Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.
Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)
Example 1:
[[0,0,1,0,0,0,0,1,0,0,0,0,0],
[0,0,0,0,0,0,0,1,1,1,0,0,0],
[0,1,1,0,1,0,0,0,0,0,0,0,0],
[0,1,0,0,1,1,0,0,1,0,1,0,0],
[0,1,0,0,1,1,0,0,1,1,1,0,0],
[0,0,0,0,0,0,0,0,0,0,1,0,0],
[0,0,0,0,0,0,0,1,1,1,0,0,0],
[0,0,0,0,0,0,0,1,1,0,0,0,0]]
Given the above grid, return 6. Note that answer is not 11, because island must be connected 4-directionally.
Example 2:
[[0,0,0,0,0,0,0,0]]
Given the above grid, return 0.
Note: The length of each dimension in the given grid does not exceed 50.
The idea is to use DFS to traverse all the cells one by one, and while traversing each cell, recursively traversing all the neighboring cells from that cell.
If the cell represents water we will immediately return 0; otherwise If the cell represents a land, then we will mark it water (so that we don't traverse that cell again) and then traverse all its neighbours.
This way, we will be able to traverse the one entire island and since we have to find out the area of the island, we return the sum of traversal of all the neighbours.
Also we keep updating the max area, after each island's traversal. And finally we return max area after traversing all the islands.
e.g. below 8*13 matrix have max area of 6 (note that we can only move horizontally, vertically but not diagonally)
public static int maxAreaOfIsland(int[][] grid) {
if(grid == null || grid.length == 0) {
return 0;
}
int rows = grid.length;
int columns = grid[0].length;
int maxRegion = 0;
for(int i = 0; i < rows; i++){
for(int j = 0; j < columns; j++){
if(grid[i][j] == 1){
maxRegion = Math.max(maxRegion, traverseIsland(grid, i, j));
}
}
}
return maxRegion;
}
public static int traverseIsland(int[][] grid, int row, int column){
if(row < 0 || row >= grid.length || column < 0 || column >= grid[0].length ||
grid[row][column] == 0){
return 0;
}
grid[row][column] = 0;
return 1 + traverseIsland(grid, row, column+1) +
traverseIsland(grid, row, column-1) +
traverseIsland(grid, row+1, column) +
traverseIsland(grid, row-1, column);
}public static int maxAreaOfIsland(int[][] grid) {
if(grid == null || grid.length == 0) {
return 0;
}
int rows = grid.length;
int columns = grid[0].length;
int maxRegion = 0;
for(int i = 0; i < rows; i++){
for(int j = 0; j < columns; j++){
if(grid[i][j] == 1){
maxRegion = Math.max(maxRegion, 1 + traverseIsland(grid, i, j)); // <------
}
}
}
return maxRegion;
}
public static int traverseIsland(int[][] grid, int row, int column){
if(row < 0 || row >= grid.length || column < 0 || column >= grid[0].length ||
grid[row][column] == 0){
return 0;
}
grid[row][column] = 0;
return traverseIsland(grid, row, column+1) + // <-------
traverseIsland(grid, row, column-1) +
traverseIsland(grid, row+1, column) +
traverseIsland(grid, row-1, column);
}The above implementation is wrong and will not work.
Case 2 : When we can move horizontally (left, right), vertically (top, down) and diagonally (top-right, bottom-left, bottom-right, top-left)
e.g. below 8*13 matrix have max area of 11 (note that now we can move horizontally, vertically and diagonally)
public static int maxAreaOfIsland(int[][] grid) {
if(grid == null || grid.length == 0) {
return 0;
}
int rows = grid.length;
int columns = grid[0].length;
int maxRegion = 0;
for(int i = 0; i < rows; i++){
for(int j = 0; j < columns; j++){
if(grid[i][j] == 1){
maxRegion = Math.max(maxRegion, traverseIsland(grid, i, j));
}
}
}
return maxRegion;
}
public static int traverseIsland(int[][] grid, int row, int column){
if(row < 0 || row >= grid.length || column < 0 || column >= grid[0].length ||
grid[row][column] == 0){
return 0;
}
grid[row][column] = 0;
return 1 + traverseIsland(grid, row, column+1) +
traverseIsland(grid, row, column-1) +
traverseIsland(grid, row+1, column) +
traverseIsland(grid, row-1, column) +
traverseIsland(grid, row-1, column+1) +
traverseIsland(grid, row+1, column-1) +
traverseIsland(grid, row-1, column-1) +
traverseIsland(grid, row+1, column+1);
}class Solution {
public int maxAreaOfIsland(int[][] grid) {
if(grid == null || grid.length == 0)
return 0;
int maxArea = 0;
Queue<int[]> q = new LinkedList<>();
int[][] directions = {{0, 1} , {0, -1} , {-1, 0} , {1, 0}};
for(int i = 0; i < grid.length; i++) {
for(int j = 0; j < grid[0].length; j++) {
if(grid[i][j] == 1) {
grid[i][j] = 0;
int area = 0;
q.add(new int[]{i,j});
while(!q.isEmpty()) {
area++;
int[] pos = q.remove();
for(int[] direction : directions) {
int row = pos[0] + direction[0];
int col = pos[1] + direction[1];
if(row >= 0 && row < grid.length && col >= 0
&& col < grid[0].length && grid[row][col] == 1) {
q.add(new int[]{row, col});
grid[row][col] = 0;
}
}
}
maxArea = Math.max(maxArea, area);
}
}
}
return maxArea;
}
}
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