Print matrix elements using DFS traversal
Given a matrix grid[][] with dimension M × N of integers, the task is to print the matrix elements using DFS traversal.
Examples:
Input: mat[][] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16}}
Output: 1 2 3 4 8 12 16 15 11 7 6 10 14 13 9 5
Explanation: The matrix elements in the order of their Depth First Search traversal are 1 2 3 4 8 12 16 15 11 7 6 10 14 13 9 5.Input: mat[][] = {{0, 1, 9, 4}, {1, 2, 3, 4}, {0, 0, -1, -1}, {-1, -1, 0, 1}}
Output: 0 1 9 4 4 -1 1 0 -1 3 2 0 -1 -1 0 1
Recursive Approach: The idea is to use a Recursive depth-first search for traversing the matrix and printing its elements. Follow the steps below to solve the problem:
- Initialize a 2D boolean vector, say vis[][], to keep track of already visited and unvisited indices.
- Define a function, say isValid(i, j), to check if the position (i, j) is valid or not i.e (i, j) should be inside the matrix and not visited.
- Define a recursive function DFS(i, j):
- On each call, mark the current position (i, j) visited and print the element at that position.
- Make the recursive call for all the adjacent sides, i.e DFS(i + 1, j), DFS(i, j + 1), DFS(i – 1, j) and DFS(i, j – 1) if respective positions are valid i.e., not visited and are within the matrix.
- Finally, call the function DFS(0, 0) to start the DFS Traversal to print the matrix.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Direction vectorsint dRow[] = { -1, 0, 1, 0 };int dCol[] = { 0, 1, 0, -1 };// Function to check if current// position is valid or notbool isValid(vector<vector<bool> >& vis, int row, int col, int COL, int ROW){ // Check if the cell is out of bounds if (row < 0 || col < 0 || col > COL - 1 || row > ROW - 1) return false; // Check if the cell is visited or not if (vis[row][col] == true) return false; return true;}// Utility function to print matrix// elements using DFS Traversalvoid DFSUtil(int row, int col, vector<vector<int> > grid, vector<vector<bool> >& vis, int M, int N){ // Mark the current cell visited vis[row][col] = true; // Print the element at the cell cout << grid[row][col] << " "; // Traverse all four adjacent // cells of the current element for (int i = 0; i < 4; i++) { int x = row + dRow[i]; int y = col + dCol[i]; // Check if x and y is // valid index or not if (isValid(vis, x, y, M, N)) DFSUtil(x, y, grid, vis, M, N); }}// Function to print the matrix elementsvoid DFS(int row, int col, vector<vector<int> > grid, int M, int N){ // Initialize a visiting matrix vector<vector<bool> > vis( M + 1, vector<bool>(N + 1, false)); // Function call to print matrix // elements by DFS traversal DFSUtil(0, 0, grid, vis, M, N);}// Driver Codeint main(){ // Given matrix vector<vector<int> > grid{ { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; // Row of the matrix int M = grid.size(); // Column of the matrix int N = grid[0].size(); DFS(0, 0, grid, M, N); return 0;} |
Java
// Java program to implement // the above approach import java.util.*;class GFG{// Direction vectorsstatic int dRow[] = { -1, 0, 1, 0 };static int dCol[] = { 0, 1, 0, -1 };// Function to check if current// position is valid or notstatic boolean isValid(boolean[][] vis, int row, int col, int COL, int ROW){ // Check if the cell is out of bounds if (row < 0 || col < 0 || col > COL - 1 || row > ROW - 1) return false; // Check if the cell is visited or not if (vis[row][col] == true) return false; return true;}// Utility function to print matrix// elements using DFS Traversalstatic void DFSUtil(int row, int col, int[][] grid, boolean[][] vis, int M, int N){ // Mark the current cell visited vis[row][col] = true; // Print the element at the cell System.out.print(grid[row][col] + " "); // Traverse all four adjacent // cells of the current element for (int i = 0; i < 4; i++) { int x = row + dRow[i]; int y = col + dCol[i]; // Check if x and y is // valid index or not if (isValid(vis, x, y, M, N)) DFSUtil(x, y, grid, vis, M, N); }}// Function to print the matrix elementsstatic void DFS(int row, int col, int[][] grid, int M, int N){ // Initialize a visiting matrix boolean[][] vis = new boolean[M + 1][N + 1]; for(int i = 0; i < M + 1; i++) { for(int j = 0; j < N + 1; j++) { vis[i][j] = false; } } // Function call to print matrix // elements by DFS traversal DFSUtil(0, 0, grid, vis, M, N);}// Driver Codepublic static void main(String args[]){ // Given matrix int[][] grid = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; // Row of the matrix int M = grid.length; // Column of the matrix int N = grid[0].length; DFS(0, 0, grid, M, N);}}// This code is contributed by susmitakundugoaldanga. |
Python3
# Python3 program for the above approach# Direction vectorsdRow = [-1, 0, 1, 0]dCol = [0, 1, 0, -1]# Function to check if current# position is valid or notdef isValid(row, col, COL, ROW): global vis # Check if the cell is out of bounds if (row < 0 or col < 0 or col > COL - 1 or row > ROW - 1): return False # Check if the cell is visited or not if (vis[row][col] == True): return False return True# Utility function to print matrix# elements using DFS Traversaldef DFSUtil(row, col,grid, M, N): global vis # Mark the current cell visited vis[row][col] = True # Print element at the cell print(grid[row][col], end = " ") # Traverse all four adjacent # cells of the current element for i in range(4): x = row + dRow[i] y = col + dCol[i] # Check if x and y is # valid index or not if (isValid(x, y, M, N)): DFSUtil(x, y, grid, M, N)# Function to print matrix elementsdefdef DFS(row, col,grid, M, N): global vis # Initialize a visiting matrix # Function call to print matrix # elements by DFS traversal DFSUtil(0, 0, grid, M, N)# Driver Codeif __name__ == '__main__': # Given matrix grid = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ] ] # Row of the matrix M = len(grid) # Column of the matrix N = len(grid[0]) vis = [[False for i in range(M)] for i in range(N)] DFS(0, 0, grid, M, N) # This code is contributed by mohit kumar 29. |
C#
// C# program to implement // the above approach using System;public class GFG{ // Direction vectorsstatic int []dRow = { -1, 0, 1, 0 };static int []dCol = { 0, 1, 0, -1 };// Function to check if current// position is valid or notstatic bool isValid(bool[,] vis, int row, int col, int COL, int ROW){ // Check if the cell is out of bounds if (row < 0 || col < 0 || col > COL - 1 || row > ROW - 1) return false; // Check if the cell is visited or not if (vis[row,col] == true) return false; return true;}// Utility function to print matrix// elements using DFS Traversalstatic void DFSUtil(int row, int col, int[,] grid, bool[,] vis, int M, int N){ // Mark the current cell visited vis[row,col] = true; // Print the element at the cell Console.Write(grid[row,col] + " "); // Traverse all four adjacent // cells of the current element for (int i = 0; i < 4; i++) { int x = row + dRow[i]; int y = col + dCol[i]; // Check if x and y is // valid index or not if (isValid(vis, x, y, M, N)) DFSUtil(x, y, grid, vis, M, N); }} // Function to print the matrix elementsstatic void DFS(int row, int col, int[,] grid, int M, int N){ // Initialize a visiting matrix bool[,] vis = new bool[M + 1,N + 1]; for(int i = 0; i < M + 1; i++) { for(int j = 0; j < N + 1; j++) { vis[i,j] = false; } } // Function call to print matrix // elements by DFS traversal DFSUtil(0, 0, grid, vis, M, N);}// Driver Codepublic static void Main(String []args){ // Given matrix int[,] grid = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; // Row of the matrix int M = grid.GetLength(0); // Column of the matrix int N = grid.GetLength(1); DFS(0, 0, grid, M, N);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// javascript program of the above approach// Direction vectorslet dRow = [ -1, 0, 1, 0 ];let dCol = [ 0, 1, 0, -1 ];// Function to check if current// position is valid or notfunction isValid(vis, row, col, COL, ROW){ // Check if the cell is out of bounds if (row < 0 || col < 0 || col > COL - 1 || row > ROW - 1) return false; // Check if the cell is visited or not if (vis[row][col] == true) return false; return true;}// Utility function to print matrix// elements using DFS Traversalfunction DFSUtil(row, col, grid, vis, M, N){ // Mark the current cell visited vis[row][col] = true; // Print the element at the cell document.write(grid[row][col] + " "); // Traverse all four adjacent // cells of the current element for (let i = 0; i < 4; i++) { let x = row + dRow[i]; let y = col + dCol[i]; // Check if x and y is // valid index or not if (isValid(vis, x, y, M, N)) DFSUtil(x, y, grid, vis, M, N); }}// Function to print the matrix elementsfunction DFS(row, col, grid, M, N){ // Initialize a visiting matrix let vis = new Array(M + 1); // Loop to create 2D array using 1D array for (var i = 0; i < vis.length; i++) { vis[i] = new Array(2); } for(let i = 0; i < M + 1; i++) { for(let j = 0; j < N + 1; j++) { vis[i][j] = false; } } // Function call to print matrix // elements by DFS traversal DFSUtil(0, 0, grid, vis, M, N);} // Driver Code // Given matrix let grid = [[ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ]]; // Row of the matrix let M = grid.length; // Column of the matrix let N = grid[0].length; DFS(0, 0, grid, M, N);</script> |
Output
1 2 3 4 8 12 16 15 11 7 6 10 14 13 9 5
Time Complexity: O(N*M)
Auxiliary Space: O(N*M)
Iterative Approach: The idea is to traverse the matrix using iterative depth-first search and print the matrix elements. Follow the steps below to solve the problem:
- Define a function, say isValid(i, j), to check if the position (i, j) is valid or not, i.e (i, j) lies inside the matrix and not visited.
- Initialize a 2d boolean vector, say vis[][], for keeping track of a position say (i, j) whether it has been already visited or not.
- Initialize a stack<pair<int, int>> say S to implement the DFS traversal.
- First push the first cell (0, 0) in the stack S marking the cell visited.
- Iterate while stack S is not empty:
- In each iteration, mark the top element of stack say (i, j) visited and print the element at that position and remove the top element from the stack S.
- Push the adjacent cells i.e (i + 1, j), (i, j + 1), (i – 1, j) and (i, j – 1) into the stack if respective positions are valid i.e., not visited and are within the matrix.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Direction vectorsint dRow[] = { -1, 0, 1, 0 };int dCol[] = { 0, 1, 0, -1 };// Function to check if curruent// position is valid or notbool isValid(vector<vector<bool> >& vis, int row, int col, int COL, int ROW){ // Check if the cell is out // of bounds if (row < 0 || col < 0 || col > COL - 1 || row > ROW - 1) return false; // Check if the cell is visited if (vis[row][col] == true) return false; return true;}// Function to print the matrix elementsvoid DFS_iterative(vector<vector<int> > grid, int M, int N){ // Stores if a position in the // matrix been visited or not vector<vector<bool> > vis( M + 5, vector<bool>(N + 5, false)); // Initialize stack to implement DFS stack<pair<int, int> > st; // Push the first position of grid[][] // in the stack st.push({ 0, 0 }); // Mark the cell (0, 0) visited vis[0][0] = true; while (!st.empty()) { // Stores top element of stack pair<int, int> p = st.top(); // Delete the top() element // of stack st.pop(); int row = p.first; int col = p.second; // Print element at the cell cout << grid[row][col] << " "; // Traverse in all four adjacent // sides of current positions for (int i = 0; i < 4; i++) { int x = row + dRow[i]; int y = col + dCol[i]; // Check if x and y is valid // position and then push // the position of current // cell in the stack if (isValid(vis, x, y, M, N)) { // Push the current cell st.push({ x, y }); // Mark current cell visited vis[x][y] = true; } } }}// Driver Codeint main(){ // Given matrix vector<vector<int> > grid{ { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; // Row of the matrix int M = grid.size(); // Column of the matrix int N = grid[0].size(); DFS_iterative(grid, M, N); return 0;} |
Java
// Java program for the above approachimport java.util.*;public class Main{ public static class Pair { int Item1, Item2; public Pair(int Item1, int Item2) { this.Item1 = Item1; this.Item2 = Item2; } } // Direction vectors static int[] dRow = { -1, 0, 1, 0 }; static int[] dCol = { 0, 1, 0, -1 }; static Vector<Vector<Boolean>> vis; // Function to check if curruent // position is valid or not static boolean isValid(int row, int col, int COL, int ROW) { // Check if the cell is out // of bounds if (row < 0 || col < 0 || col > COL - 1 || row > ROW - 1) return false; // Check if the cell is visited if (vis.get(row).get(col) == true) return false; return true; } // Function to print the matrix elements static void DFS_iterative(int[][] grid, int M, int N) { // Stores if a position in the // matrix been visited or not vis = new Vector<Vector<Boolean>>(); for(int i = 0; i < M + 5; i++) { vis.add(new Vector<Boolean>()); for(int j = 0; j < N + 5; j++) { vis.get(i).add(false); } } // Initialize stack to implement DFS Vector<Pair> st = new Vector<Pair>(); // Push the first position of grid[][] // in the stack st.add(new Pair(0, 0)); // Mark the cell (0, 0) visited vis.get(0).set(0, true); while (st.size() > 0) { // Stores top element of stack Pair p = st.get(st.size() - 1); // Delete the top() element // of stack st.remove(st.size() - 1); int row = p.Item1; int col = p.Item2; // Print element at the cell System.out.print(grid[row][col] + " "); // Traverse in all four adjacent // sides of current positions for (int i = 0; i < 4; i++) { int x = row + dRow[i]; int y = col + dCol[i]; // Check if x and y is valid // position and then push // the position of current // cell in the stack if (isValid(x, y, M, N)) { // Push the current cell st.add(new Pair(x, y)); // Mark current cell visited vis.get(x).set(y, true); } } } } public static void main(String[] args) { // Given matrix int[][] grid = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; // Row of the matrix int M = 4; // Column of the matrix int N = 4; DFS_iterative(grid, M, N); }}// This code is contributed by suresh07. |
Python3
# Python3 program for the above approach# Direction vectorsdRow = [ -1, 0, 1, 0 ]dCol = [ 0, 1, 0, -1 ] vis = []# Function to check if curruent# position is valid or notdef isValid(row, col, COL, ROW): global vis # Check if the cell is out # of bounds if (row < 0 or col < 0 or col > COL - 1 or row > ROW - 1): return False # Check if the cell is visited if (vis[row][col] == True): return False return True# Function to print the matrix elementsdef DFS_iterative(grid, M, N): global vis # Stores if a position in the # matrix been visited or not vis = [] for i in range(M+5): vis.append([]) for j in range(N + 5): vis[i].append(False) # Initialize stack to implement DFS st = [] # Push the first position of grid[][] # in the stack st.append([ 0, 0 ]) # Mark the cell (0, 0) visited vis[0][0] = True while (len(st) > 0): # Stores top element of stack p = st[-1] # Delete the top() element # of stack st.pop() row = p[0] col = p[1] # Print element at the cell print(grid[row][col], "", end = "") # Traverse in all four adjacent # sides of current positions for i in range(4): x = row + dRow[i] y = col + dCol[i] # Check if x and y is valid # position and then push # the position of current # cell in the stack if (isValid(x, y, M, N)): # Push the current cell st.append([ x, y ]) # Mark current cell visited vis[x][y] = True# Given matrixgrid = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ] ]# Row of the matrixM = len(grid)# Column of the matrixN = len(grid[0])DFS_iterative(grid, M, N)# This code is contributed by mukesh07. |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG { // Direction vectors static int[] dRow = { -1, 0, 1, 0 }; static int[] dCol = { 0, 1, 0, -1 }; static List<List<bool>> vis; // Function to check if curruent // position is valid or not static bool isValid(int row, int col, int COL, int ROW) { // Check if the cell is out // of bounds if (row < 0 || col < 0 || col > COL - 1 || row > ROW - 1) return false; // Check if the cell is visited if (vis[row][col] == true) return false; return true; } // Function to print the matrix elements static void DFS_iterative(int[,] grid, int M, int N) { // Stores if a position in the // matrix been visited or not vis = new List<List<bool>>(); for(int i = 0; i < M + 5; i++) { vis.Add(new List<bool>()); for(int j = 0; j < N + 5; j++) { vis[i].Add(false); } } // Initialize stack to implement DFS List<Tuple<int,int>> st = new List<Tuple<int,int>>(); // Push the first position of grid[][] // in the stack st.Add(new Tuple<int,int>(0, 0)); // Mark the cell (0, 0) visited vis[0][0] = true; while (st.Count > 0) { // Stores top element of stack Tuple<int,int> p = st[st.Count - 1]; // Delete the top() element // of stack st.RemoveAt(st.Count - 1); int row = p.Item1; int col = p.Item2; // Print element at the cell Console.Write(grid[row,col] + " "); // Traverse in all four adjacent // sides of current positions for (int i = 0; i < 4; i++) { int x = row + dRow[i]; int y = col + dCol[i]; // Check if x and y is valid // position and then push // the position of current // cell in the stack if (isValid(x, y, M, N)) { // Push the current cell st.Add(new Tuple<int,int>(x, y)); // Mark current cell visited vis[x][y] = true; } } } } static void Main() { // Given matrix int[,] grid = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; // Row of the matrix int M = 4; // Column of the matrix int N = 4; DFS_iterative(grid, M, N); }}// This code is contributed by divyesh072019. |
Javascript
<script> // Javascript program for the above approach // Direction vectors let dRow = [ -1, 0, 1, 0 ]; let dCol = [ 0, 1, 0, -1 ]; let vis; // Function to check if curruent // position is valid or not function isValid(row, col, COL, ROW) { // Check if the cell is out // of bounds if (row < 0 || col < 0 || col > COL - 1 || row > ROW - 1) return false; // Check if the cell is visited if (vis[row][col] == true) return false; return true; } // Function to print the matrix elements function DFS_iterative(grid, M, N) { // Stores if a position in the // matrix been visited or not vis = []; for(let i = 0; i < M + 5; i++) { vis.push([]); for(let j = 0; j < N + 5; j++) { vis[i].push(false); } } // Initialize stack to implement DFS let st = []; // Push the first position of grid[][] // in the stack st.push([ 0, 0 ]); // Mark the cell (0, 0) visited vis[0][0] = true; while (st.length > 0) { // Stores top element of stack let p = st[st.length - 1]; // Delete the top() element // of stack st.pop(); let row = p[0]; let col = p[1]; // Print element at the cell document.write(grid[row][col] + " "); // Traverse in all four adjacent // sides of current positions for (let i = 0; i < 4; i++) { let x = row + dRow[i]; let y = col + dCol[i]; // Check if x and y is valid // position and then push // the position of current // cell in the stack if (isValid(x, y, M, N)) { // Push the current cell st.push([ x, y ]); // Mark current cell visited vis[x][y] = true; } } } } // Given matrix let grid = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ] ]; // Row of the matrix let M = grid.length; // Column of the matrix let N = grid[0].length; DFS_iterative(grid, M, N);// This code is contributed by decode2207.</script> |
Output:
1 5 9 13 14 15 16 12 8 7 3 4 11 10 6 2
Time Complexity: O(N*M)
Auxiliary Space: O(N*M)
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