Given a grid with different colors in a different cell, each color represented by a different number. The task is to find out the largest connected component on the grid. Largest component grid refers to a maximum set of cells such that you can move from any cell to any other cell in this set by only moving between side-adjacent cells from the set.
Examples:
Input :
Grid of different colors
Output : 9
Largest connected component of grid
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Approach :
The approach is to visualize the given grid as a graph with each cell representing a separate node of the graph and each node connected to four other nodes which are to immediately up, down, left, and right of that grid. Now doing a BFS search for every node of the graph, find all the nodes connected to the current node with same color value as the current node.
Here is the graph for above example :
Graph representation of grid
.
At every cell (i, j), a BFS can be done. The possible moves from a cell will be either to right, left, top or bottom. Move to only those cells which are in range and are of the same color. It the same nodes have been visited previously, then the largest component value of the grid is stored in result[][] array. Using memoization, reduce the number of BFS on any cell. visited[][] array is used to mark if the cell has been visited previously and count stores the count of the connected component when a BFS is done for every cell. Store the maximum of the count and print the resultant grid using result[][] array.
Below is the illustration of the above approach:
C++
// CPP program to print the largest
// connected component in a grid
#include <bits/stdc++.h>
using namespace std;
const int n = 6;
const int m = 8;
// stores information about which cell
// are already visited in a particular BFS
int visited[n][m];
// result stores the final result grid
int result[n][m];
// stores the count of cells in the largest
// connected component
int COUNT;
// Function checks if a cell is valid i.e it
// is inside the grid and equal to the key
bool is_valid(int x, int y, int key, int input[n][m])
{
if (x < n && y < m && x >= 0 && y >= 0) {
if (visited[x][y] == false && input[x][y] == key)
return true;
else
return false;
}
else
return false;
}
// BFS to find all cells in
// connection with key = input[i][j]
void BFS(int x, int y, int i, int j, int input[n][m])
{
// terminating case for BFS
if (x != y)
return;
visited[i][j] = 1;
COUNT++;
// x_move and y_move arrays
// are the possible movements
// in x or y direction
int x_move[] = { 0, 0, 1, -1 };
int y_move[] = { 1, -1, 0, 0 };
// checks all four points connected with input[i][j]
for (int u = 0; u < 4; u++)
if (is_valid(i + y_move[u], j + x_move[u], x, input))
BFS(x, y, i + y_move[u], j + x_move[u], input);
}
// called every time before a BFS
// so that visited array is reset to zero
void reset_visited()
{
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
visited[i][j] = 0;
}
// If a larger connected component
// is found this function is called
// to store information about that component.
void reset_result(int key, int input[n][m])
{
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (visited[i][j] && input[i][j] == key)
result[i][j] = visited[i][j];
else
result[i][j] = 0;
}
}
}
// function to print the result
void print_result(int res)
{
cout << "The largest connected "
<< "component of the grid is :" << res << "\n";
// prints the largest component
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (result[i][j])
cout << result[i][j] << " ";
else
cout << ". ";
}
cout << "\n";
}
}
// function to calculate the largest connected
// component
void computeLargestConnectedGrid(int input[n][m])
{
int current_max = INT_MIN;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
reset_visited();
COUNT = 0;
// checking cell to the right
if (j + 1 < m)
BFS(input[i][j], input[i][j + 1], i, j, input);
// updating result
if (COUNT >= current_max) {
current_max = COUNT;
reset_result(input[i][j], input);
}
reset_visited();
COUNT = 0;
// checking cell downwards
if (i + 1 < n)
BFS(input[i][j], input[i + 1][j], i, j, input);
// updating result
if (COUNT >= current_max) {
current_max = COUNT;
reset_result(input[i][j], input);
}
}
}
print_result(current_max);
}
// Drivers Code
int main()
{
int input[n][m] = { { 1, 4, 4, 4, 4, 3, 3, 1 },
{ 2, 1, 1, 4, 3, 3, 1, 1 },
{ 3, 2, 1, 1, 2, 3, 2, 1 },
{ 3, 3, 2, 1, 2, 2, 2, 2 },
{ 3, 1, 3, 1, 1, 4, 4, 4 },
{ 1, 1, 3, 1, 1, 4, 4, 4 } };
// function to compute the largest
// connected component in the grid
computeLargestConnectedGrid(input);
return 0;
}
Java
// Java program to print the largest
// connected component in a grid
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
static final int n = 6;
static final int m = 8;
// stores information about which cell
// are already visited in a particular BFS
static final int visited[][] = new int [n][m];
// result stores the final result grid
static final int result[][] = new int [n][m];
// stores the count of
// cells in the largest
// connected component
static int COUNT;
// Function checks if a cell
// is valid i.e it is inside
// the grid and equal to the key
static boolean is_valid(int x, int y,
int key,
int input[][])
{
if (x < n && y < m &&
x >= 0 && y >= 0)
{
if (visited[x][y] == 0 &&
input[x][y] == key)
return true;
else
return false;
}
else
return false;
}
// BFS to find all cells in
// connection with key = input[i][j]
static void BFS(int x, int y, int i,
int j, int input[][])
{
// terminating case for BFS
if (x != y)
return;
visited[i][j] = 1;
COUNT++;
// x_move and y_move arrays
// are the possible movements
// in x or y direction
int x_move[] = { 0, 0, 1, -1 };
int y_move[] = { 1, -1, 0, 0 };
// checks all four points
// connected with input[i][j]
for (int u = 0; u < 4; u++)
if ((is_valid(i + y_move[u],
j + x_move[u], x, input)) == true)
BFS(x, y, i + y_move[u],
j + x_move[u], input);
}
// called every time before
// a BFS so that visited
// array is reset to zero
static void reset_visited()
{
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
visited[i][j] = 0;
}
// If a larger connected component
// is found this function is
// called to store information
// about that component.
static void reset_result(int key,
int input[][])
{
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
if (visited[i][j] ==1 &&
input[i][j] == key)
result[i][j] = visited[i][j];
else
result[i][j] = 0;
}
}
}
// function to print the result
static void print_result(int res)
{
System.out.println ("The largest connected " +
"component of the grid is :" +
res );
// prints the largest component
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
if (result[i][j] != 0)
System.out.print(result[i][j] + " ");
else
System.out.print(". ");
}
System.out.println();
}
}
// function to calculate the
// largest connected component
static void computeLargestConnectedGrid(int input[][])
{
int current_max = Integer.MIN_VALUE;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
reset_visited();
COUNT = 0;
// checking cell to the right
if (j + 1 < m)
BFS(input[i][j], input[i][j + 1],
i, j, input);
// updating result
if (COUNT >= current_max)
{
current_max = COUNT;
reset_result(input[i][j], input);
}
reset_visited();
COUNT = 0;
// checking cell downwards
if (i + 1 < n)
BFS(input[i][j],
input[i + 1][j], i, j, input);
// updating result
if (COUNT >= current_max)
{
current_max = COUNT;
reset_result(input[i][j], input);
}
}
}
print_result(current_max);
}
// Driver Code
public static void main(String args[])
{
int input[][] = {{1, 4, 4, 4, 4, 3, 3, 1},
{2, 1, 1, 4, 3, 3, 1, 1},
{3, 2, 1, 1, 2, 3, 2, 1},
{3, 3, 2, 1, 2, 2, 2, 2},
{3, 1, 3, 1, 1, 4, 4, 4},
{1, 1, 3, 1, 1, 4, 4, 4}};
// function to compute the largest
// connected component in the grid
computeLargestConnectedGrid(input);
}
}
// This code is contributed by Subhadeep
Python3
# Python3 program to print the largest
# connected component in a grid
n = 6;
m = 8;
# stores information about which cell
# are already visited in a particular BFS
visited = [[0 for j in range(m)]for i in range(n)]
# result stores the final result grid
result = [[0 for j in range(m)]for i in range(n)]
# stores the count of cells in the largest
# connected component
COUNT = 0
# Function checks if a cell is valid i.e it
# is inside the grid and equal to the key
def is_valid(x, y, key, input):
if (x < n and y < m and x >= 0 and y >= 0):
if (visited[x][y] == 0 and input[x][y] == key):
return True;
else:
return False;
else:
return False;
# BFS to find all cells in
# connection with key = input[i][j]
def BFS(x, y, i, j, input):
global COUNT
# terminating case for BFS
if (x != y):
return;
visited[i][j] = 1;
COUNT += 1
# x_move and y_move arrays
# are the possible movements
# in x or y direction
x_move = [ 0, 0, 1, -1 ]
y_move = [ 1, -1, 0, 0 ]
# checks all four points connected with input[i][j]
for u in range(4):
if (is_valid(i + y_move[u], j + x_move[u], x, input)):
BFS(x, y, i + y_move[u], j + x_move[u], input);
# called every time before a BFS
# so that visited array is reset to zero
def reset_visited():
for i in range(n):
for j in range(m):
visited[i][j] = 0
# If a larger connected component
# is found this function is called
# to store information about that component.
def reset_result(key, input):
for i in range(n):
for j in range(m):
if (visited[i][j] != 0 and input[i][j] == key):
result[i][j] = visited[i][j];
else:
result[i][j] = 0;
# function to print the result
def print_result(res):
print("The largest connected "+
"component of the grid is :" + str(res));
# prints the largest component
for i in range(n):
for j in range(m):
if (result[i][j] != 0):
print(result[i][j], end = ' ')
else:
print('. ',end = '')
print()
# function to calculate the largest connected
# component
def computeLargestConnectedGrid(input):
global COUNT
current_max = -10000000000
for i in range(n):
for j in range(m):
reset_visited();
COUNT = 0;
# checking cell to the right
if (j + 1 < m):
BFS(input[i][j], input[i][j + 1], i, j, input);
# updating result
if (COUNT >= current_max):
current_max = COUNT;
reset_result(input[i][j], input);
reset_visited();
COUNT = 0;
# checking cell downwards
if (i + 1 < n):
BFS(input[i][j], input[i + 1][j], i, j, input);
# updating result
if (COUNT >= current_max):
current_max = COUNT;
reset_result(input[i][j], input);
print_result(current_max);
# Drivers Code
if __name__=='__main__':
input = [ [ 1, 4, 4, 4, 4, 3, 3, 1 ],
[ 2, 1, 1, 4, 3, 3, 1, 1 ],
[ 3, 2, 1, 1, 2, 3, 2, 1 ],
[ 3, 3, 2, 1, 2, 2, 2, 2 ],
[ 3, 1, 3, 1, 1, 4, 4, 4 ],
[ 1, 1, 3, 1, 1, 4, 4, 4 ] ];
# function to compute the largest
# connected component in the grid
computeLargestConnectedGrid(input);
# This code is contributed by pratham76
C#
// C# program to print the largest
// connected component in a grid
using System;
class GFG
{
public const int n = 6;
public const int m = 8;
// stores information about which cell
// are already visited in a particular BFS
public static readonly int[][] visited =
RectangularArrays.ReturnRectangularIntArray(n, m);
// result stores the final result grid
public static readonly int[][] result =
RectangularArrays.ReturnRectangularIntArray(n, m);
// stores the count of cells in the
// largest connected component
public static int COUNT;
// Function checks if a cell is valid i.e
// it is inside the grid and equal to the key
internal static bool is_valid(int x, int y,
int key, int[][] input)
{
if (x < n && y < m &&
x >= 0 && y >= 0)
{
if (visited[x][y] == 0 &&
input[x][y] == key)
{
return true;
}
else
{
return false;
}
}
else
{
return false;
}
}
// BFS to find all cells in
// connection with key = input[i][j]
public static void BFS(int x, int y, int i,
int j, int[][] input)
{
// terminating case for BFS
if (x != y)
{
return;
}
visited[i][j] = 1;
COUNT++;
// x_move and y_move arrays
// are the possible movements
// in x or y direction
int[] x_move = new int[] {0, 0, 1, -1};
int[] y_move = new int[] {1, -1, 0, 0};
// checks all four points
// connected with input[i][j]
for (int u = 0; u < 4; u++)
{
if ((is_valid(i + y_move[u],
j + x_move[u], x, input)) == true)
{
BFS(x, y, i + y_move[u],
j + x_move[u], input);
}
}
}
// called every time before
// a BFS so that visited
// array is reset to zero
internal static void reset_visited()
{
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
visited[i][j] = 0;
}
}
}
// If a larger connected component is
// found this function is called to
// store information about that component.
internal static void reset_result(int key,
int[][] input)
{
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
if (visited[i][j] == 1 &&
input[i][j] == key)
{
result[i][j] = visited[i][j];
}
else
{
result[i][j] = 0;
}
}
}
}
// function to print the result
internal static void print_result(int res)
{
Console.WriteLine("The largest connected " +
"component of the grid is :" + res);
// prints the largest component
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
if (result[i][j] != 0)
{
Console.Write(result[i][j] + " ");
}
else
{
Console.Write(". ");
}
}
Console.WriteLine();
}
}
// function to calculate the
// largest connected component
public static void computeLargestConnectedGrid(int[][] input)
{
int current_max = int.MinValue;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
reset_visited();
COUNT = 0;
// checking cell to the right
if (j + 1 < m)
{
BFS(input[i][j], input[i][j + 1],
i, j, input);
}
// updating result
if (COUNT >= current_max)
{
current_max = COUNT;
reset_result(input[i][j], input);
}
reset_visited();
COUNT = 0;
// checking cell downwards
if (i + 1 < n)
{
BFS(input[i][j], input[i + 1][j],
i, j, input);
}
// updating result
if (COUNT >= current_max)
{
current_max = COUNT;
reset_result(input[i][j], input);
}
}
}
print_result(current_max);
}
public static class RectangularArrays
{
public static int[][] ReturnRectangularIntArray(int size1,
int size2)
{
int[][] newArray = new int[size1][];
for (int array1 = 0; array1 < size1; array1++)
{
newArray[array1] = new int[size2];
}
return newArray;
}
}
// Driver Code
public static void Main(string[] args)
{
int[][] input = new int[][]
{
new int[] {1, 4, 4, 4, 4, 3, 3, 1},
new int[] {2, 1, 1, 4, 3, 3, 1, 1},
new int[] {3, 2, 1, 1, 2, 3, 2, 1},
new int[] {3, 3, 2, 1, 2, 2, 2, 2},
new int[] {3, 1, 3, 1, 1, 4, 4, 4},
new int[] {1, 1, 3, 1, 1, 4, 4, 4}
};
// function to compute the largest
// connected component in the grid
computeLargestConnectedGrid(input);
}
}
// This code is contributed by Shrikant13
Output
The largest connected component of the grid is :9 . . . . . . . . . 1 1 . . . . . . . 1 1 . . . . . . . 1 . . . . . . . 1 1 . . . . . . 1 1 . . .
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