Flood fill Algorithm – how to implement fill() in paint?
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.
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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},
};This question can be solved using either Recursion or BFS. Both the solutions are discussed below
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
floodFil(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' andvoid 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 codeint 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 algorithmclass GFG{// Dimensions of paint screenstatic 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' andstatic 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 codepublic 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 screenM = 8N = 8# A recursive function to replace# previous color 'prevC' at '(x, y)'# and all surrounding pixels of (x, y)# with new color 'newC' anddef 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 Codescreen = [[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 = 4y = 4newC = 3floodFill(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 algorithmusing System; class GFG{// Dimensions of paint screenstatic 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' andstatic 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 codepublic 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
Method 2: Using the BFS approach
Algorithm for BFS based approach :
- Create a queue of pairs.
- Insert an initial index given in the queue.
- Mark initial index as visited in vis[][] matrix.
- 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[][]
- 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 coordinateint 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 bfsvoid bfs(int n, int m, int data[][8], int x, int y, int color){ // Visiting array int vis[101][101]; // Initialing 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 Codeint 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 pairsclass 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]; // Initialing 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()) { // Extrating 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 |
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
References:
http://en.wikipedia.org/wiki/Flood_fill
This article is contributed by Anmol. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
