
{
// If a cell with value 1 is not
if (M[i][j] && !visited[i][j])
{
// visited yet, then new region found
int count = 1 ;
DFS(M, i, j, visited , count);

// maximum region
result = max(result , count);
}
}
}
return result ;
}

// Driver program to test above function
int main()
{
int M[][COL] = { {0, 0, 1, 1, 0},
{1, 0, 1, 1, 0},
{0, 1, 0, 0, 0},
{0, 0, 0, 0, 1}};

cout << largestRegion(M);

return 0;
}
```

Output:

```6
```

Time complexity: O(ROW x COL)

This article is contributed by Nishant Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
2.7 Average Difficulty : 2.7/5.0
Based on 30 vote(s)