How to print element of 2d matrix in spiral?

Question

The input tree is as shown below
            40
            / \
        20      60
        / \       \
    10      30      80
      \            /  \ 
       15        70    90

Output: 15 30 60 80 90

Answer 1

See the following code

#define ROWS 5
#define COLS 5

int main()
{
    int arr[ROWS][COLS] =  {{ 1, 2, 3, 4, 17},
                            { 5, 6, 7, 8, 18},
                            { 9,10,11,12, 19},
                            {13,14,15,16, 20},
                            {21,22,23,24, 25}};

    // Four direction counters of current movement
    // Horizontal right, vertical bottom, horizontal left and vertical top respectively
    int hr, vb, hl, vt;

    // levl indicates current depth of our imaginary rectangle into array. Starting value is zero
    // since we are looping on the boundaries and ending value is the inner most rectangle
    int levl;
    for (levl=0; levl < COLS-levl; levl++)
    {
        for(hr=levl; hr < COLS-levl; hr++)   // go right
            printf("%d ,", arr[levl][hr]);

        for(vb=levl+1; vb < COLS-levl; vb++) // go down
            printf("%d ,", arr[vb][hr-1]);

        for(hl=vb-1; hl-1 >= levl; hl--)  // go left
            printf("%d ,", arr[vb-1][hl-1]);

        for(vt=vb-1; vt-1 > levl; vt--)  // go up
            printf("%d ,", arr[vt-1][hl]);
    }

    return 0;
}

Output

1 ,2 ,3 ,4 ,17 ,18 ,19 ,20 ,25 ,24 ,23 ,22 ,21 ,13 ,9 ,5 ,6 ,7 ,8 ,12 ,16 ,15 ,14 ,10 ,11 ,

Comments

What is the complexity of this code ??

Answer 2

void main()
{
    int r;      // no of row
    int c;      //no of column
    int x;      
    int z;
    int l=0,n=0;
    int a[20][20];

    printf("Enter the row => ");
    scanf("%d",&r);
    printf("Enter the column => ");
    scanf("%d",&c);

    mat1(a,r,c,l,n,c*r);        //  This function will print matrix

    printf("\n MATRIX => \n");
    for(x=0;x<r;x++)
    {
        printf("\n\n\n");
        for(z=0;z<c;z++)
            printf("    %d",a[x][z]);
    }
    getch();
}


int mat1(int a[][20],int r,int c,int l,int n,int m)
{
    if(n<m)
    {
        int x,z;
        for(x=l;x<c;x++)
            a[l][x]=++n;
        for(z=l+1;z<r;z++)
            a[z][c-1]=++n;
        c--;
        mat2(a,r,c,l,n,m);
    }
    else
        return 1;
}

int mat2(int a[][20],int r,int c,int l,int n,int m)
{
    int x,z;
    if(n<m)
    {
        for(x=c-1;x>=l;x--)
            a[r-1][x]=++n;
        for(z=r-2;z>=l+1;z--)
            a[z][l]=++n;
        r--;
        l++;
        mat1(a,r,c,l,n,m);
    }
    else
        return 1;
}

complexity O(no of element in matrix).

Answer 3

Thanks.

Assume 4*4

You have to walk as this, Row-Column[Forward] then Row-Column[Reverse]

1. Ro ==> C0 to C2
   C3 ==> R0 to R2

2. R3 ==> C3 to C1
   C0 ==> R3 to R1

With above you will be able to print outer layer. You have to increment all the value[ie index] by one and try for first inner later.

Hope this 2 steps will be unique and will work for N*N matrix also.

I let yo to write code for whole. Let me know if you find difficulties and let others also comment. with their ideas..

Comments

Good Idea :)
Can you please write code too ??