DFS: Connected Cell in a Grid
Problem Statement :
Consider a matrix 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. In the diagram below, the two colored regions show cells connected to the filled cells. Black on white are not connected.
If one or more filled cells are also connected, they form a region. Note that each cell in a region is connected to at least one other cell in the region but is not necessarily directly connected to all the other cells in the region.
Function Description
Complete the function maxRegion in the editor below. It must return an integer value, the size of the largest region.
maxRegion has the following parameter(s):
grid: a two dimensional array of integers
Input Format
The first line contains an integer, n, the number of rows in the matrix, grid.
The second line contains an integer, m, the number of columns in the matrix.
Each of the following n lines contains a row of m space-separated integers, grid[ i ][ j ].
Constraints
0 <= n, m <= 10
grid[ i ][ j ] e { 0,1 }
Output Format
Print the number of cells in the largest region in the given matrix.
Sample Input
4
4
1 1 0 0
0 1 1 0
0 0 1 0
1 0 0 0
Sample Output
5
Solution :
Solution in C :
In C :
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int max=0,count=1;
int a[]={-1,-1,-1,0,0,1,1,1};
int b[]={-1,0,1,-1,1,-1,0,1};
int safe(int r,int c,int row,int col,int g[][10],int v[][10])
{
if(r>=0 && r<row && c>=0 && c<col && g[r][c] && !v[r][c])
return 1;
return 0;
}
void dfs(int grid[][10],int visited[][10],int r,int c,int row,int col)
{
//int a[]={-1,-1,-1,0,0,1,1,1},i,j;
//int b[]={-1,0,1,-1,1,-1,0,1};
int i;
visited[r][c]=1;
for(i=0;i<8;i++)
{
if(safe(r+a[i],c+b[i],row,col,grid,visited))
{
count++;
dfs(grid,visited,r+a[i],c+b[i],row,col);
}
}
if(count>max)
max=count;
}
int main(){
int n,visited[10][10],i,j,c=1;
scanf("%d",&n);
int m;
scanf("%d",&m);
int grid[10][10];
for(int grid_i = 0; grid_i < n; grid_i++){
for(int grid_j = 0; grid_j < m; grid_j++){
visited[grid_i][grid_j]=0;
scanf("%d",&grid[grid_i][grid_j]);
}
}
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
if(grid[i][j]==1 && !visited[i][j])
{
dfs(grid,visited,i,j,n,m);
count=1;
}
}
}
/*for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
printf("%d ",visited[i][j]);
}
printf("\n");
}*/
printf("%d",max);
return 0;
}
Solution in C++ :
In C++ :
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
int count_region(vector<vector<int>> &grid, int i, int j) {
if (i < 0 || j < 0 || i >= grid.size() || j >= grid[0].size())
return 0;
if (grid[i][j] != 1)
return 0;
grid[i][j] = 2;
return 1 + count_region(grid, i + 1, j + 0)
+ count_region(grid, i - 1, j + 0)
+ count_region(grid, i + 0, j + 1)
+ count_region(grid, i + 0, j - 1)
+ count_region(grid, i + 1, j + 1)
+ count_region(grid, i - 1, j - 1)
+ count_region(grid, i + 1, j - 1)
+ count_region(grid, i - 1, j + 1);
}
int get_biggest_region(vector< vector<int> > grid) {
int res = 0;
for (int i = 0; i < grid.size(); ++i) {
for (int j = 0; j < grid[i].size(); ++j) {
if (grid[i][j] == 1) {
int size = count_region(grid, i, j);
res = max(res, size);
}
}
}
return res;
}
int main(){
int n;
cin >> n;
int m;
cin >> m;
vector< vector<int> > grid(n,vector<int>(m));
for(int grid_i = 0;grid_i < n;grid_i++){
for(int grid_j = 0;grid_j < m;grid_j++){
cin >> grid[grid_i][grid_j];
}
}
cout << get_biggest_region(grid) << endl;
return 0;
}
Solution in Java :
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static int getRegion(int[][] matrix, Point pt, int N, int M) {
// int ptr_x = x;
// int ptr_y = y;
Stack<Point> stack = new Stack<>();
ArrayList<Point> visited = new ArrayList<>();
//System.out.printf("Debug: start: x: %d, y : %d\n", pt.x, pt.y);
stack.push(pt);
visited.add(pt);
int area = 1;
while(!stack.isEmpty()) {
Point popPt = stack.pop();
if(popPt == null) {
continue;
}
/*
x,y
x-1,x,y-1,y
*/
Point[] nearby = new Point[] {
new Point(popPt.x-1, popPt.y-1),
new Point(popPt.x-1, popPt.y),
new Point(popPt.x-1, popPt.y+1),
new Point(popPt.x, popPt.y-1),
new Point(popPt.x, popPt.y+1),
new Point(popPt.x+1, popPt.y-1),
new Point(popPt.x+1, popPt.y),
new Point(popPt.x+1, popPt.y+1)
};
for(Point pt2 : nearby) {
int x = pt2.x;
int y = pt2.y;
if(x < 0 || x >= N || y < 0 || y >= M
|| matrix[x][y] == 0
|| visited.contains(pt2))
continue;
//System.out.printf("Debug: x: %d, y: %d, val: %d\n", x, y, matrix[x][y]);
stack.push(pt2);
visited.add(pt2);
area += 1;
}
}
//System.out.printf("Debug:area %d\n", area);
return area;
}
public static int getBiggestRegion(int[][] matrix) {
int N = matrix.length;
int M = matrix[0].length;
//System.out.printf("Debug: length: %d, %d\n", N, M);
int maxRegion = 0;
for(int i = 0; i < N; i += 1) {
for(int j = 0; j < M; j += 1) {
if(matrix[i][j] == 1) {
//System.out.printf("Debug: %d, %d\n", i, j);
int region = getRegion(matrix, new Point(i, j), N, M);
//System.out.printf("Debug: %d, %d\n", i, j);
if(region > maxRegion)
maxRegion = region;
}
}
}
return maxRegion;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int m = in.nextInt();
int grid[][] = new int[n][m];
for(int grid_i=0; grid_i < n; grid_i++){
for(int grid_j=0; grid_j < m; grid_j++){
grid[grid_i][grid_j] = in.nextInt();
}
}
System.out.println(getBiggestRegion(grid));
}
}
class Point {
public int x;
public int y;
public Point(int a, int b) {
x =a;
y = b;
}
public int hashcode() {
int hash = 37;
hash *= x + 17;
hash *= y + 17;
return hash;
}
public boolean equals(Object o) {
if(o == null)
return false;
if(o == this)
return true;
Point o2 = (Point) o;
return o2.x == this.x && o2.y == this.y;
}
}
Solution in Python :
In Python3 :
def get_biggest_region(grid):
grid_rows = len(grid)
grid_cols = len(grid[0])
# Mark all cells unexplored.
explored = [[False] * grid_cols for row in range(grid_rows)]
# Add all cells == 1 to the queue.
queue = []
for row in range(grid_rows):
for col in range(grid_cols):
if grid[row][col] == 1:
queue.append((row, col))
biggest_region = 0
while len(queue) > 0:
row, col = queue.pop()
if explored[row][col]:
continue
region_size = get_region_size(grid, row, col, explored)
biggest_region = max(region_size, biggest_region)
return biggest_region
def get_region_size(grid, row, col, explored):
if explored[row][col] or grid[row][col] == 0:
return 0
explored[row][col] = True
size = 1
for neighbor in get_neighbors(grid, row, col):
nrow, ncol = neighbor
size += get_region_size(grid, nrow, ncol, explored)
return size
def get_neighbors(grid, row, col):
for nrow in range(row - 1, row + 2):
for ncol in range(col - 1, col + 2):
valid_row = nrow >= 0 and nrow < len(grid)
valid_col = ncol >= 0 and ncol < len(grid[0])
not_self = nrow != row or ncol != col
if valid_row and valid_col and not_self:
yield (nrow, ncol)
n = int(input().strip())
m = int(input().strip())
grid = []
for grid_i in range(n):
grid_t = list(map(int, input().strip().split(' ')))
grid.append(grid_t)
print(get_biggest_region(grid))
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