Interview Question for Software Trainees Applications Developers


Country: India
Interview Type: Written Test




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6
of 6 vote

The value in position (i,j) is 1 plus the maximum of the horizontal distance from the center and the vertical distance from the center. In Java a function to print this would look like:

public static void printMatrix(int N) {		 
		for(int i = 0; i < 2 * N - 1; i++) {
			for(int j = 0; j < 2 * N - 1; j++) {
				System.out.print(1+Math.max(Math.abs(N-i-1), Math.abs(N-j-1)) );
			}
			System.out.println();
		}
	}

- Guy March 08, 2014 | Flag Reply
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0
of 0 votes

Beautiful finding on the pattern.

- Anonymous March 09, 2014 | Flag
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0
of 0 votes

nice...

- svrussu March 09, 2014 | Flag
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0
of 0 votes

well done

- Anonymous March 10, 2014 | Flag
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0
of 0 votes

One of the best solution.

- Sharma March 22, 2014 | Flag
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0
of 0 votes

Wow .. awesome

- Rajesh March 30, 2014 | Flag
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0
of 0 votes

U are awesome !
Hats off !

- Cody March 31, 2014 | Flag
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0
of 0 votes

Now my solution look kind of stupid.. the way you did

public static void printInPattern(int n){
		
		if(n<=0){
			System.out.println("NO NO Mr");
			return;
		}
		int length = 2*n-1;
		int printArr[][] = new int[length][length];
		
		for(int i=0;i<n;i++)
			fillArr(printArr, i);
		
		for(int i=0;i<length;i++){
			for(int j=0;j<length;j++)
				System.out.print(""+printArr[i][j]+" ");
			
			System.out.print("\n");
		}
			
	}

	public static void fillArr(int[][] printArr, int i){
		int N = printArr.length;
		int k=i;
		int val = (N+1)/2 -k;
		//int i=0;
		int j=0;
		//start at daigonal a[i][i] 
		// from j=i to N-i-1
		for(j=i;j<N-k;j++){
			printArr[i][j]= val;
		}
		j--;
		i++;
		for(;i<N-k;i++){
			printArr[i][j]= val;
		}
		i--;
		j--;
		
		for(;j>=k;j--)
			printArr[i][j]= val;
		
		j++;
		i--;
		for(;i>k;i--)
			printArr[i][j]= val;
	}

- sumit May 02, 2014 | Flag
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4
of 4 vote

private static void print(int n) {
		int len = 2 * n - 1;
		for (int i = 0; i < len; i++) {
			for (int j = 0; j < len; j++) {
				System.out.print(n - min(i, j, len - 1 - i, len - 1 - j));
			}
			System.out.println();
		}
	}

- myk March 08, 2014 | Flag Reply
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1
of 1 vote

#include <iostream>
#include <vector>

using namespace std;

int main()
{
    int N = 9;

    for (int i = 0; i < 2 * N - 1; i++) {
        int k = i < N ? i : 2 * (N - 1) - i; 

        for (int j = 0; j < k; j++)
            cout << N - j;

        for (int j = 0; j < 2 * (N - k) - 1; j++)
            cout << N - k;

        for (int j = k - 1; j >= 0; j--)
            cout << N - j;

        cout << endl;
    }
    return 0;
}

/**** output for N = 9
99999999999999999
98888888888888889
98777777777777789
98766666666666789
98765555555556789
98765444444456789
98765433333456789
98765432223456789
98765432123456789
98765432223456789
98765433333456789
98765444444456789
98765555555556789
98766666666666789
98777777777777789
98888888888888889
99999999999999999
***/

- Westlake March 08, 2014 | Flag Reply
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1
of 1 vote

Java code.

public static void  draw(int N)  {
	if (N > 0) { 
	    int m = 2 * (N - 1) + 1;
	    int mid = N - 1;
	    for (int i = 0; i < m; i++) { 
		for (int j =  0; j < m; j++) { 
		    int d  = Math.max(Math.abs(i -mid), Math.abs(j - mid)) + 1;
		    System.out.print(d);
		}
		System.out.print("\n");
	    }
	}
}

- Ehsan March 08, 2014 | Flag Reply
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0
of 0 vote

A simple and brute force approach by storing values to print in a 2D matrix .Though we can optimize it.

public static void printPattern(int x)
{
int len=x*2-1;
int leftCorner=0,rightCorner=len,upCorner=0,downCorner=len;
int a[][]=new int[len][len];

while(true)
{
if(leftCorner>rightCorner)
break;
else
for(int i=leftCorner;i<rightCorner;i++)
a[upCorner][i]=x;

upCorner++;// up row is visited

if(upCorner>downCorner)
break;
else
for(int i=upCorner;i<downCorner;i++)
a[i][rightCorner-1]=x;

rightCorner--;//right colum is visited
if(leftCorner>rightCorner)
break;
else
for(int i=rightCorner;i>=leftCorner;i--)
a[downCorner-1][i]=x;

downCorner--;//down row is visited
if(upCorner>downCorner)
break;
else
for(int i=downCorner;i>=upCorner;i--)
a[i][leftCorner]=x;

leftCorner++;//left column is visited

x--;//save next pattern
}

/*
* Print pattern
*/
for(int i=0;i<len;i++)
{
for(int j=0;j<len;j++)
{
System.out.print(a[i][j]);
}
System.out.println("");
}

}

- braj March 08, 2014 | Flag Reply
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0
of 0 vote

Java code.

public static void  draw(int N)  {
	int N = Integer.parseInt(args[0]);
	if (N > 0) { 
	    int m = 2 * (N - 1) + 1;
	    int mid = N - 1;
	    for (int i = 0; i < m; i++) { 
		for (int j =  0; j < m; j++) { 
		    int d  = Math.max(Math.abs(i -mid), Math.abs(j - mid)) + 1;
		    System.out.print(d);
		}
		System.out.print("\n");
	    }
	}
}

- Ehsan March 08, 2014 | Flag Reply
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0
of 0 vote

Here is the C++ code.

Basically, you need the print the maximum of the x-distance or y-distance to the center position

#include <iostream>
using namespace std;

int abs(int i) {
  return (i < 0 ? -i : i);
}

void print(int n) {
  for (int i = 0; i<2*n-1; ++i) {
    for (int j = 0; j < 2*n-1; ++j)
      printf("%d", (abs(i-n+1) > abs(j-n+1) ? abs(i-n+1) : abs(j-n+1))+1);
    printf("\n");
  }
}
int main() {
  print(7);
  return 0;
}

- gokayhuz March 09, 2014 | Flag Reply
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0
of 0 votes

fail on 1

- suddy April 12, 2019 | Flag
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0
of 0 vote

Here is my little solution:

package other;

public class PrintSomeNumbers {
	
	private static int[] array = null;
	private static int number = -1;
	private static int direction = 1;
	
	public static void main(String[] args){

		printThem(9);
		
	}
	
	public static void printThem(int n){
		
		if(n < 1)
			return;
		
		if(n == 1){
			System.out.println(1);
		}
		
		number = n;
		int size = (n*2) - 1;
		array = new int[size];
		for(int i =0; i < array.length; i++){
			array[i] = number;
		}
		int index = 0;
		printThemRec(index);
	}
	
	public static void printThemRec(int ind){
		
		int rightIndex = (array.length - 1) - ind;
		int index = ind;
		
		
		//Case when the left crosses right
		//which is time to reverse.
		if(index > rightIndex){
			direction = -1;
		}
		
		//for going backwards
		if(ind < 0)
			return;
		
		//current number to print
		int curNum = number - index;
		
		//print within the range
		while(index <= rightIndex){
			//System.out.println(curNum);
			array[index] = curNum;
			index++;
		}
		
		//omits the two extra printings before the middle one.
		if(ind != number -1)
			printIt();
		
		printThemRec(ind+(1*direction));
	}
	
	private static void printIt(){
		
		for(int i : array){
			System.out.print("|"+i);
		}
		System.out.println("|");
	}
}

- svrussu March 09, 2014 | Flag Reply
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0
of 0 vote

There will be always a center point that has value of 1 (Let's call it Layer 1).
Then, surround it with number 2 in all directions (Let's call it Layer 2)
Then, surround all with number 3 in all directions (Let's call it Layer 3)
and so on...

Basically, n will represent how many Layers are there...

- Ahmsayat March 09, 2014 | Flag Reply
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0
of 0 vote

Java code... Without nested loops...

private static void print2(int n) {

		int length = n * 2 - 1;
		// long trim = Long.parseLong(Long.toBinaryString((long) (Math.pow(2, length) - 1)), 10);
		// long highTrim = (long) Math.pow(10, length - 1);
		long trim = 1, lowTrim = 1, highTrim = 1;
		for (int x = 0; x < length - 1; x++) {
			trim = trim * 10 + 1;
			highTrim = highTrim * 10;
		}
		// trim = 1111111
		// highTrim = 1000000
		long number = trim * n;
		for (int i = 0; i < length; i++) {
			System.out.println(number);
			// remove 1 from left, 1 from right and subtract 1 in between ;)
			trim = trim - highTrim - lowTrim;
			number = number - trim;
			lowTrim = lowTrim * 10;
			highTrim = highTrim / 10;
		}
	}

Print lowTrim, highTirm and trim inside the loop to better understand the logic.

- mad piranha March 09, 2014 | Flag Reply
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0
of 0 vote

Python version:

#generate first line
line=str(n)*(2*n-1)

#array for the unique numbers/lines 
rows=[]

# generate the unique rows
for i in range(len(line)/2+1):
	line=line[0:i]+str(n-i)*(2*(n-i)-1)+line[2*n-1-i:2*n-1]
	rows.append(line)

#print the matrix
for i in range(len(rows)):
	print rows[i]
for i in range(len(rows)-1):
	print rows[len(rows)-2-i]

- Kemal March 10, 2014 | Flag Reply
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0
of 0 vote

void palinPrint(int N)
{
	if(N==0) return;
	cout<<N;
	palinPrint(N-1);
	if(N==1) return;
	cout<<N;
}

- lazyGuy March 11, 2014 | Flag Reply
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0
of 0 vote

#include "stdlib.h"
#include "stdio.h"

int main(int argc, char* argv[]) {

        int i, j, n, k;
        unsigned int abs_i, abs_j;

        if(argc != 2) {
                return 0;
        }

        n = atoi(argv[1]);

        for(i = -(n-1); i <  n; i++) {
                for(j = -(n-1); j < n; j++) {
                        abs_i = (i < 0) ? (-i) : (i) ;
                        abs_j = (j < 0) ? (-j) : (j) ;
                        k = (abs_i > abs_j) ? (abs_i) : (abs_j);
                        printf("%d ",k+1);
                }
                printf("\n");
        }
        return 0;
}

- Vamshi Krishna Ramaka March 25, 2014 | Flag Reply
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0
of 0 vote

public class testStrings {
	
	public static void main(String[] args) 
	{
		/*input N = 4, print below pattern
		4444444 
		4333334 
		4322234 
		4321234 
		4322234 
		4333334 
		4444444 */
		int max = 4;
		int n = 4;
		
		List<String> list = new ArrayList<String>();
		
		while(n > 0)
		{
			String s = "";
			for (int i = (n * 2) - 1; i > 0; i--)
			{
				s += n;
			}
			
			if (list.size() == 0)
			{
				list.add(s);
			}
			else
			{
				String source = list.get(list.size() - 1);
				char sourceArr[] = source.toCharArray();
				char destArr[] = s.toCharArray();
				int j = 0;
				int pos = max - n;
				int len = sourceArr.length;
				
				for (int i = pos; i < (len - pos); i++)
				{
					sourceArr[i] = destArr[j];
					j++;
				}
				
				source = "";
				for (int k = 0; k < len; k++)
				{
					source += sourceArr[k];
				}
				list.add(source);
			}
			
			n--;
		}
		
		List<String> finalList = new ArrayList<String>();
		finalList.addAll(list);
		
		for (int i = list.size() - 2; i >= 0; i--)
		{
			finalList.add(list.get(i));
		}
		list = null;
		
		for (String s: finalList)
		{
			System.out.println(s);
		}
		
	}

}

- Bharath paturi March 28, 2014 | Flag Reply
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0
of 0 vote

#include <stdio.h>

int main(){
    int i, j, k, n, flag = 0;
    int number_to_print, number_to_repeat, next_number_to_repeat;

    printf("Please enter the number\n");
    scanf("%d", &n);
    number_to_print = n;

    for(i = 0; i < n; i++){
        //--next_number_to_repeat;
        number_to_print = n;

        for(j =0 ; j< (2*n-1); j++){

            printf("%d ", number_to_print);
            if( j < i   )   
                --number_to_print;
            else if( j >  ( (n*2)- i -3  ) ) 
                ++number_to_print;

        }   
        printf("\n");
    }   
#if 1
    for(i = n-2; i >= 0; i--){
        //--next_number_to_repeat;
        number_to_print = n;

        for(j =0 ; j < (2*n-1); j++){

            printf("%d ", number_to_print);
            if( j < i   )   
                --number_to_print;
            else if( j >  ( (n*2)- i -3  ) ) 
                ++number_to_print;

        }   
        printf("\n");
    }   
        
#endif
    return 0;

}

- Ganesh Wani April 15, 2014 | Flag Reply
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0
of 0 vote

I solved it essentially looking at the closest side and calculating the proper value to se.

public static void printPattern (int n)
    {
        int width               = n*2 - 1;
        int height              = n*2 - 1;
        StringBuilder pattern   = new StringBuilder();
        
        System.out.println("Pattern of "+n);
        
        for (int row=0; row<height; row++)
        {
            for (int col=0; col<width; col++)
        	{
        		int left = col;
                	int right = width-col-1;
        		int top = row;
        		int bottom = height-row-1;
        		
        		int lr = Math.min(left, right);
        		int tb = Math.min(top, bottom);
        		int closest = Math.min(lr, tb);
                
        		pattern.append(""+(n-closest));
        	}
        	
        	pattern.append("\n");
        }
        
        System.out.println(pattern.toString());
    }

- Jona April 30, 2014 | Flag Reply
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0
of 0 vote

public class CombinationPrint {
public static void main(String[] args) {
Scanner sc =new Scanner(System.in);
System.out.println("Enter the no.");
int n = sc.nextInt();
int arr[] =new int[(2*n)-1];
int j=0;
int k=(2*n)-1;
int l=2;
for(int i = 0; i < ((2 * n) - 1); i++) {
if((n-i)>0){

for(j=0;j<k;j++){
arr[j+i]=n-i;
}
k=k-2;
}else{
if(k==-1){
k=3;
}
for(j=0;j<k;j++){
arr[j+i-l]=i+2-n;
}
k=k+2;
l=l+2;
}

print(arr);
j=0;
System.out.println("");
}
}

private static void print(int[] arr) {
for(int x=0;x<arr.length;x++){
System.out.print(arr[x]);
}
}
}

- Rajesh May 08, 2014 | Flag Reply
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0
of 0 vote

public class CombinationPrint {
    public static void main(String[] args) {
        Scanner sc =new Scanner(System.in);
        System.out.println("Enter the no.");
        int n = sc.nextInt();
        int arr[] =new int[(2*n)-1];
        int j=0;
        int k=(2*n)-1;
        int l=2;
        for(int i = 0; i < ((2 * n) - 1); i++) {
			if((n-i)>0){
                            
                              for(j=0;j<k;j++){
                                  arr[j+i]=n-i;
                              }
                              k=k-2;
                        }else{
                            if(k==-1){
                                k=3;
                            }
                            for(j=0;j<k;j++){
                                  arr[j+i-l]=i+2-n;
                              }
                             k=k+2;
                             l=l+2;
                        }
                        
                        print(arr);
                        j=0;
                        System.out.println("");
        }
    }

    private static void print(int[] arr) {
        for(int x=0;x<arr.length;x++){
            System.out.print(arr[x]);
        }
    }

}

- Anonymous May 08, 2014 | Flag Reply
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0
of 0 vote

package com.practice;

public class Matrix {

public static void main(String[] args) {
Matrix m = new Matrix();

int[][] a = m.fillValues(3);

for (int[] is : a) {
for (int i : is) {
System.out.print(i + " ");
}
System.out.println();
}
}


private int[][] fillValues(int n){

int first = 0;
int last = 2*n - 1;
int i = first;
int j = first;
int[][] a = new int[last][last];

while(n>0){
i = first;
j = first;

for (; i < last; i++) {
a[i][j] = n;
}
i--;
for (; j < last; j++) {
a[i][j] = n;
}
j--;
for (; i > first; i--) {
a[i][j] = n;
}
for (; j > first; j--) {
a[i][j] = n;
}
n--;
first++;
last--;
}

return a;
}



}

- Sravana kumar pulivendula(ARICENT) May 11, 2014 | Flag Reply
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0
of 0 vote

public class Matrix {


public static void main(String[] args) {

Scanner s = new Scanner(System.in);

int value = s.nextInt();

int actuallength = value * 2;
int matrixLength = actuallength - 1;
int[][] a = new int[matrixLength][matrixLength];

for (int i = 0; i < actuallength / 2; i++) {

for (int j = i; j < (matrixLength - i); j++) {
a[i][j] = value;
a[j][i] = value;
a[matrixLength - i - 1][j] = value;
a[j][matrixLength - i - 1] = value;

}

value--;
}

for (int i = 0; i < matrixLength; i++) {

for (int j = 0; j < matrixLength; j++) {

System.out.print(a[i][j] + " ");
}
System.out.println();
}

}
}

- Anonymous May 12, 2014 | Flag Reply
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0
of 0 vote

import java.io.*;
import java.util.*;
public class Trial {
    public static void main(String [] args)
	{
		int n, row, col,i,j;
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		row = 2*n - 1;
		col = row;
		int [][] matrix = new int [row][col];
		for(i=0; i<row; i++)
		{
			for( j=i; j<col-i; j++)
			{
				matrix[i][j] = n;
			}
			for(j=i+1; j<col-i; j++)
			{
				matrix[j][col-i-1] = n;
			}
			for(j=i; j<col-i-1; j++)
			{
				matrix[col-i-1][j] = n;
			}
			for(j=i+1; j<col-i-1; j++)
			{
				matrix[j][i] = n;
			}
			n--;
		}
		
		for(i=0; i<row; i++)
		{
			for(j=0;j<col; j++)
				System.out.print(matrix[i][j] +" ");
			System.out.println();
		}
	}

}

- Anonymous May 18, 2014 | Flag Reply
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0
of 0 vote

public class PrintNums {

public static void main(String[] args) {
// TODO Auto-generated method stub
int N = 9, k = 0;
int [][] matrix = new int [2*N-1][2*N-1];
int counter = 0, counterLast = N-2;

for(int i = 0; i<matrix.length; i++){
for(int j = 0; j<matrix[0].length ; j++){
if(i < (matrix.length + 1)/2){
if( i == 0){
matrix[i][j] = N;
}
else if(j < k || j > (matrix[0].length-1) - k){
matrix[i][j] = matrix[i-1][j];
}
else{
matrix[i][j] = N - k;
}

}
else{
matrix[i][j] = matrix[counterLast][j];
}
}
k++;
if(i >= (matrix.length + 1)/2){
counterLast--;
}
}

for(int i = 0; i<matrix.length; i++){
System.out.println();
for(int j = 0; j<matrix.length; j++){
System.out.print(matrix[i][j] + " ");
}
}
}

}

- Will B July 03, 2014 | Flag Reply
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0
of 0 vote

#include<stdio.h>
#include<conio.h>
int main()
{
int i,j,k,row,col,x,n,tmp,temp;
printf("Enter the number:");
scanf("%d",&n);
row=n-1;
col=n+row;
x=1-row;
temp=col-1;
int a[col][col];
for(tmp=0;tmp<col;tmp++)
{
for(i=0;i<col;i++)
{
for(j=0;j<col;j++)
{
if(i==tmp || j==tmp || i==temp || j==temp)
a[i][j]=x;
}
}
temp--;
x++;
}
for(i=0;i<col;i++)
{
for(j=0;j<col;j++)
{
printf("%d",a[i][j]);
}
printf("\n");
}
getch();
}

- Anonymous July 06, 2014 | Flag Reply
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of 0 vote

#include<stdio.h>

void main()
{

int i,j,n=0,k=0,l=0;

printf("\n Enter n \n");
scanf("%d",&n);

printf("\n\n");
for(i=0;i<(2*n)-1;i++)
{
for(j=0;j<(2*n)-1;j++)
{

if(j>=n)
k=(2*n)-j-2;
else
k=j;

if(i>=n)
l=(2*n)-i-2;
else
l=i;

if((j==0) || (j==(2*n)-2) || (i==0) || (i==(2*n)-2))
printf("%d",n);
else if(k==l)
printf("%d",n-k);
else if(l<k)
printf("%d",n-l);
else
printf("%d",n-k);
}
printf("\n");

}

}

- Kunal Bansal July 18, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

write a program to print following pattern
0
101
21012
3210123
432101234
54321012345
432101234
3210123
21012
101
0

- Anonymous August 11, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include<stdio.h>
int main()
{ int a[7],j,z,k,m,o,i,p,l;
m=4;
i=0;
z=6;
p=4;
for(j=0;j<=6;j++)
{if(j<=3)
{for(k=j;k<=z;k++)
{ a[k]=m;
}
for(o=0;o<=6;o++)
{printf("%d",a[o]);}
m--;
z=z-1;
}

else
{
for(l=p;l<=j;l++)
{a[p-1]=a[p];}
for(o=0;o<=6;o++)
{printf("%d",a[o]);}

}
printf("\n");
p=p-1;
}
return 0;
}

- Anonymous February 02, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include<stdio.h>
int main()
{ int a[7],j,z,k,m,o,i,p,l;
m=4;
i=0;
z=6;
p=4;
for(j=0;j<=6;j++)
{if(j<=3)
{for(k=j;k<=z;k++)
{ a[k]=m;
}
for(o=0;o<=6;o++)
{printf("%d",a[o]);}
m--;
z=z-1;
}

else
{
for(l=p;l<=j;l++)
{a[p-1]=a[p];}
for(o=0;o<=6;o++)
{printf("%d",a[o]);}

}
printf("\n");
p=p-1;
}
return 0;
}

- Anonymous February 02, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include<stdio.h>
int main()
{ int a[7],j,z,k,m,o,i,p,l;
m=4;
i=0;
z=6;
p=4;
for(j=0;j<=6;j++)
{if(j<=3)
{for(k=j;k<=z;k++)
{ a[k]=m;
}
for(o=0;o<=6;o++)
{printf("%d",a[o]);}
m--;
z=z-1;
}

else
{
for(l=p;l<=j;l++)
{a[p-1]=a[p];}
for(o=0;o<=6;o++)
{printf("%d",a[o]);}

}
printf("\n");
p=p-1;
}
return 0;
}

- Anonymous February 02, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class Beta {
	public static void main(String args[])
	{
		
		int i=0,j=0,N=9,k=0,n=2*N-1;
		int a[][]=new int[17][17];
		
		for(k=0;k<=n/2;k++)
		{
		
		for(i=k;i<n-k;i++)
		{
			for(j=k;j<n-k;j++)
			{
				a[i][j]=N;
			}
			
		}
		
		N=N-1;
		}
		
		for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++)
			{
				System.out.print(a[i][j]);				
			}
			System.out.print("\n");
		}
		
		
	}
 }

- Anonymous April 02, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class Beta {
	public static void main(String args[])
	{
		
		int i=0,j=0,N=9,k=0,n=2*N-1;
		int a[][]=new int[17][17];
		
		for(k=0;k<=n/2;k++)
		{
		
		for(i=k;i<n-k;i++)
		{
			for(j=k;j<n-k;j++)
			{
				a[i][j]=N;
			}
			
		}
		
		N=N-1;
		}
		
		for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++)
			{
				System.out.print(a[i][j]);				
			}
			System.out.print("\n");
		}
		
		
	}
 }

- Anonymous April 02, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class Beta {
public static void main(String args[])
{

int i=0,j=0,N=9,k=0,n=2*N-1;
int a[][]=new int[17][17];

for(k=0;k<=n/2;k++)
{

for(i=k;i<n-k;i++)
{
for(j=k;j<n-k;j++)
{
a[i][j]=N;
}

}

N=N-1;
}

for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
System.out.print(a[i][j]);
}
System.out.print("\n");
}


}
}

- Anonymous April 02, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

Use printf!

- Anonymous March 09, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

ascii a stupid question, get a stupid ansi.

- Anonymous March 09, 2014 | Flag


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