Problem of the day
You are given a 2-dimensional array/list having N rows and M columns, which is filled with ones(1) and zeroes(0). 1 signifies land, and 0 signifies water.
A cell is said to be connected to another cell, if one cell lies immediately next to the other cell, in any of the eight directions (two vertical, two horizontal, and four diagonals).
A group of connected cells having value 1 is called an island. Your task is to find the number of such islands present in the matrix.
The first line of input contains two integer values, 'N' and 'M', separated by a single space. They represent the 'rows' and 'columns' respectively, for the two-dimensional array/list.
The second line onwards, the next 'N' lines or rows represent the ith row values.
Each of the i-th row constitutes 'M' column values separated by a single space.
Output Format :
The only line of output prints the number of islands present in the 2-dimensional array.
Note :
You are not required to print anything explicitly, it has already been taken care of. Implement the function and return the desired output.
1 <= N <= 10^3
1 <= M <= 10^3
0 <= ARR[i][j] <= 1
Time limit: 1sec
4 5
0 1 1 0 0
1 0 0 1 0
0 0 1 0 0
1 0 0 0 1
3
The first island of connected 1s is signified by: {0, 1}, {0, 2}, {1, 0}, {1, 3}, {2, 2}.
The second island being: {3, 0}.
The third island being: {3, 4}.
4 4
1 0 0 1
0 1 1 0
0 1 1 0
1 0 0 1
1
Think of how we can apply a BFS/DFS like approach to traverse the matrix. Also, think about an efficient way to check all the 8 directions.
We can use the flood fill algorithm to check for all connected 1s.
O(N * M), where N and M are the dimensions of the array.
Traversing each cell of the 2D array once requires O(N * M) time.
O(N * M), where N and M are the dimensions of the array.
We have to maintain a boolean 2D array, which keeps track of whether a cell has been visited before or not.