Print the Pattern
Ninja is given a pattern. Now he is asked to print the same pattern for any given ‘N’ number of rows.

Note:

There is only one space between the values of each column in a row. For example, Pattern for ‘N’ = 5 will be. 1 2 3 4 5 11 12 13 14 15 21 22 23 24 25 16 17 18 19 20 6 7 8 9 10

Input Format:

The first line contains an integer 'T' which denotes the number of test cases or queries to be run. The first line of each test case contains a single integer ‘N’ denoting the number of rows for which pattern needs to be printed.

Output Format:

The output of each test case will be 'N' strings denoting the pattern printed for the given ‘N’ number of rows. The output of each test case will be printed on a separate line.

Constraints:

1

Question

Print the Pattern

Ninja is given a pattern. Now he is asked to print the same pattern for any given ‘N’ number of rows.

Note:

There is only one space between the values of each column in a row.

For example, Pattern for ‘N’ = 5 will be.
1 2 3 4 5 
11 12 13 14 15 
21 22 23 24 25 
16 17 18 19 20 
6 7 8 9 10

Input Format:

The first line contains an integer 'T' which denotes the number of test cases or queries to be run.

The first line of each test case contains a single integer ‘N’ denoting the number of rows for which pattern needs to be printed.

Output Format:

The output of each test case will be 'N' strings denoting the pattern printed for the given ‘N’ number of rows.

The output of each test case will be printed on a separate line.

Constraints:

1 <=T <= 5 
1 <= N <= 100

Where ‘T’ is the total number of test cases and ‘N’ is the given number of rows for which pattern needs to be printed.

Time limit: 1 sec

Accepted Answer

printPattern()
{

declare variables i,j,k,sp
initialise space to 4

for (i = 1 to i<=5) do :
{

// For printing the space
for (sp = space to sp >= 1) do :
{
print space
}

// For printing the $
for (j=1 to j<=i) do :
{
print '$'
}

for (k = 1 to k=1) do :
{
for (sp = space to sp >= 1) do :
{
print space
}

for (j=1 to j<=i) do :
{
print '$'
}

for (k = 1 to k < i) do :
{
print '$'
}

space++;
print new line
}

reset space to 3;

for (i = 2 to i<=5) do :
{
if (i is odd)
{
for (sp = space to sp>=1) do :
{
print space
}

for (j=1 to j<=i) do :
{
print '$'
}
}

if (i is odd)
{
print new line
space--;
}
}

}

Accepted Answer

Observation

We can observe that there are ‘N’ numbers of rows (where ‘N’ is the total number of rows to be printed). Each row contains a different value in the first column and its value increments by 1 for each subsequent column.

Algorithm:

Declare an array of strings for storing the pattern.
Declare a variable ‘startValue’ and initialize its value to 1.
Declare a variable ‘row’ for keeping track of the number of rows
Run an outer loop from ‘row’ = ‘1 to ‘row’ = ‘N’.
- Declare an empty string for storing the values in the current row.
- Run an inner loop for printing values from ‘col’ = ‘startvalue’ to ‘col’= ‘startvalue + N - 1’.
- For every row, we need to set the value of ‘startvalue‘ according to the following conditions:
  - If ‘row’ == (‘N’ + 1) / 2
    - If ‘N’ % 2 ! = 0
      - Set ‘startValue’ = ‘N’ * (‘N’ - 2) + 1
    - Else
      - Set ‘startValue’ = ‘N’ * (‘N’ - 1) + 1
  - Else If ‘row’ > (‘N’ + 1) / 2
    - Set ‘startValue’ = ‘startValue’ - 2 * ‘N’
  - Else
    - Set ‘startValue’ = ‘startValue’ + 2 * ‘N’
- After the loop, we append the string to our new row and move to a new row of our array.

Space Complexity: O(n^2)Explanation:

O(N ^ 2), where ‘N’ is the given number of rows.

As we are using an array of strings that takes O(N ^ 2) space.

Time Complexity: O(n^2)Explanation:

O(N ^ 2), where ‘N’ is the given number of rows.

As we will be traversing the given ‘N’ number of rows and for each row, we will be traversing it for ‘N’ no of columns so our complexity will be O(N ^ 2).

Java (SE 1.8)


/*
	Time Complexity: O(N ^ 2)
	Space Complexity: O(N ^ 2)
	
	Where N is the given number of rows.
*/
public class Solution {

	public static String[] NumberPattern(int N) {

		// Declaring a array of string for storing the pattern
		String result[]= new String[N];
		
		// Result index marker
		int index = 0;
		
		// Keeping track of the start value of each row
		int startValue = 1;

		for (int row = 1; row <= N; row++) {

			// Declaring a string to store our row in the pattern
			String ans = "";

			// Appending value of each column of each row in ans
			for (int col = startValue; col < startValue + N; col++) {
				ans += col + " ";
			}

			// 'col' represents our current columnn in our current row
			if (row == (N + 1) / 2) {
				if (N % 2 != 0) {
					startValue = N * (N - 2) + 1;
				} else {
					startValue = N * (N - 1) + 1;
				}
			} else if (row > (N + 1) / 2) {
				startValue = startValue - 2 * N;
			} else {
				startValue = startValue + 2 * N;
			}

			result[index++]=ans;
		}

		return result;
	}
}

Python (3.5)

'''

    Time Complexity: O(N ^ 2)

    Space Complexity: O(N ^ 2)



    Where N is the given number of rows.

'''



def printPattern(n):



	# Declaring a vector of string for storing the pattern

	result = []

	

	# Keeping track of the start value of each row

	 

	startValue = 1



	for row in range(1, n + 1):

		

		# Declaring a string to store our row in the pattern

		ans = ""

		

		# Appending value of each column of each row in ans



		for col in range(startValue, startValue + n):

			

			ans += str(col) + " "

			

		# 'col' represents our current columnn in our current row

		

		if (row == (n + 1) // 2):

			if (n % 2 != 0):

				

				startValue = n * (n - 2) + 1

				

			else:

				startValue = n * (n - 1) + 1

				

		elif (row > (n + 1) // 2):

			startValue = startValue - 2 * n

			

		else:

			startValue = startValue + 2 * n

			

			

		result.append(ans)

		

		

	return result

C++ (g++ 5.4)

/*

    Time Complexity: O(N ^ 2)

    Space Complexity: O(N ^ 2)



    Where N is the given number of rows.

*/



vector printPattern(int N)

{



    //Declaring a vector of string for storing the pattern



    vector result;



    //Keeping track of the start value of each row



    int startValue = 1;



    for (int row = 1; row <= N; row++)

    {



        //Declaring a string to store our row in the pattern



        string ans = "";



        // Appending value of each column of each row in ans



        for(int col = startValue; col < startValue + N; col++)

        {

            ans += to_string(col) + " ";

        }



        //'col' represents our current columnn in our current row



        if (row == (N + 1) / 2)

        {

            if (N % 2 != 0)

            {

                startValue = N * (N - 2) + 1;

            }

            else

            {

                startValue = N * (N - 1) + 1;

            }

        }

        else if (row > (N + 1) / 2)

        {

            startValue = startValue - 2 * N;

        }

        else

        {

            startValue = startValue + 2 * N;

        }



        result.push_back(ans);

    }



    return result;

}

Ninja is given a pattern. Now he is asked to print the same pattern for any given ‘N’ number of rows.

Note:

Input Format:

Output Format:

Constraints:

Top UBS ENO interview questions & answers

Top HR questions asked in UBS ENO