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Flood fill Algorithm – how to implement fill() in paint?

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  • Difficulty Level : Medium
  • Last Updated : 24 Jun, 2022
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In MS-Paint, when we take the brush to a pixel and click, the color of the region of that pixel is replaced with a new selected color. Following is the problem statement to do this task. 

Given a 2D screen, location of a pixel in the screen and a color, replace color of the given pixel and all adjacent same colored pixels with the given color.

Example: 

Input:
screen[M][N] = {{1, 1, 1, 1, 1, 1, 1, 1},
               {1, 1, 1, 1, 1, 1, 0, 0},
               {1, 0, 0, 1, 1, 0, 1, 1},
               {1, 2, 2, 2, 2, 0, 1, 0},
               {1, 1, 1, 2, 2, 0, 1, 0},
               {1, 1, 1, 2, 2, 2, 2, 0},
               {1, 1, 1, 1, 1, 2, 1, 1},
               {1, 1, 1, 1, 1, 2, 2, 1},
               };
    x = 4, y = 4, newColor = 3
The values in the given 2D screen
  indicate colors of the pixels.
x and y are coordinates of the brush,
   newColor is the color that
should replace the previous color on 
   screen[x][y] and all surrounding
pixels with same color.

Output:
Screen should be changed to following.
screen[M][N] = {{1, 1, 1, 1, 1, 1, 1, 1},
               {1, 1, 1, 1, 1, 1, 0, 0},
               {1, 0, 0, 1, 1, 0, 1, 1},
               {1, 3, 3, 3, 3, 0, 1, 0},
               {1, 1, 1, 3, 3, 0, 1, 0},
               {1, 1, 1, 3, 3, 3, 3, 0},
               {1, 1, 1, 1, 1, 3, 1, 1},
               {1, 1, 1, 1, 1, 3, 3, 1},
               };
Recommended Practice

This question can be solved using either Recursion or BFS. Both the solutions are discussed below 

Method 1 (Using Recursion): The idea is simple, we first replace the color of the current pixel, then recur for 4 surrounding points. The following is a detailed algorithm. 

// A recursive function to replace 
// previous color 'prevC' at  '(x, y)' 
// and all surrounding pixels of (x, y)
// with new color 'newC' and
floodFill(screen[M][N], x, y, prevC, newC)
1) If x or y is outside the screen, then return.
2) If color of screen[x][y] is not same as prevC, then return
3) Recur for north, south, east and west.
    floodFillUtil(screen, x+1, y, prevC, newC);
    floodFillUtil(screen, x-1, y, prevC, newC);
    floodFillUtil(screen, x, y+1, prevC, newC);
    floodFillUtil(screen, x, y-1, prevC, newC); 

The following is the implementation of the above algorithm.  

C++




// A C++ program to implement flood fill algorithm
#include<iostream>
using namespace std;
 
// Dimensions of paint screen
#define M 8
#define N 8
 
// A recursive function to replace previous color 'prevC' at  '(x, y)'
// and all surrounding pixels of (x, y) with new color 'newC' and
void floodFillUtil(int screen[][N], int x, int y, int prevC, int newC)
{
    // Base cases
    if (x < 0 || x >= M || y < 0 || y >= N)
        return;
    if (screen[x][y] != prevC)
        return;
    if (screen[x][y] == newC)
        return;
 
    // Replace the color at (x, y)
    screen[x][y] = newC;
 
    // Recur for north, east, south and west
    floodFillUtil(screen, x+1, y, prevC, newC);
    floodFillUtil(screen, x-1, y, prevC, newC);
    floodFillUtil(screen, x, y+1, prevC, newC);
    floodFillUtil(screen, x, y-1, prevC, newC);
}
 
// It mainly finds the previous color on (x, y) and
// calls floodFillUtil()
void floodFill(int screen[][N], int x, int y, int newC)
{
    int prevC = screen[x][y];
      if(prevC==newC) return;
    floodFillUtil(screen, x, y, prevC, newC);
}
 
// Driver code
int main()
{
    int screen[M][N] = {{1, 1, 1, 1, 1, 1, 1, 1},
                      {1, 1, 1, 1, 1, 1, 0, 0},
                      {1, 0, 0, 1, 1, 0, 1, 1},
                      {1, 2, 2, 2, 2, 0, 1, 0},
                      {1, 1, 1, 2, 2, 0, 1, 0},
                      {1, 1, 1, 2, 2, 2, 2, 0},
                      {1, 1, 1, 1, 1, 2, 1, 1},
                      {1, 1, 1, 1, 1, 2, 2, 1},
                     };
    int x = 4, y = 4, newC = 3;
    floodFill(screen, x, y, newC);
 
    cout << "Updated screen after call to floodFill: \n";
    for (int i=0; i<M; i++)
    {
        for (int j=0; j<N; j++)
           cout << screen[i][j] << " ";
        cout << endl;
    }
}
//Updated by Arun Pandey


Java




// Java program to implement flood fill algorithm
class GFG
{
 
// Dimensions of paint screen
static int M = 8;
static int N = 8;
 
// A recursive function to replace previous color 'prevC' at '(x, y)'
// and all surrounding pixels of (x, y) with new color 'newC' and
static void floodFillUtil(int screen[][], int x, int y,
                                    int prevC, int newC)
{
    // Base cases
    if (x < 0 || x >= M || y < 0 || y >= N)
        return;
    if (screen[x][y] != prevC)
        return;
 
    // Replace the color at (x, y)
    screen[x][y] = newC;
 
    // Recur for north, east, south and west
    floodFillUtil(screen, x+1, y, prevC, newC);
    floodFillUtil(screen, x-1, y, prevC, newC);
    floodFillUtil(screen, x, y+1, prevC, newC);
    floodFillUtil(screen, x, y-1, prevC, newC);
}
 
// It mainly finds the previous color on (x, y) and
// calls floodFillUtil()
static void floodFill(int screen[][], int x, int y, int newC)
{
    int prevC = screen[x][y];
      if(prevC==newC) return;
    floodFillUtil(screen, x, y, prevC, newC);
}
 
// Driver code
public static void main(String[] args)
{
    int screen[][] = {{1, 1, 1, 1, 1, 1, 1, 1},
                    {1, 1, 1, 1, 1, 1, 0, 0},
                    {1, 0, 0, 1, 1, 0, 1, 1},
                    {1, 2, 2, 2, 2, 0, 1, 0},
                    {1, 1, 1, 2, 2, 0, 1, 0},
                    {1, 1, 1, 2, 2, 2, 2, 0},
                    {1, 1, 1, 1, 1, 2, 1, 1},
                    {1, 1, 1, 1, 1, 2, 2, 1},
                    };
    int x = 4, y = 4, newC = 3;
    floodFill(screen, x, y, newC);
 
    System.out.println("Updated screen after call to floodFill: ");
    for (int i = 0; i < M; i++)
    {
        for (int j = 0; j < N; j++)
        System.out.print(screen[i][j] + " ");
        System.out.println();
    }
    }
}
 
// This code has been contributed by 29AjayKumar
// Updated by Arun Pandey


Python3




# Python3 program to implement
# flood fill algorithm
 
# Dimensions of paint screen
M = 8
N = 8
 
# A recursive function to replace
# previous color 'prevC' at '(x, y)'
# and all surrounding pixels of (x, y)
# with new color 'newC' and
def floodFillUtil(screen, x, y, prevC, newC):
     
    # Base cases
    if (x < 0 or x >= M or y < 0 or
        y >= N or screen[x][y] != prevC or
        screen[x][y] == newC):
        return
 
    # Replace the color at (x, y)
    screen[x][y] = newC
 
    # Recur for north, east, south and west
    floodFillUtil(screen, x + 1, y, prevC, newC)
    floodFillUtil(screen, x - 1, y, prevC, newC)
    floodFillUtil(screen, x, y + 1, prevC, newC)
    floodFillUtil(screen, x, y - 1, prevC, newC)
 
# It mainly finds the previous color on (x, y) and
# calls floodFillUtil()
def floodFill(screen, x, y, newC):
    prevC = screen[x][y]
    if(prevC==newC):
      return
    floodFillUtil(screen, x, y, prevC, newC)
 
# Driver Code
screen = [[1, 1, 1, 1, 1, 1, 1, 1],
          [1, 1, 1, 1, 1, 1, 0, 0],
          [1, 0, 0, 1, 1, 0, 1, 1],
          [1, 2, 2, 2, 2, 0, 1, 0],
          [1, 1, 1, 2, 2, 0, 1, 0],
          [1, 1, 1, 2, 2, 2, 2, 0],
          [1, 1, 1, 1, 1, 2, 1, 1],
          [1, 1, 1, 1, 1, 2, 2, 1]]
 
x = 4
y = 4
newC = 3
floodFill(screen, x, y, newC)
 
print ("Updated screen after call to floodFill:")
for i in range(M):
    for j in range(N):
        print(screen[i][j], end = ' ')
    print()
 
# This code is contributed by Ashutosh450
# Updated by Arun Pandey


C#




// C# program to implement
// flood fill algorithm
using System;
     
class GFG
{
 
// Dimensions of paint screen
static int M = 8;
static int N = 8;
 
// A recursive function to replace
// previous color 'prevC' at '(x, y)'
// and all surrounding pixels of (x, y)
// with new color 'newC' and
static void floodFillUtil(int [,]screen,
                          int x, int y,
                          int prevC, int newC)
{
    // Base cases
    if (x < 0 || x >= M ||
        y < 0 || y >= N)
        return;
    if (screen[x, y] != prevC)
        return;
 
    // Replace the color at (x, y)
    screen[x, y] = newC;
 
    // Recur for north, east, south and west
    floodFillUtil(screen, x + 1, y, prevC, newC);
    floodFillUtil(screen, x - 1, y, prevC, newC);
    floodFillUtil(screen, x, y + 1, prevC, newC);
    floodFillUtil(screen, x, y - 1, prevC, newC);
}
 
// It mainly finds the previous color
// on (x, y) and calls floodFillUtil()
static void floodFill(int [,]screen, int x,
                      int y, int newC)
{
    int prevC = screen[x, y];
      if(prevC == newC)
      return;
    floodFillUtil(screen, x, y, prevC, newC);
}
 
// Driver code
public static void Main(String[] args)
{
    int [,]screen = {{1, 1, 1, 1, 1, 1, 1, 1},
                     {1, 1, 1, 1, 1, 1, 0, 0},
                     {1, 0, 0, 1, 1, 0, 1, 1},
                     {1, 2, 2, 2, 2, 0, 1, 0},
                     {1, 1, 1, 2, 2, 0, 1, 0},
                     {1, 1, 1, 2, 2, 2, 2, 0},
                     {1, 1, 1, 1, 1, 2, 1, 1},
                     {1, 1, 1, 1, 1, 2, 2, 1}};
    int x = 4, y = 4, newC = 3;
    floodFill(screen, x, y, newC);
 
    Console.WriteLine("Updated screen after" +
                      "call to floodFill: ");
    for (int i = 0; i < M; i++)
    {
        for (int j = 0; j < N; j++)
        Console.Write(screen[i, j] + " ");
        Console.WriteLine();
    }
    }
}
 
// This code is contributed by PrinciRaj1992
// Updated by Arun Pandey


Javascript




<script>
      // JavaScript program to implement
      // flood fill algorithm
      // Dimensions of paint screen
      var M = 8;
      var N = 8;
 
      // A recursive function to replace
      // previous color 'prevC' at '(x, y)'
      // and all surrounding pixels of (x, y)
      // with new color 'newC' and
      function floodFillUtil(screen, x, y, prevC, newC)
      {
       
        // Base cases
        if (x < 0 || x >= M || y < 0 || y >= N) return;
        if (screen[x][y] != prevC) return;
 
        // Replace the color at (x, y)
        screen[x][y] = newC;
 
        // Recur for north, east, south and west
        floodFillUtil(screen, x + 1, y, prevC, newC);
        floodFillUtil(screen, x - 1, y, prevC, newC);
        floodFillUtil(screen, x, y + 1, prevC, newC);
        floodFillUtil(screen, x, y - 1, prevC, newC);
      }
 
      // It mainly finds the previous color
      // on (x, y) and calls floodFillUtil()
      function floodFill(screen, x, y, newC) {
        var prevC = screen[x][y];
        if (prevC == newC) return;
        floodFillUtil(screen, x, y, prevC, newC);
      }
 
      // Driver code
      var screen = [
        [1, 1, 1, 1, 1, 1, 1, 1],
        [1, 1, 1, 1, 1, 1, 0, 0],
        [1, 0, 0, 1, 1, 0, 1, 1],
        [1, 2, 2, 2, 2, 0, 1, 0],
        [1, 1, 1, 2, 2, 0, 1, 0],
        [1, 1, 1, 2, 2, 2, 2, 0],
        [1, 1, 1, 1, 1, 2, 1, 1],
        [1, 1, 1, 1, 1, 2, 2, 1],
      ];
      var x = 4,
        y = 4,
        newC = 3;
      floodFill(screen, x, y, newC);
 
      document.write("Updated screen after" + "call to floodFill: <br>");
      for (var i = 0; i < M; i++) {
        for (var j = 0; j < N; j++) document.write(screen[i][j] + " ");
        document.write("<br>");
      }
       
      // This code is contributed by rdtank.
    </script>


Output

Updated screen after call to floodFill: 
1 1 1 1 1 1 1 1 
1 1 1 1 1 1 0 0 
1 0 0 1 1 0 1 1 
1 3 3 3 3 0 1 0 
1 1 1 3 3 0 1 0 
1 1 1 3 3 3 3 0 
1 1 1 1 1 3 1 1 
1 1 1 1 1 3 3 1 

Time complexity: O(M x N).
Auxiliary Space: O(M x N), as implicit stack is created due to recursion

Method 2 (Using the BFS approach):

Algorithm for BFS based approach :

  1. Create a queue of pairs.
  2. Insert an initial index given in the queue.
  3. Mark initial index as visited in vis[][] matrix.
  4. Until the queue is not empty repeat step 4.1 to 4.6
    • Take the front element from the queue
    • Pop from the queue
    • Store current value/color at coordinate taken out from queue (precolor)
    • Update the value/color of the current index which is taken out from the queue
    • Check for all 4 direction i.e (x+1,y),(x-1,y),(x,y+1),(x,y-1) is valid or not and if valid then check that value at that coordinate should be equal to precolor and value of that coordinate in vis[][] is 0.
    • If all the above condition is true push the corresponding coordinate in queue and mark as 1 in vis[][]
  5. Print the matrix.

Below is the implementation of the above approach:  

C++




// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check valid coordinate
int validCoord(int x, int y, int n, int m)
{
 
    if (x < 0 || y < 0) {
        return 0;
    }
    if (x >= n || y >= m) {
        return 0;
    }
    return 1;
}
 
// Function to run bfs
void bfs(int n, int m, int data[][8],
                     int x, int y, int color)
{
   
  // Visiting array
  int vis[101][101];
   
  // Initializing all as zero
  memset(vis, 0, sizeof(vis));
   
  // Creating queue for bfs
  queue<pair<int, int> > obj;
   
  // Pushing pair of {x, y}
  obj.push({ x, y });
   
  // Marking {x, y} as visited
  vis[x][y] = 1;
   
  // Until queue is empty
  while (obj.empty() != 1)
  {
 
    // Extracting front pair
    pair<int, int> coord = obj.front();
    int x = coord.first;
    int y = coord.second;
    int preColor = data[x][y];
 
    data[x][y] = color;
     
    // Popping front pair of queue
    obj.pop();
 
    // For Upside Pixel or Cell
    if (validCoord(x + 1, y, n, m)
        && vis[x + 1][y] == 0
        && data[x + 1][y] == preColor)
    {
      obj.push({ x + 1, y });
      vis[x + 1][y] = 1;
    }
     
    // For Downside Pixel or Cell
    if (validCoord(x - 1, y, n, m)
        && vis[x - 1][y] == 0
        && data[x - 1][y] == preColor)
    {
      obj.push({ x - 1, y });
      vis[x - 1][y] = 1;
    }
     
    // For Right side Pixel or Cell
    if (validCoord(x, y + 1, n, m)
        && vis[x][y + 1] == 0
        && data[x][y + 1] == preColor)
    {
      obj.push({ x, y + 1 });
      vis[x][y + 1] = 1;
    }
     
    // For Left side Pixel or Cell
    if (validCoord(x, y - 1, n, m)
        && vis[x][y - 1] == 0
        && data[x][y - 1] == preColor)
    {
      obj.push({ x, y - 1 });
      vis[x][y - 1] = 1;
    }
  }
   
  // Printing The Changed Matrix Of Pixels
  for (int i = 0; i < n; i++)
  {
    for (int j = 0; j < m; j++)
    {
      cout << data[i][j] << " ";
    }
    cout << endl;
  }
  cout << endl;
}
 
// Driver Code
int main()
{
  int n, m, x, y, color;
  n = 8;
  m = 8;
 
  int data[8][8] = {
    { 1, 1, 1, 1, 1, 1, 1, 1 },
    { 1, 1, 1, 1, 1, 1, 0, 0 },
    { 1, 0, 0, 1, 1, 0, 1, 1 },
    { 1, 2, 2, 2, 2, 0, 1, 0 },
    { 1, 1, 1, 2, 2, 0, 1, 0 },
    { 1, 1, 1, 2, 2, 2, 2, 0 },
    { 1, 1, 1, 1, 1, 2, 1, 1 },
    { 1, 1, 1, 1, 1, 2, 2, 1 },
  };
 
  x = 4, y = 4, color = 3;
   
  // Function Call
  bfs(n, m, data, x, y, color);
  return 0;
}


Java




// Java program for above approachimport java.io.*;
import java.util.*;
 
// Class to store the pairs
class Pair implements Comparable<Pair> {
    int first;
    int second;
 
    public Pair(int first, int second) {
        this.first = first;
        this.second = second;
    }
 
    @Override
    public int compareTo(Pair o) {
        return second - o.second;
    }
}
 
class GFG {
    public static int validCoord(int x, int y, int n, int m)
    {
        if (x < 0 || y < 0) {
            return 0;
        }
        if (x >= n || y >= m) {
            return 0;
        }
        return 1;
    }
    // Function to run bfs
    public static void bfs(int n, int m, int data[][],int x, int y, int color)
    {
   
        // Visiting array
         int vis[][]=new int[101][101];
         
        // Initializing all as zero
        for(int i=0;i<=100;i++){
            for(int j=0;j<=100;j++){
                vis[i][j]=0;
            }
        }
         
        // Creating queue for bfs
         Queue<Pair> obj = new LinkedList<>();
   
        // Pushing pair of {x, y}
         Pair pq=new Pair(x,y);
         obj.add(pq);
          
        // Marking {x, y} as visited
         vis[x][y] = 1;
   
        // Until queue is empty
        while (!obj.isEmpty())
        {
            // Extracting front pair
            Pair coord = obj.peek();
            int x1 = coord.first;
            int y1 = coord.second;
            int preColor = data[x1][y1];
             
            data[x1][y1] = color;
     
            // Popping front pair of queue
            obj.remove();
 
            // For Upside Pixel or Cell
            if ((validCoord(x1 + 1, y1, n, m)==1) && vis[x1 + 1][y1] == 0 && data[x1 + 1][y1] == preColor)
            {
                Pair p=new Pair(x1 +1, y1);
                obj.add(p);
                vis[x1 + 1][y1] = 1;
            }
     
            // For Downside Pixel or Cell
            if ((validCoord(x1 - 1, y1, n, m)==1) && vis[x1 - 1][y1] == 0 && data[x1 - 1][y1] == preColor)
            {
                Pair p=new Pair(x1-1,y1);
                obj.add(p);
                vis[x1- 1][y1] = 1;
            }
     
            // For Right side Pixel or Cell
            if ((validCoord(x1, y1 + 1, n, m)==1) && vis[x1][y1 + 1] == 0 && data[x1][y1 + 1] == preColor)
            {
                Pair p=new Pair(x1,y1 +1);
                obj.add(p);
                vis[x1][y1 + 1] = 1;
            }
     
            // For Left side Pixel or Cell
            if ((validCoord(x1, y1 - 1, n, m)==1) && vis[x1][y1 - 1] == 0 && data[x1][y1 - 1] == preColor)
            {
                Pair p=new Pair(x1,y1 -1);
                obj.add(p);
                vis[x1][y1 - 1] = 1;
            }
        }
   
         // Printing The Changed Matrix Of Pixels
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {  
             System.out.print(data[i][j]+" ");
            }
            System.out.println();
        }
        System.out.println();
    }
    public static void main (String[] args) {
        int nn, mm, xx, yy, colorr;
        nn = 8;
        mm = 8;
 
        int data[][] = {{ 1, 1, 1, 1, 1, 1, 1, 1 },
                          { 1, 1, 1, 1, 1, 1, 0, 0 },
                          { 1, 0, 0, 1, 1, 0, 1, 1 },
                          { 1, 2, 2, 2, 2, 0, 1, 0 },
                          { 1, 1, 1, 2, 2, 0, 1, 0 },
                          { 1, 1, 1, 2, 2, 2, 2, 0 },
                          { 1, 1, 1, 1, 1, 2, 1, 1 },
                          { 1, 1, 1, 1, 1, 2, 2, 1 },};
 
        xx = 4; yy = 4; colorr = 3;
   
        // Function Call
        bfs(nn, mm, data, xx, yy, colorr);
    }
}
// This code is contributed by Manu Pathria


Python3




# Python3 program for above approach
 
# Function to check valid coordinate
def validCoord(x, y, n, m):
    if x < 0 or y < 0:
        return 0
    if x >= n or y >= m:
        return 0
    return 1
 
# Function to run bfs
def bfs(n, m, data, X, Y, color):
   
  # Visiting array
  vis = [[0 for i in range(101)] for j in range(101)]
     
  # Creating queue for bfs
  obj = []
     
  # Pushing pair of {x, y}
  obj.append([X, Y])
     
  # Marking {x, y} as visited
  vis[X][Y] = 1
     
  # Until queue is empty
  while len(obj) > 0:
     
    # Extracting front pair
    coord = obj[0]
    x = coord[0]
    y = coord[1]
    preColor = data[x][y]
   
    data[x][y] = color
       
    # Popping front pair of queue
    obj.pop(0)
   
    # For Upside Pixel or Cell
    if validCoord(x + 1, y, n, m) == 1 and vis[x + 1][y] == 0 and data[x + 1][y] == preColor:
      obj.append([x + 1, y])
      vis[x + 1][y] = 1
       
    # For Downside Pixel or Cell
    if validCoord(x - 1, y, n, m) == 1 and vis[x - 1][y] == 0 and data[x - 1][y] == preColor:
      obj.append([x - 1, y])
      vis[x - 1][y] = 1
       
    # For Right side Pixel or Cell
    if validCoord(x, y + 1, n, m) == 1 and vis[x][y + 1] == 0 and data[x][y + 1] == preColor:
      obj.append([x, y + 1])
      vis[x][y + 1] = 1
       
    # For Left side Pixel or Cell
    if validCoord(x, y - 1, n, m) == 1 and vis[x][y - 1] == 0 and data[x][y - 1] == preColor:
      obj.append([x, y - 1])
      vis[x][y - 1] = 1
     
  # Printing The Changed Matrix Of Pixels
  for i in range(n):
    for j in range(m):
      print(data[i][j], end = " ")
    print()
  print()
 
n = 8
m = 8
 
data = [
  [ 1, 1, 1, 1, 1, 1, 1, 1 ],
  [ 1, 1, 1, 1, 1, 1, 0, 0 ],
  [ 1, 0, 0, 1, 1, 0, 1, 1 ],
  [ 1, 2, 2, 2, 2, 0, 1, 0 ],
  [ 1, 1, 1, 2, 2, 0, 1, 0 ],
  [ 1, 1, 1, 2, 2, 2, 2, 0 ],
  [ 1, 1, 1, 1, 1, 2, 1, 1 ],
  [ 1, 1, 1, 1, 1, 2, 2, 1 ],
]
 
x, y, color = 4, 4, 3
 
# Function Call
bfs(n, m, data, x, y, color)
 
# This code is contributed by decode2207.


C#




// C# program for above approach
using System;
using System.Collections.Generic;
class GFG {
     
    // Function to check valid coordinate
    static int validCoord(int x, int y, int n, int m)
    {
      
        if (x < 0 || y < 0) {
            return 0;
        }
        if (x >= n || y >= m) {
            return 0;
        }
        return 1;
    }
      
    // Function to run bfs
    static void bfs(int n, int m, int[,] data, int X, int Y, int color)
    {
        
      // Visiting array
      int[,] vis = new int[101,101];
        
      // Creating queue for bfs
      List<Tuple<int, int>> obj = new List<Tuple<int, int>>();
        
      // Pushing pair of {x, y}
      obj.Add(new Tuple<int,int>( X, Y ));
        
      // Marking {x, y} as visited
      vis[X,Y] = 1;
        
      // Until queue is empty
      while (obj.Count > 0)
      {
      
        // Extracting front pair
        Tuple<int, int> coord = obj[0];
        int x = coord.Item1;
        int y = coord.Item2;
        int preColor = data[x,y];
      
        data[x,y] = color;
          
        // Popping front pair of queue
        obj.RemoveAt(0);
      
        // For Upside Pixel or Cell
        if (validCoord(x + 1, y, n, m) == 1
            && vis[x + 1,y] == 0
            && data[x + 1,y] == preColor)
        {
          obj.Add(new Tuple<int,int>(x + 1, y));
          vis[x + 1,y] = 1;
        }
          
        // For Downside Pixel or Cell
        if (validCoord(x - 1, y, n, m) == 1
            && vis[x - 1,y] == 0
            && data[x - 1,y] == preColor)
        {
          obj.Add(new Tuple<int,int>(x - 1, y));
          vis[x - 1,y] = 1;
        }
          
        // For Right side Pixel or Cell
        if (validCoord(x, y + 1, n, m) == 1
            && vis[x,y + 1] == 0
            && data[x,y + 1] == preColor)
        {
          obj.Add(new Tuple<int,int>(x, y + 1));
          vis[x,y + 1] = 1;
        }
          
        // For Left side Pixel or Cell
        if (validCoord(x, y - 1, n, m) == 1
            && vis[x,y - 1] == 0
            && data[x,y - 1] == preColor)
        {
          obj.Add(new Tuple<int,int>(x, y - 1));
          vis[x,y - 1] = 1;
        }
      }
        
      // Printing The Changed Matrix Of Pixels
      for (int i = 0; i < n; i++)
      {
        for (int j = 0; j < m; j++)
        {
          Console.Write(data[i,j] + " ");
        }
        Console.WriteLine();
      }
      Console.WriteLine();
    }
 
  static void Main() {
      int n, m, x, y, color;
      n = 8;
      m = 8;
      
      int[,] data = {
        { 1, 1, 1, 1, 1, 1, 1, 1 },
        { 1, 1, 1, 1, 1, 1, 0, 0 },
        { 1, 0, 0, 1, 1, 0, 1, 1 },
        { 1, 2, 2, 2, 2, 0, 1, 0 },
        { 1, 1, 1, 2, 2, 0, 1, 0 },
        { 1, 1, 1, 2, 2, 2, 2, 0 },
        { 1, 1, 1, 1, 1, 2, 1, 1 },
        { 1, 1, 1, 1, 1, 2, 2, 1 },
      };
      
      x = 4;
      y = 4;
      color = 3;
        
      // Function Call
      bfs(n, m, data, x, y, color);
  }
}
 
// This code is contributed by divyeshrabadiya07.


Javascript




<script>
    // Javascript program for above approach
     
    // Function to check valid coordinate
    function validCoord(x, y, n, m)
    {
        if (x < 0 || y < 0) {
            return 0;
        }
        if (x >= n || y >= m) {
            return 0;
        }
        return 1;
    }
     
    // Function to run bfs
    function bfs(n, m, data, X, Y, color)
    {
         
      // Visiting array
      let vis = new Array(101);
      for(let i = 0; i < 101; i++)
      {
          vis[i] = new Array(101);
        for(let j = 0; j < 101; j++)
        {
            vis[i][j] = 0;
        }
      }
         
      // Creating queue for bfs
      let obj = [];
         
      // Pushing pair of {x, y}
      obj.push([X, Y]);
         
      // Marking {x, y} as visited
      vis[X][Y] = 1;
         
      // Until queue is empty
      while (obj.length > 0)
      {
       
        // Extracting front pair
        let coord = obj[0];
        let x = coord[0];
        let y = coord[1];
        let preColor = data[x][y];
       
        data[x][y] = color;
           
        // Popping front pair of queue
        obj.shift();
       
        // For Upside Pixel or Cell
        if (validCoord(x + 1, y, n, m) == 1
            && vis[x + 1][y] == 0
            && data[x + 1][y] == preColor)
        {
          obj.push([x + 1, y]);
          vis[x + 1][y] = 1;
        }
           
        // For Downside Pixel or Cell
        if (validCoord(x - 1, y, n, m) == 1
            && vis[x - 1][y] == 0
            && data[x - 1][y] == preColor)
        {
          obj.push([x - 1, y]);
          vis[x - 1][y] = 1;
        }
           
        // For Right side Pixel or Cell
        if (validCoord(x, y + 1, n, m) == 1
            && vis[x][y + 1] == 0
            && data[x][y + 1] == preColor)
        {
          obj.push([x, y + 1]);
          vis[x][y + 1] = 1;
        }
           
        // For Left side Pixel or Cell
        if (validCoord(x, y - 1, n, m) == 1
            && vis[x][y - 1] == 0
            && data[x][y - 1] == preColor)
        {
          obj.push([x, y - 1]);
          vis[x][y - 1] = 1;
        }
      }
         
      // Printing The Changed Matrix Of Pixels
      for (let i = 0; i < n; i++)
      {
        for (let j = 0; j < m; j++)
        {
          document.write(data[i][j] + " ");
        }
        document.write("</br>");
      }
      document.write("</br>");
    }
     
    let n, m, x, y, color;
    n = 8;
    m = 8;
 
    let data = [
      [ 1, 1, 1, 1, 1, 1, 1, 1 ],
      [ 1, 1, 1, 1, 1, 1, 0, 0 ],
      [ 1, 0, 0, 1, 1, 0, 1, 1 ],
      [ 1, 2, 2, 2, 2, 0, 1, 0 ],
      [ 1, 1, 1, 2, 2, 0, 1, 0 ],
      [ 1, 1, 1, 2, 2, 2, 2, 0 ],
      [ 1, 1, 1, 1, 1, 2, 1, 1 ],
      [ 1, 1, 1, 1, 1, 2, 2, 1 ],
    ];
 
    x = 4, y = 4, color = 3;
 
    // Function Call
    bfs(n, m, data, x, y, color);
 
// This code is contributed by mukesh07.
</script>


Output

1 1 1 1 1 1 1 1 
1 1 1 1 1 1 0 0 
1 0 0 1 1 0 1 1 
1 3 3 3 3 0 1 0 
1 1 1 3 3 0 1 0 
1 1 1 3 3 3 3 0 
1 1 1 1 1 3 1 1 
1 1 1 1 1 3 3 1 

Time complexity: O(M x N).
Auxiliary Space: O(M x N),


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