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Easy
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Average time to solve is 10m
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Problem statement

Ninja’s younger sister got a project to print the pattern for the given number of rows, say ‘N’. She finds it difficult to write such a pattern for every ‘N’.

Ninja decided to help his sister and decided to write a code for this but wasn't able to write the code for the same. Ninja wants help to write a code and asks you for help. Help Ninja.

Example:

Pattern for N = 2
    0
   101
  21012
Detailed explanation ( Input/output format, Notes, Images )
Input format : 
The first line of input contains an integer ‘T’ denoting the number of test cases. And each test case follows:

Each test case contains an integer ‘N’ representing the number of rows.
Output format : 
Each test case print 'N' strings denoting the required pattern of integers for a given integer ‘N’.

The output of each test case will be printed on a separate line
Constraints : 
1 <= T <= 5
1 <= N <= 100

Time Limit: 1 sec.

Sample Input 1 :

2
1
2

Sample Output 1 :

 0
101
  0
 101
21012
Explanation for Sample Input 1 :
Test Case 1:
 For ‘N’  = 1 , we need to print 2 rows - in first row 0, and second row 1 0 1.
Test Case 2:
The Number of rows for ‘N’ = 2 will be 3.

Sample Input 2 :

2
3
4

Sample Output 2 :

    0
   101
  21012
 3210123
432101234
     0
    101
   21012
  3210123
 432101234
54321012345
Hint

Observe the pattern and try using nested iterations.

Approaches (1)
Nested Iterations

The idea is to use nested iterations for generating spaces and the digits inside the pattern.

 

Approach :

 

  • First, initialize two integer variables, say ‘number1’ and ‘number2’ to print the digits in the pattern and initialize both the variables with value = 1;
  • Make an iteration with iterator pointer ‘i’ to print the number of rows which will iterate from 0 to ‘N’.
    • Make a nested iteration to generate the spaces in the pattern with an iterator pointer ‘j’ which will iterate from ‘N’ - 1 to ‘i’ and print the spaces.
    • Make an iteration to print the digits inside the pattern with an iterator pointer ’k’ from 1 to ‘number1’and print the value = (‘k’ - ‘number2’).
    • Increment ‘number1’ with value 2.
    • Increment ‘number2’ with value 1.
    • Leave a line after that.
Time Complexity

O(N ^ 2), where ‘N’ is the given integer.

 

As, we are making an iteration ‘N’ number of times to print different rows in the pattern and at max (2 * (N + 1)) columns for the pattern with nested iterations. Therefore, overall time complexity will be O(N ^ 2).

Space Complexity

O(1)

 

As we are using constant space. Therefore, space complexity will be O(1).

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