Counts “islands” of 1s in sea of 0s (2d array matrix)
$begingroup$
This solution surpassed 100% of submissions for efficiency! My method was to recursively check surrounding values and change any contiguous "land" to "water". Is there a better way to write what I wrote?
Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
class Solution {
public int numIslands(char grid) {
int islands = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[i].length; j++) {
if (grid[i][j] == '1') {
islands++;
destroyIsland(grid, i, j);
}
}
}
return islands;
}
public void destroyIsland(char grid, int i, int j) {
grid[i][j] = '0';
if (i < grid.length - 1 && grid[i+1][j] == '1') {
destroyIsland(grid, i+1, j);
}
if (i > 0 && grid[i-1][j] == '1') {
destroyIsland(grid, i-1, j);
}
if (j < grid[i].length - 1 && grid[i][j+1] == '1') {
destroyIsland(grid, i, j+1);
}
if (j > 0 && grid[i][j-1] == '1') {
destroyIsland(grid, i, j-1);
}
}
}
java
asked 4 hours ago
Ned RedmondNed Redmond
61
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add a comment |
$begingroup$
This solution surpassed 100% of submissions for efficiency! My method was to recursively check surrounding values and change any contiguous "land" to "water". Is there a better way to write what I wrote?
Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
class Solution {
public int numIslands(char grid) {
int islands = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[i].length; j++) {
if (grid[i][j] == '1') {
islands++;
destroyIsland(grid, i, j);
}
}
}
return islands;
}
public void destroyIsland(char grid, int i, int j) {
grid[i][j] = '0';
if (i < grid.length - 1 && grid[i+1][j] == '1') {
destroyIsland(grid, i+1, j);
}
if (i > 0 && grid[i-1][j] == '1') {
destroyIsland(grid, i-1, j);
}
if (j < grid[i].length - 1 && grid[i][j+1] == '1') {
destroyIsland(grid, i, j+1);
}
if (j > 0 && grid[i][j-1] == '1') {
destroyIsland(grid, i, j-1);
}
}
}
java
asked 4 hours ago
Ned RedmondNed Redmond
61
New contributor
Ned Redmond is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
This solution surpassed 100% of submissions for efficiency! My method was to recursively check surrounding values and change any contiguous "land" to "water". Is there a better way to write what I wrote?
Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
class Solution {
public int numIslands(char grid) {
int islands = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[i].length; j++) {
if (grid[i][j] == '1') {
islands++;
destroyIsland(grid, i, j);
}
}
}
return islands;
}
public void destroyIsland(char grid, int i, int j) {
grid[i][j] = '0';
if (i < grid.length - 1 && grid[i+1][j] == '1') {
destroyIsland(grid, i+1, j);
}
if (i > 0 && grid[i-1][j] == '1') {
destroyIsland(grid, i-1, j);
}
if (j < grid[i].length - 1 && grid[i][j+1] == '1') {
destroyIsland(grid, i, j+1);
}
if (j > 0 && grid[i][j-1] == '1') {
destroyIsland(grid, i, j-1);
}
}
}
java
asked 4 hours ago
Ned RedmondNed Redmond
61
New contributor
Ned Redmond is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
This solution surpassed 100% of submissions for efficiency! My method was to recursively check surrounding values and change any contiguous "land" to "water". Is there a better way to write what I wrote?
Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
class Solution {
public int numIslands(char grid) {
int islands = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[i].length; j++) {
if (grid[i][j] == '1') {
islands++;
destroyIsland(grid, i, j);
}
}
}
return islands;
}
public void destroyIsland(char grid, int i, int j) {
grid[i][j] = '0';
if (i < grid.length - 1 && grid[i+1][j] == '1') {
destroyIsland(grid, i+1, j);
}
if (i > 0 && grid[i-1][j] == '1') {
destroyIsland(grid, i-1, j);
}
if (j < grid[i].length - 1 && grid[i][j+1] == '1') {
destroyIsland(grid, i, j+1);
}
if (j > 0 && grid[i][j-1] == '1') {
destroyIsland(grid, i, j-1);
}
}
}
java
java
asked 4 hours ago
Ned RedmondNed Redmond
61
New contributor
Ned Redmond is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Ned RedmondNed Redmond
61
New contributor
Ned Redmond is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Ned RedmondNed Redmond
61
New contributor
Ned Redmond is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Ned RedmondNed Redmond
61
asked 4 hours ago
Ned RedmondNed Redmond
61
61
New contributor
Ned Redmond is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ned Redmond is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ned Redmond is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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$begingroup$
The code seems correct. The only missing part is
You may assume all four edges of the grid are all surrounded by water.
which means that numIslands may iterate
for (int i = 1; i < grid.length - 1; i++) {
for (int j = 1; j < grid[i].length - 1; j++) {
and do not bother destroyIsland with validating the surroundings.
answered 1 hour ago
vnpvnp
39k13098
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add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
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oldest
votes
$begingroup$
The code seems correct. The only missing part is
You may assume all four edges of the grid are all surrounded by water.
which means that numIslands may iterate
for (int i = 1; i < grid.length - 1; i++) {
for (int j = 1; j < grid[i].length - 1; j++) {
and do not bother destroyIsland with validating the surroundings.
answered 1 hour ago
vnpvnp
39k13098
$endgroup$
add a comment |
$begingroup$
The code seems correct. The only missing part is
You may assume all four edges of the grid are all surrounded by water.
which means that numIslands may iterate
for (int i = 1; i < grid.length - 1; i++) {
for (int j = 1; j < grid[i].length - 1; j++) {
and do not bother destroyIsland with validating the surroundings.
answered 1 hour ago
vnpvnp
39k13098
$endgroup$
add a comment |
$begingroup$
The code seems correct. The only missing part is
You may assume all four edges of the grid are all surrounded by water.
which means that numIslands may iterate
for (int i = 1; i < grid.length - 1; i++) {
for (int j = 1; j < grid[i].length - 1; j++) {
and do not bother destroyIsland with validating the surroundings.
answered 1 hour ago
vnpvnp
39k13098
$endgroup$
The code seems correct. The only missing part is
You may assume all four edges of the grid are all surrounded by water.
which means that numIslands may iterate
for (int i = 1; i < grid.length - 1; i++) {
for (int j = 1; j < grid[i].length - 1; j++) {
and do not bother destroyIsland with validating the surroundings.
answered 1 hour ago
vnpvnp
39k13098
answered 1 hour ago
vnpvnp
39k13098
answered 1 hour ago
vnpvnp
39k13098
answered 1 hour ago
vnpvnp
39k13098
39k13098
add a comment |
add a comment |
Ned Redmond is a new contributor. Be nice, and check out our Code of Conduct.
Ned Redmond is a new contributor. Be nice, and check out our Code of Conduct.
Ned Redmond is a new contributor. Be nice, and check out our Code of Conduct.
Ned Redmond is a new contributor. Be nice, and check out our Code of Conduct.
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