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

You are given a 2-D array. We have to traverse each and every cell of 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 shows next following cells 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 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.

/* 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));
    }
}

Output:

true

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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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