Last Updated: 27 Nov, 2020

Diamond of numbers

Moderate

Asked in company

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Problem statement

Students at Ninja School were bored with their daily routine and wanted to try something new. So, they thought about observing patterns. While observing the patterns, they found a very interesting pattern which they wished to explore. Your task is to help the students to visualize the pattern for any given ‘N’.

Given an integer ‘N’ you need to return the corresponding pattern for it.

For example, pattern for N = 15 will be :

          1 
         123 
        12345 
       1234567 
      123456789 
     12345678912 
    1234567891234 
   123456789123456 
    1234567891234 
     12345678912 
      123456789 
       1234567 
        12345 
         123 
          1

Input format :

The first line contains a single integer ‘T’ denoting the number of test cases to be run. Then the test cases follow.
The first and the only line of each test case contains a single integer 'N' denoting the number of rows.

Output format :

For each test case return the corresponding pattern in the next ‘N’ lines.

Note :

You are not required to print anything, it has already been taken care of. Just implement the function.

Constraints :

1 <= T <= 10    
1 <= N <= 500

Time Limit : 1 sec

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Approaches

01 Approach

In this pattern, each row starts with 1 and also ends with 1 and it is a diamond pattern so every row contains odd items.

Algorithm :

Declare the 2-D vector ‘VECT’ and push N empty vectors in it.
Initialize variable ‘COUNT’ to (N/2 + 1), variable ‘I’ to 0, and variable ‘M’ to ‘I’.
Iterate from 1 to COUNT (say, iterator = ‘i’).
1. Iterate from 1 to (‘COUNT’ - ‘i’) (say, iterator = ‘j’).
2. Push space (“ ”) in the 2-D vector.
3. Initialize ‘TEMP’ to ‘1’.
4. Iterate from 1 to ( 2 * i - 1 ) (say, iterator = ‘k’).
  1. Push the value of ‘TEMP’ in ‘VECT’.
  2. Increment ‘TEMP’ by 1.
  3. If the value of ‘TEMP’ equals ‘ : ’ , then , initialize ‘TEMP’ to ‘1’.
  4. Increment the ‘M’.
Iterate from (‘COUNT’-1) to 1 (say, iterator = ‘i’).
1. Iterate from (‘COUNT’ - 1) to ‘i’ (say, iterator = ‘j’).
  1. Push space (“ ”) in the ‘VECT’.
  2. Initialize ‘TEMP’ to ‘1’.
  3. Iterate from 1 to ( 2 *i - 1 ) (say, iterator = ‘k’).
  4. Push the value of ‘TEMP’ in the ‘VECT’.
  5. Increment ‘TEMP’ by 1.
  6. If the value of ‘TEMP’ equals ‘ : ’ , then , initialize ‘TEMP’ to ‘1’.
  7. Increment the ‘M’.
Return ‘VECT’.

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