CareerCup

def spiralPrint(dArray,x1,y1,x2,y2):

    #base condition
    if x2<x1 or y2<y1:
        return
        
    #print top row
    for i in range(0,x2-x1+1):
        #print y1,i
        print dArray[y1][x1+i]

    #print right col
    for i in range(1,y2-y1+1):
        print dArray[y1+i][x2]

    #print bottom col
    for i in range(1,x2-x1+1):
        print dArray[y2][x2-i]

    #print left col
    for i in range(1,y2-y1):
        print dArray[y2-i][x1]

    spiralPrint(dArray,x1+1,y1+1,x2-1,y2-1)

    return

sample input:
dArray1 = [[1,2,3],[8,9,4],[7,6,5]]
dArray2 = [[10,11],[13,12]]
dArray3 = [[14]]
dArray4 = [[1,2,3,4],[12,13,14,5],[11,16,15,6],[10,9,8,7]]

spiralPrint(dArray1,0,0,2,2)
spiralPrint(dArray2,0,0,1,1)
spiralPrint(dArray3,0,0,0,0)
spiralPrint(dArray4,0,0,3,3)

I found the solution in "Print 2D array in spiral order" of the book "Elements of Programming Interviews: 300 Questions and Solutions". They have code in the entry "Print 2D array in spiral order" at code.google.com/p/elements-of-programming-interviews/wiki/Programs

They have recursive and iterative implementations.

The trick is to have another 2d array with same m rows and rows to record whether the element has been visited. The C++ code for the same,

#include<iostream>

using namespace std;

bool allvisited(int visited[][3])
{
	//bool visited = false;
        for(int i=0 ; i<3 ; i++)
        {
	   for(int j=0 ; j <3; j++)
             if(visited[i][j] == 0)
                 return false;
	}
        return true;
}
int main()
{
  int mat[3][3] = { 1,2,3,
                4,5,6,
                7,8,9 };

  int visited[3][3] = {0};
  int currenti,currentj,tempi,tempj,lefttoright,toptobottom;
  int m=3 , n=3;
  currenti=currentj=lefttoright=toptobottom = tempi = tempj =0;
  while(!allvisited(visited))
  {
	if(!lefttoright)
	{	tempj = currentj;
		while(currentj < n)
		{
			if(visited[currenti][currentj] == 0)
			{
				visited[currenti][currentj] = 1;
                                cout << mat[currenti][currentj] << " ";
				tempj = currentj;
			}
			currentj ++ ;
		}
                currentj = tempj ; //resetting
	}
	else
        {
		tempj = currentj;
		while(currentj >= 0)
		{
			if(visited[currenti][currentj] == 0)
			{
				visited[currenti][currentj] = 1;
                                cout << mat[currenti][currentj] << " ";
				tempj = currentj;
			}
			currentj -- ;
		}
                currentj = tempj ; //resetting		
	}
	if(!toptobottom)
	{	tempi = currenti;
		while(currenti < m)
		{
			if(visited[currenti][currentj] == 0)
			{
				visited[currenti][currentj] = 1;
                                cout << mat[currenti][currentj] << " ";
				tempi = currenti;
			}
			currenti ++ ;
		}
                currenti = tempi ; //resetting
	}
	else
        {	tempi = currenti;
		while(currenti >= 0)
		{
			if(visited[currenti][currentj] == 0)
			{
				visited[currenti][currentj] = 1;
                                 cout << mat[currenti][currentj] << " ";
				tempi = currenti;
			}
			currenti -- ;
		}
                currenti = tempi ; //resetting		
	}		
        lefttoright = !lefttoright;
	toptobottom = !toptobottom;			
  }
  cout << endl << "All visited\n"; 
  return 0;

}

import java.util.Scanner;

public class spiralNumber {
    public static void main(String[] args)
    {
    	System.out.print("Enter the width: ");
    	Scanner in = new Scanner(System.in);
    	int width = in.nextInt();
    	
    	System.out.print("Enter the height: ");
    	int height = in.nextInt();
    	
    	int value[][] = new int[height][width];
      
    	int direction = 1;	// left to right
    	int row=0;
    	int col=0;
    	for(int i=0;i<width*height;i++){
    		if(value[row][col]==0){
    			value[row][col]=i+1;
    			if(direction==1){
    				if(col+1<width && value[row][col+1]==0){
    					col++;
    				}else{
    					direction=2;	// top to bottom
    					row++;
    				}
    			}else if(direction==2){
    				if(row+1<height && value[row+1][col]==0){
    					row++;
    				}else{
    					direction=3;	// right to left
    					col--;
    				}
    			}else if(direction==3){
    				if(col-1>=0 && value[row][col-1]==0){
    					col--;
    				}else{
    					direction=4;	// bottom to top
    					row--;
    				}
    			}else{
    				if(row-1>=0 && value[row-1][col]==0){
    					row--;
    				}else{
    					direction=1;
    					col++;
    				}
    			}
    		}
    	}
    	
    	String string = "%" + String.valueOf(width*height).length() + "d ";
    	for(int i=0;i<height;i++){
    		for(int j=0;j<width;j++)
    			System.out.printf(string, value[i][j]);
    		System.out.println();             
    	}
   	}
}

Here h t t p://stackoverflow.com/questions/7840389/print-2-d-array-in-spiral-order is a good solution

#include<iostream>
using namespace std;

int main()
{
int n;
cout<<endl<<"enter n:";
cin>>n;

int a[n][n],i,j,m;
for (i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
a[i][j]=(i*n)+(j+1);
}
}

if(n%2==0)
m=n/2;
else
m = (int) n/2+1;


for(int k=0;k<=m;k++)

{
cout<<endl;
for (j=k;j<n-k;j++)
{
i=k;
cout<<"\t"<<a[i][j];
}
cout<<endl;
for(i=k+1;i<n-k;i++)
{
j=n-1-k;
cout<<"\t"<<a[i][j];
}
cout<<endl;
for(j=n-2-k;j>=0+k;j--)
{
i=n-1-k;
cout<<"\t"<<a[i][j];
}
cout<<endl;
for(i=n-2-k;i>=1+k;i--)
{
j=0+k;
cout<<"\t"<<a[i][j];
}

}

}

Java Code


int[][] arr = {{1,2,3,4,5}, {6,7,8,9,10}, {11,12,13,14,15}, {16,17,18,19,20},{21,22,23,24,25}};
int size = arr.length;
int width = arr[0].length;
for (int l = 0; l < size / 2; l++)
{
int min = l;
int max = size - 1 - l;
int max1 = width-1 -l;
for (int i = min; i < max; i++)
System.out.print("\t" + arr[i][min]);
for (int j = min; j < max1; j++)
System.out.print("\t" + arr[max][j]);
for (int i = max; i > min; i--)
System.out.print("\t" + arr[i][max1]);
for (int j = max1; j > min; j--)
System.out.print("\t" + arr[min][j]);
}
// centre is special case: avoiding printing it 4 times.
if (size % 2 == 1)
System.out.print("\t" + arr[size / 2][size / 2]);

I think the matrix can be split into levels, such as 1, 2, 3,
4, 5, 6,
7, 8, 9,
we can takes it as two levels: 1, 2, 3, 6, 9, 8, 7, 4 and 5 respetively. and print in level, here are my codes, there is no need extra space and time complexicity is O(n*n), or O(n).
#include<iostream>
using namespace std;
const int n = 5;

int a[n][n] = {1, 2, 3, 4, 5, 16, 17, 18, 19, 6, 15, 24, 25, 20, 7, 14, 23, 22, 21, 8, 13, 12, 11, 10, 9};
void spiral_traverse() {
int level;
int i;
for(level = 1; level <= n / 2; level++) {
for(i = level - 1; i < n - level; i++) {
cout << a[level - 1][i] << " ";
}
for(i = level - 1; i < n - level; i++) {
cout << a[i][n - level] << " ";
}
for(i = n - level; i >= level; i--) {
cout << a[n - level][i] << " ";
}
for(i = n - level; i >= level; i--) {
cout << a[i][level - 1] << " ";
}
}
if(n % 2 == 1) {
cout << a[n / 2][n / 2];
}
cout << endl;
}

void main() {
spiral_traverse();
getchar();
}

This is my code an it works well.
sort the matrix in an array and then map the pattern traversal in it u will come to know about it.

#include<stdio.h>
#include<conio.h>
#include<Windows.h>

int main()
{
	int i=0,j=0,k=0,b=0,l=0,m=0,n=0;
	int giv[25][25],in[25],out[25];
	
	puts("\nEnter the order of the matrix->");
	scanf_s("%d",&m);

   puts("Enter the matrix ->");
   for(i=0;i<m;i++)
	 {  for(j=0;j<m;j++)
		   scanf_s("%d",&giv[i][j]);
        puts("");
     }
	for(i=0;i<m;i++)
	{	for(j=0;j<m;j++)
		 {	in[k]=giv[i][j];
	        k++;
	     }
	}
	system("cls");

		puts("\nEntered matix is\n ");
		for(i=0;i<m;i++)
	 {  for(j=0;j<m;j++)
		printf("%4d",giv[i][j]);
        puts("");
     } 
    	b=(m*m)/2;
		out[0]=in[b];
		j=1;
/*this is the main procedure try to map it using an array of no.s*/
while(l<m)
{     for(k=1;k<=m;k++)
        {  n=k;
    
	      if(n%2==0)
	       {
	         for(i=1;i<=n;i++)
	           {
	             out[j]=in[b+m];
	        	 j++;
		         b=b+m;
	           }
            for(i=1;i<=n;i++)
	           { 
		         out[j]=in[b+1];
				 j++;
		         b=b+1;
		         
	           }l++;
	      }
	 
	  
	    if(n%2==1)
	     {                  
		   for(i=1;i<=n;i++)
	         { 
	           out[j]=in[b-m];
 	     	   j++;
		       b=b-m;
	   
	         }	
	     for(i=1;i<=n;i++)
	        { out[j]=in[b-1];
		      j++;
	          b=b-1;
	        }l++;
	   }
		
	}
   
  }
 		
	printf("\noutput is-> ");
	printf("%d",out[0]);
	for(i=1;i<(m*m);i++)
		{   printf("-");
			printf("%d",out[i]);
	     
	    }
	
	_getch();
	return 0;
}

Recursive call might be cleaner and easier.

public class SpiralPrint{
    
    //print the elements of matrix in the spiral order.
    //my idea is to use recursive, for each outer loop
    
    public static void printSpiral(int[][] mat, int layer){
        int up = layer;
        int buttom = mat.length - layer - 1;
        int left = layer;
        int right = mat[0].length - layer - 1;
        if(up > buttom+1 || left > right + 1)
            return; // termination condition
        
        //traverse the other frame, 
        //print up
        for(int i = left; i <= right; i ++){
            System.out.print( mat[up][i]+ " " );
        }
        //print right
        for(int i = up + 1; i <=buttom; i ++){
            System.out.print(mat[i][right] + " ");
        }

        //print buttom
        for(int i = right - 1; i >= left; i --){
            System.out.print(mat[buttom][i] + " ");
        }

        //print left
        for(int i = buttom - 1; i > up; i --){
            System.out.print(mat[i][left] + " ");
        }

        //recursive call for the next level
        printSpiral(mat, layer + 1);
    }

    public static void main(String[] args){
        
        int[][] mat = {{1,2,3,4}, {5,6,7,8}, {9,10,11,12}, {13,14,15,16}};
        int[][] mat2 = {{1,2,3}, {4,5,6}, {7,8,9}, {10,11,12}};
        SpiralPrint.printSpiral(mat2,0);
        return;
    }
}