Star Pattern
Print the following pattern

Pattern for N = 4

The dots represent spaces.

Input Format:

The first line contains a single integer ‘T’ representing the number of test cases. The first line of each test case will contain an integer ‘N’ that is the total number of rows.

Output Format:

For each test case, print the pattern of N lines. Output for every test case will be printed in a separate line.

Constraints :

1

Question

Star Pattern

Print the following pattern

Pattern for N = 4

The dots represent spaces.

Input Format:

The first line contains a single integer ‘T’ representing the number of test cases. 

The first line of each test case will contain an integer ‘N’ that is the total number of rows.

Output Format:

For each test case, print the pattern of N lines.

Output for every test case will be printed in a separate line.

Constraints :

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

Accepted Answer

Tip 1 : A simple program of nested for loop. Display the star symbols at locations where row no. is equal to column no.

Accepted Answer

Brute Force

The basic idea of this approach is to print the pattern row-wise. The pattern consists of ’N’ rows. Each row contains 2 * ‘N’ - 1 star. In addition to these stars, there are leading spaces. Each row contains ‘N’ - ‘Row’ spaces (where ‘ROW’ is the current row number).

Here is the algorithm:

Run a loop for ‘ROW’ = 0 to ‘N’:
- To print spaces, create a variable ‘SPACES’ and run a loop till ’N’ - ‘ROW.’
- To print the stars ‘*’, run another loop from 1 to 2 * ‘ROW’ - 1:
  - Inside the loop print the star.
- Print a new line.

Space Complexity: O(1)Explanation:

O(1),

Because we are not using any extra space for finding our answer.

Time Complexity: O(n^2)Explanation:

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

Because we are using two loops, the outer loop runs for the number of rows given to us, and the inner loop prints the value for every column. Hence the time complexity is O(N^2).

Java (SE 1.8)

/*

    Time Complexity : O(N^2)
    Space Complexity : O(1)

    Where N denotes the total number of rows.


*/


public class Solution {

    public static void printPattern(int n) {
        
        // Outer loop determines the current row.
        for(int row = 1; row <= n; row++) {
            
            // The inner loop to print the spaces.
            int spaces = 0;
            while(spaces < n - row) {
                System.out.print(" ");
                spaces += 1;
            }
            
            // The loop to print the stars
            for(int col = 1; col <= 2 * row - 1; col++) {
                
                // Printing the value at every column, here value is 1.
                System.out.print("*");
            }
            
            System.out.println(" ");
        }
    }
}

Python (3.5)

'''

    Time Complexity : O(N^2)
    Space Complexity : O(1)

    Where N denotes the total number of rows.


'''


def printPattern(n):
    
    # Outer loop determines the current row.
    for row in range(1, n + 1):
        
        # The inner loop to print the spaces.
        spaces = 0
        
        while(spaces < (n - row)):
            print(" ", end ="")
            spaces += 1
            
        # The loop to print the stars.
        for col in range(1, (2*row-1) + 1, 1):
            
            # Printing the value at every column, here value is 1.
            print("*", end = "")
            
        print()

C++ (g++ 5.4)

/*

    Time Complexity : O(N^2)
    Space Complexity : O(1)

    Where N denotes the total number of rows.


*/

void printPattern(int n) {
            // Outer loop determines the current row.
        for(int row = 1; row <= n; row++)
        {
            // The inner loop to print the spaces.
            int spaces = 0;
            while(spaces < n - row) {
                cout << " ";
                spaces += 1;
            }
            
            // The loop to print the stars
            for(int col = 1; col <= 2 * row - 1; col++)
            {
                // Printing the value at every column, here value is 1.
                cout << "*";
            }
            cout << endl;
        }
}

Print the following pattern

Pattern for N = 4

The dots represent spaces.

Input Format:

Output Format:

Constraints :

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