1992. Find All Groups of Farmland | leetcode daily challenge | DSA | Hindi | shashwat

[@yashhingu7955]

I used same approach 😊

[@aadityapatel4546]

my dfs approach :
class Solution {
int row;
int col;
ArrayList ansal;
public int[][] findFarmland(int[][] land) {
row=land.length;
col=land[0].length;
ansal=new ArrayList();
for(int i=0;i

[@AdityaKumar-wq3nb]

class Solution {
public:
void findcorner(vector& land,int i,int j,vector& ans){
int n=land.size();
int m=land[0].size();
vectortemp(4,-1);
temp[0]=i;
temp[1]=j;
while(i

[@AdityaKumar-wq3nb]

without dfs ithink this is more optimal than dfs anyone tel me?

[@AdityaKumar-wq3nb]

lol i didnt watched the video the video have same appproch

[@ankitpatel0101]

class Solution {
public static int startX = 0;
public static int startY = 0;
public static int endX = 0;
public static int endY = 0;

public int[][] findFarmland(int[][] land) {
ArrayList ansList = new ArrayList();
for (int i = 0; i < land.length; i++) {
for (int j = 0; j < land[0].length; j++) {
if (land[i][j] == 1) {
startX = i;
startY = j;
endX = i;
endY = j;
dfs(land, i, j);
ansList.add(new int[] { startX, startY, endX, endY });
}
}
}
return ansList.toArray(new int[ansList.size()][]);
}
private static void dfs(int[][] land, int i, int j) {
int n = land.length;
int m = land[0].length;
// out of bounds
if (i < 0 || i >= n || j < 0 || j >= m) {
return;
}
// visited for forest land
if (land[i][j] == 0 || land[i][j] == -1) {
return;
}
// work
endX = Math.max(endX, i);
endY = Math.max(endY, j);
land[i][j] = -1;
// check again
dfs(land, i - 1, j);
dfs(land, i, j + 1);
dfs(land, i + 1, j);
dfs(land, i, j - 1);
}
}