Number Pattern
A high-security meeting has been arranged. Tables for the delegates and security personnel have been arranged. A total of ‘N’ rows of tables has been set up. The first row has one table, the second row has two, and so on. To ensure maximum security, the tables on either end of each row have been assigned for security personnel each. If there is only one table in a row, it will be assigned to a security personnel. The tables assigned for security personnel will host exactly one security personnel each. All the other tables will host two guests each.

You are given an integer ‘N’, which denotes the number of rows., You are supposed to print the table pattern indicating the number of guests or security personnel at each table. In other words, print the number of people in each table.

For example, if the number of rows are 4, the table pattern is as follows:

1 11 121 1221

Input Format:

The first line contains ‘T’, denoting the number of test cases. Each test case contains a single integer ‘N’, denoting the number of rows.

Output Format:

For each test case, print 'N' strings denoting the pattern.

Note:

You are not required to print the expected output. It has already been taken care of. Just implement the function.

Constraints:

1

Question

Number Pattern

A high-security meeting has been arranged. Tables for the delegates and security personnel have been arranged. A total of ‘N’ rows of tables has been set up. The first row has one table, the second row has two, and so on. To ensure maximum security, the tables on either end of each row have been assigned for security personnel each. If there is only one table in a row, it will be assigned to a security personnel. The tables assigned for security personnel will host exactly one security personnel each. All the other tables will host two guests each.

You are given an integer ‘N’, which denotes the number of rows., You are supposed to print the table pattern indicating the number of guests or security personnel at each table. In other words, print the number of people in each table.

For example, if the number of rows are 4, the table pattern is as follows:

Input Format:

The first line contains ‘T’, denoting the number of test cases.

Each test case contains a single integer ‘N’, denoting the number of rows.

Output Format:

For each test case, print 'N' strings denoting the pattern.

Note:

You are not required to print the expected output. It has already been taken care of. Just implement the function.

Constraints:

1 <= T <= 10
1 <= N <= 10^3

Where ‘N’ is the number of lines.

Time Limit: 1 sec

Accepted Answer

Step 1 : For rows of rectangle ran the outer loop from 1 to rows.
Step 2 : For column of rectangle run the inner loop from 1 to columns.
Step 3 : Print star for first or last row or for first or last column, otherwise print blank space.
Step 4 : After printing all columns of a row, print new line after inner loop.
There was no restriction of language on this question

Accepted Answer

Brute Force

The idea here is to traverse all the rows and, for each row, print the suitable character at any particular table.

The steps are as follows:

We define a ‘RESULT’ matrix.
We traverse from ‘i’ = 1 to ‘N’, indicating the row number
- We then traverse from ‘j’ = 1 to ‘i’, indicating the tables of each line.
  - If it is the first or the last character, that is ‘j’ == 1 or ‘j’ == ‘i’, we insert ‘1’ in the ‘RESULT’ matrix.
  - Else we insert ‘2’ in the ‘RESULT’ matrix.

Space Complexity: O(n^2)Explanation:

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

We are using a matrix of ‘N’ rows and ‘N’ columns. Hence, the overall space complexity becomes O(N^2).

Time Complexity: O(n^2)Explanation:

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

In this approach, we are inserting every table detail of each row one at a time. The first row contains one table, the second row two, and so on. Hence time required to print N rows is 1 + 2 + 3 + … + N = (N*(N+1))/2, which is equivalent to O(N^2).

Hence, the overall complexity is O(N^2).

C++ (g++ 5.4)

/*
    Time Complexity: O(N^2)
    Space Complexity: O(N^2)

    where 'N' is the number of rows.
*/

vector> numberPattern(int n)
{
    vector> result;
    result.resize(n);

    for (int i = 0; i < n; i++)
    {
        result[i].resize(i + 1);
        for (int j = 0; j <= i; j++)
        {
            // If end positions, then insert 1.
            if (j == 0 || j == i)
            {
                result[i][j] = 1;
            }
            // All other tables will have 2 guests.
            else
            {
                result[i][j] = 2;
            }
        }
    }

    return result;
}

Java (SE 1.8)

/*
    Time Complexity: O(N^2)
    Space Complexity: O(N^2)

    where 'N' is the number of rows.
*/

import java.util.ArrayList;

public class Solution {

    public static ArrayList> numberPattern(int n) {
        ArrayList> result = new ArrayList>(n);

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

            result.add(new ArrayList(i + 1));

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

                int currentAns;

                // If end positions, then insert 1.
                if (j == 0 || j == i) {
                    currentAns = 1;
                }
                // All other tables will have 2 guests.
                else {
                    currentAns = 2;
                }
                result.get(i).add(currentAns);
            }
        }

        return result;

    }
}

Python (3.5)

'''
    Time Complexity: O(N^2)
    Space Complexity: O(N^2)

    where 'N' is the number of rows.
'''

def numberPattern(n):

    result = [[] for k in range(n)]

    for i in range(n):

        for j in range(i + 1):

            # If end positions, insert 1.
            if (j == 0 or j == i):
                currentAns = 1

            # All other tables will have 2 guests.
            else:
                currentAns = 2

            result[i].append(currentAns)

    return result