Find if a 2-D array is completely traversed or not by following the cell values

0 0 6 minutes read

You are given a 2-D array. We have to traverse each and every cell of the given array by following the cell locations then return true else false. The value of each cell is given by (x, y) where (x, y) is also shown next following cell position. Eg. (0, 0) shows starting cell. And ‘null’ shows our final destination after traversing every cell.

Examples:

Input : { 0, 1   1, 2   1, 1 
          0, 2   2, 0   2, 1 
          0, 0   1, 0   null }
Output : false

Input : { 0, 1   2, 0 
          null  1, 0
          2, 1   1, 1 }
Output : true

We take a visited array if we visit a cell then make its value true in the visited array so that we can capture the cycle in our grid for next time when we visit it again. And if we find null before completing the loop then it means we didn’t traversed all the cell of given array.

Implementation:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// function which tells all cells are visited or not
bool isAllCellTraversed(vector<vector<pair<int,int>>>grid, int n, int m)
{
    bool visited[n][m];
    int total = n*m;
   
    // starting cell values
    int startx = grid[0][0].first;
    int starty = grid[0][0].second;
 
    for (int i = 0; i < total - 2; i++) 
    {
 
        // if we get {0,0} before the end of loop
        // then returns false. Because it means we
        // didn't traverse all the cells
        if (grid[startx][starty].first == -1 and 
            grid[startx][starty].second == -1)
            return false;
 
        // If found cycle then return false
        if (visited[startx][starty] == true)
            return false;
 
        visited[startx][starty] = true;
        int x = grid[startx][starty].first;
        int y = grid[startx][starty].second;
 
        // Update startx and starty values to next
        // cell values
        startx = x;
        starty = y;
    }
 
    // finally if we reach our goal then returns true
    if (grid[startx][starty].first == -1 and 
        grid[startx][starty].second == -1)
        return true;
 
    return false;
}
 
// Driver code
int main()
{
  vector<vector<pair<int,int>>> cell(3, vector<pair<int,int>> (2));
  cell[0][0] = {0, 1};
  cell[0][1] = {2, 0};
  cell[1][0] = {-1,-1};
  cell[1][1] = {1, 0};
  cell[2][0] = {2, 1};
  cell[2][1] = {1, 1};
 
  if(!isAllCellTraversed(cell, 3, 2)) 
    cout << "true";
  else
    cout << "false";
 
  return 0;
}
 
// This code is contributed by mohit kumar 29.

Java

/* Java program to Find a 2-D array is completely
traversed or not by following the cell values */
import java.io.*;
 
class Cell {
    int x;
    int y;
 
    // Cell class constructor
    Cell(int x, int y)
    {
        this.x = x;
        this.y = y;
    }
}
 
public class MoveCellPerCellValue {
 
    // function which tells all cells are visited or not
    public boolean isAllCellTraversed(Cell grid[][])
    {
        boolean[][] visited = 
              new boolean[grid.length][grid[0].length];
 
        int total = grid.length * grid[0].length;
        // starting cell values
        int startx = grid[0][0].x;
        int starty = grid[0][0].y;
 
        for (int i = 0; i < total - 2; i++) {
 
            // if we get null before the end of loop 
            // then returns false. Because it means we 
            // didn't traverse all the cells
            if (grid[startx][starty] == null) 
                return false;
             
            // If found cycle then return false
            if (visited[startx][starty] == true) 
                return false;
             
            visited[startx][starty] = true;
            int x = grid[startx][starty].x;
            int y = grid[startx][starty].y;
 
            // Update startx and starty values to next
            // cell values
            startx = x;
            starty = y;
        }
 
        // finally if we reach our goal then returns true
        if (grid[startx][starty] == null) 
            return true;
         
        return false;
    }
 
    /* Driver program to test above function */
    public static void main(String args[])
    {
        Cell cell[][] = new Cell[3][2];
        cell[0][0] = new Cell(0, 1);
        cell[0][1] = new Cell(2, 0);
        cell[1][0] = null;
        cell[1][1] = new Cell(1, 0);
        cell[2][0] = new Cell(2, 1);
        cell[2][1] = new Cell(1, 1);
 
        MoveCellPerCellValue mcp = new MoveCellPerCellValue();
        System.out.println(mcp.isAllCellTraversed(cell));
    }
}

Python3

# Python3 program for the above approach
 
# function which tells all cells are visited or not
def isAllCellTraversed(grid, n, m):
 
    visited = [[True for j in range(m)] for i in range(n)];
    total = n*m;
   
    # starting cell values
    startx = grid[0][0][0];
    starty = grid[0][0][1];
 
    for i in range(total-2):
 
        # if we get {0,0} before the end of loop
        # then returns False. Because it means we
        # didn't traverse all the cells
        if (grid[startx][starty][0] == -1 and
            grid[startx][starty][1] == -1):
            return False;
 
        # If found cycle then return False
        if (visited[startx][starty] == True):
            return False;
 
        visited[startx][starty] = True;
        x = grid[startx][starty][0];
        y = grid[startx][starty][1];
 
        # Update startx and starty values to next
        # cell values
        startx = x;
        starty = y;
 
    # finally if we reach our goal then returns True
    if (grid[startx][starty][0] == -1 and
        grid[startx][starty][1] == -1):
        return True;
 
    return False;
     
# Driver code
cell = [[[] for j in range(3)] for i in range(3)]
cell[0][0] = [0, 1];
cell[0][1] = [2, 0];
cell[1][0] = [-1,-1];
cell[1][1] = [1, 0];
cell[2][0] = [2, 1];
cell[2][1] = [1, 1];
 
if(not isAllCellTraversed(cell, 3, 2)):
  print("True");
else:
  print("False");
 
  # This code is contributed by rrrtnx.

C#

/* C# program to Find a 2-D array is completely
traversed or not by following the cell values */
using System;
 
public class Cell
{
    public int x;
    public int y;
 
    // Cell class constructor
    public Cell(int x, int y)
    {
        this.x = x;
        this.y = y;
    }
}
 
public class MoveCellPerCellValue 
{
 
    // function which tells all cells are visited or not
    public Boolean isAllCellTraversed(Cell [,]grid)
    {
        Boolean[,] visited = 
            new Boolean[grid.GetLength(0), 
                        grid.GetLength(1)];
 
        int total = grid.GetLength(0) * 
                    grid.GetLength(1);
                     
        // starting cell values
        int startx = grid[0, 0].x;
        int starty = grid[0, 0].y;
 
        for (int i = 0; i < total - 2; i++) 
        {
 
            // if we get null before the end of loop 
            // then returns false. Because it means we 
            // didn't traverse all the cells
            if (grid[startx, starty] == null) 
                return false;
             
            // If found cycle then return false
            if (visited[startx, starty] == true) 
                return false;
             
            visited[startx, starty] = true;
            int x = grid[startx, starty].x;
            int y = grid[startx, starty].y;
 
            // Update startx and starty values 
            // to next cell values
            startx = x;
            starty = y;
        }
 
        // finally if we reach our goal
        // then returns true
        if (grid[startx, starty] == null) 
            return true;
         
        return false;
    }
 
    // Driver Code
    public static void Main(String []args)
    {
        Cell [,]cell = new Cell[3, 2];
        cell[0, 0] = new Cell(0, 1);
        cell[0, 1] = new Cell(2, 0);
        cell[1, 0] = null;
        cell[1, 1] = new Cell(1, 0);
        cell[2, 0] = new Cell(2, 1);
        cell[2, 1] = new Cell(1, 1);
 
        MoveCellPerCellValue mcp = new MoveCellPerCellValue();
        Console.WriteLine(mcp.isAllCellTraversed(cell));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
 
/* Javascript program to 
Find a 2-D array is completely
traversed or not by following 
the cell values */
     
    class Cell
    {
    // Cell class constructor
        constructor(x,y)
        {
            this.x=x;
            this.y=y;
        }
    }
     
    // function which tells all 
    // cells are visited or not
    function isAllCellTraversed(grid)
    {
        let visited =
              new Array(grid.length);
        for(let i=0;i<visited.length;i++ )
        {
            visited[i]=new Array(grid[0].length);
        }
  
        let total = grid.length * grid[0].length;
        // starting cell values
        let startx = grid[0][0].x;
        let starty = grid[0][0].y;
  
        for (let i = 0; i < total - 2; i++) {
  
            // if we get null before the end of loop
            // then returns false. Because it means we
            // didn't traverse all the cells
            if (grid[startx][starty] == null)
                return false;
              
            // If found cycle then return false
            if (visited[startx][starty] == true)
                return false;
              
            visited[startx][starty] = true;
            let x = grid[startx][starty].x;
            let y = grid[startx][starty].y;
  
            // Update startx and starty
            // values to next
            // cell values
            startx = x;
            starty = y;
        }
  
        // finally if we reach our goal 
        // then returns true
        if (grid[startx][starty] == null)
            return true;
          
        return false;
    }
     
    /* Driver program to test above function */
    let cell=new Array(3);
    for(let i=0;i<3;i++)
    {
        cell[i]=new Array(2);
    }
    cell[0][0] = new Cell(0, 1);
    cell[0][1] = new Cell(2, 0);
    cell[1][0] = null;
    cell[1][1] = new Cell(1, 0);
    cell[2][0] = new Cell(2, 1);
    cell[2][1] = new Cell(1, 1);
  
         
    document.write(isAllCellTraversed(cell));
     
 
 
// This code is contributed by unknown2108
 
</script>

Output

true

Time Complexity : O(N)
Auxiliary Space : O(M*N)

This article is contributed by Harshit Agrawal. If you like zambiatek and would like to contribute, you can also write an article using write.zambiatek.com or mail your article to review-team@zambiatek.com. See your article appearing on the zambiatek main page and help other Geeks.

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