Given a matrix of size M x N consisting of integers, the task is to print the matrix elements using Breadth-First Search traversal.
Examples:
Input: grid[][] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16}}
Output: 1 2 5 3 6 9 4 7 10 13 8 11 14 12 15 16Input: grid[][] = {{-1, 0, 0, 1}, {-1, -1, -2, -1}, {-1, -1, -1, -1}, {0, 0, 0, 0}}
Output: -1 0 -1 0 -1 -1 1 -2 -1 0 -1 -1 0 -1 0 0
Approach: Follow the steps below to solve the problem:
- Initialize the direction vectors dRow[] = {-1, 0, 1, 0} and dCol[] = {0, 1, 0, -1} and a queue of pairs to store the indices of matrix cells.
- Start BFS traversal from the first cell, i.e. (0, 0), and enqueue the index of this cell into the queue.
- Initialize a boolean array to mark the visited cells of the matrix. Mark the cell (0, 0) as visited.
- Declare a function isValid() to check if the cell coordinates are valid or not, i.e lies within the boundaries of the given Matrix and are unvisited or not.
- Iterate while the queue is not empty and perform the following operations:
- Dequeue the cell present at the front of the queue and print it.
- Move to its adjacent cells that are not visited.
- Mark them visited and enqueue them into the queue.
Note: Direction vectors are used to traverse the adjacent cells of a given cell in a given order. For example (x, y) is a cell whose adjacent cells (x – 1, y), (x, y + 1), (x + 1, y), (x, y – 1) need to be traversed, then it can be done using the direction vectors (-1, 0), (0, 1), (1, 0), (0, -1) in the up, left, down and right order.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;#define ROW 4#define COL 4// Direction vectorsint dRow[] = { -1, 0, 1, 0 };int dCol[] = { 0, 1, 0, -1 };// Function to check if a cell// is be visited or notbool isValid(bool vis[][COL], int row, int col){ // If cell lies out of bounds if (row < 0 || col < 0 || row >= ROW || col >= COL) return false; // If cell is already visited if (vis[row][col]) return false; // Otherwise return true;}// Function to perform the BFS traversalvoid BFS(int grid[][COL], bool vis[][COL], int row, int col){ // Stores indices of the matrix cells queue<pair<int, int> > q; // Mark the starting cell as visited // and push it into the queue q.push({ row, col }); vis[row][col] = true; // Iterate while the queue // is not empty while (!q.empty()) { pair<int, int> cell = q.front(); int x = cell.first; int y = cell.second; cout << grid[x][y] << " "; q.pop(); // Go to the adjacent cells for (int i = 0; i < 4; i++) { int adjx = x + dRow[i]; int adjy = y + dCol[i]; if (isValid(vis, adjx, adjy)) { q.push({ adjx, adjy }); vis[adjx][adjy] = true; } } }}// Driver Codeint main(){ // Given input matrix int grid[ROW][COL] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; // Declare the visited array bool vis[ROW][COL]; memset(vis, false, sizeof vis); BFS(grid, vis, 0, 0); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{ static class pair{ int first, second; public pair(int first, int second) { this.first = first; this.second = second; } } static final int ROW = 4;static final int COL = 4;// Direction vectorsstatic int dRow[] = { -1, 0, 1, 0 };static int dCol[] = { 0, 1, 0, -1 };// Function to check if a cell// is be visited or notstatic boolean isValid(boolean vis[][], int row, int col){ // If cell lies out of bounds if (row < 0 || col < 0 || row >= ROW || col >= COL) return false; // If cell is already visited if (vis[row][col]) return false; // Otherwise return true;}// Function to perform the BFS traversalstatic void BFS(int grid[][], boolean vis[][], int row, int col){ // Stores indices of the matrix cells Queue<pair > q = new LinkedList<>(); // Mark the starting cell as visited // and push it into the queue q.add(new pair(row, col)); vis[row][col] = true; // Iterate while the queue // is not empty while (!q.isEmpty()) { pair cell = q.peek(); int x = cell.first; int y = cell.second; System.out.print(grid[x][y] + " "); q.remove(); // Go to the adjacent cells for(int i = 0; i < 4; i++) { int adjx = x + dRow[i]; int adjy = y + dCol[i]; if (isValid(vis, adjx, adjy)) { q.add(new pair(adjx, adjy)); vis[adjx][adjy] = true; } } }}// Driver Codepublic static void main(String[] args){ // Given input matrix int grid[][] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; // Declare the visited array boolean [][]vis = new boolean[ROW][COL]; BFS(grid, vis, 0, 0);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approachfrom collections import deque as queue# Direction vectorsdRow = [ -1, 0, 1, 0]dCol = [ 0, 1, 0, -1]# Function to check if a cell# is be visited or notdef isValid(vis, row, col): # If cell lies out of bounds if (row < 0 or col < 0 or row >= 4 or col >= 4): return False # If cell is already visited if (vis[row][col]): return False # Otherwise return True# Function to perform the BFS traversaldef BFS(grid, vis, row, col): # Stores indices of the matrix cells q = queue() # Mark the starting cell as visited # and push it into the queue q.append(( row, col )) vis[row][col] = True # Iterate while the queue # is not empty while (len(q) > 0): cell = q.popleft() x = cell[0] y = cell[1] print(grid[x][y], end = " ") #q.pop() # Go to the adjacent cells for i in range(4): adjx = x + dRow[i] adjy = y + dCol[i] if (isValid(vis, adjx, adjy)): q.append((adjx, adjy)) vis[adjx][adjy] = True# Driver Codeif __name__ == '__main__': # Given input matrix grid= [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ] ] # Declare the visited array vis = [[ False for i in range(4)] for i in range(4)] # vis, False, sizeof vis) BFS(grid, vis, 0, 0)# This code is contributed by mohit kumar 29. |
C#
// C# program for the above approachusing System;using System.Collections.Generic;public class GFG{ class pair { public int first, second; public pair(int first, int second) { this.first = first; this.second = second; } } static readonly int ROW = 4; static readonly int COL = 4; // Direction vectors static int []dRow = { -1, 0, 1, 0 }; static int []dCol = { 0, 1, 0, -1 }; // Function to check if a cell // is be visited or not static bool isValid(bool [,]vis, int row, int col) { // If cell lies out of bounds if (row < 0 || col < 0 || row >= ROW || col >= COL) return false; // If cell is already visited if (vis[row,col]) return false; // Otherwise return true; } // Function to perform the BFS traversal static void BFS(int [,]grid, bool [,]vis, int row, int col) { // Stores indices of the matrix cells Queue<pair> q = new Queue<pair>(); // Mark the starting cell as visited // and push it into the queue q.Enqueue(new pair(row, col)); vis[row,col] = true; // Iterate while the queue // is not empty while (q.Count!=0) { pair cell = q.Peek(); int x = cell.first; int y = cell.second; Console.Write(grid[x,y] + " "); q.Dequeue(); // Go to the adjacent cells for(int i = 0; i < 4; i++) { int adjx = x + dRow[i]; int adjy = y + dCol[i]; if (isValid(vis, adjx, adjy)) { q.Enqueue(new pair(adjx, adjy)); vis[adjx,adjy] = true; } } } } // Driver Code public static void Main(String[] args) { // Given input matrix int [,]grid = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; // Declare the visited array bool [,]vis = new bool[ROW,COL]; BFS(grid, vis, 0, 0); }} // This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program for the above approachvar ROW = 4;var COL = 4;// Direction vectorsvar dRow = [-1, 0, 1, 0 ];var dCol = [0, 1, 0, -1 ];// Function to check if a cell// is be visited or notfunction isValid(vis, row, col){ // If cell lies out of bounds if (row < 0 || col < 0 || row >= ROW || col >= COL) return false; // If cell is already visited if (vis[row][col]) return false; // Otherwise return true;}// Function to perform the BFS traversalfunction BFS( grid, vis,row, col){ // Stores indices of the matrix cells var q = []; // Mark the starting cell as visited // and push it into the queue q.push([row, col ]); vis[row][col] = true; // Iterate while the queue // is not empty while (q.length!=0) { var cell = q[0]; var x = cell[0]; var y = cell[1]; document.write( grid[x][y] + " "); q.shift(); // Go to the adjacent cells for (var i = 0; i < 4; i++) { var adjx = x + dRow[i]; var adjy = y + dCol[i]; if (isValid(vis, adjx, adjy)) { q.push([adjx, adjy ]); vis[adjx][adjy] = true; } } }}// Driver Code// Given input matrixvar grid = [[1, 2, 3, 4 ], [5, 6, 7, 8 ], [9, 10, 11, 12 ], [13, 14, 15, 16 ] ];// Declare the visited arrayvar vis = Array.from(Array(ROW), ()=> Array(COL).fill(false));BFS(grid, vis, 0, 0);</script> |
Output:
1 2 5 3 6 9 4 7 10 13 8 11 14 12 15 16
Time Complexity: O(N * M)
Auxiliary Space: O(N * M)
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