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

For example, Pattern for ‘N’ = 4 will be. 1 232 34545 4567654

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

Constraints:

1

Question

Pattern: Triangle of numbers

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

For example, Pattern for ‘N’ = 4 will be.
   1
  232
 34545
4567654

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.

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

In pattern based programs, two things need to be taken care of : No. of lines and If the pattern is increasing or decreasing per line.
Pseudo Code :
printPattern(n) :
num = 1;
gap = n - 1;
for (j=1 to n , j++):
{
num = j;
for (i=1 to gap, i++)
print(" ")
gap --;
for (i=1 to j, i++)
{
print num
num++;
}
num--;
num--;
for (i=1 to i {
print num
num--;
}
print new line
}

Accepted Answer

Observation

We can observe that there is an ‘N’ number of rows where ‘N’ is the total number of rows to be printed. And each row exactly contains ‘N’ - ‘row’ number of spaces where ‘row’ denotes the row number starting from topmost as 1. Also, when we go left to right the value of each element in a particular row increases by one starting from the ‘row’ number and then decrements in the same column from 2* ‘row’ - 1 till 'row'.

Algorithm:

Declare an array of strings say ‘result’ for storing the pattern.
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 spaces from ‘space’ = 1 to ‘space’ = ‘N’ - ‘row’ (where ‘row’ is the current row number).
- Initialize a variable say ‘val’ = ‘row’.
- Run an inner loop from ‘currCol’ = 1 to ‘currCol’ = ‘row' to printing values in increasing order.
  - Append ‘val’ to the current string. and increment ‘val’ by 1.
- Set ‘val’ = 2 * (‘row’ - 1)
- Run an inner loop from ‘currCol’ = ‘val’ to ‘currCol’ = ‘row' to print values in decreasing order.
  - Append ‘currCol’ to the current string.
- Add current string into ‘result’.
Return ‘result’.

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 number of rows and for each row number, we will be traversing it from 1 to that row number for printing spaces, and traversing 1 to row number, and 2* row -1 to row for printing the columns of the pattern, Hence the overall time complexity will be O(N ^ 2).

Python (3.5)

'''

    Time Complexity: O(N ^ 2)

    Space Complexity: O(N ^ 2)



    Where 'N' is the given number of rows.

'''





def trianglePattern(n):

	

	# Declaring a array of string for storing the pattern

	result = []

	

	# Keeping track of the number of rows and initializing its value to 'n'.

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

		

		# Declaring a string to store our row in the pattern 

		ans = ""

		

		# Keeping track of the number of spaces in each row

		

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

			ans+=' '

			

		

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

		val = row



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

			

			# We append the value of 'val' in the current row.  

			ans += str(val)

			val += 1

			

		val = 2 * (row - 1)



		for currCol in range(val, row - 1, - 1):



			# We append the value of 'currCol' in the current row.

			ans += str(currCol)

			

		result.append(ans)

		

	return result

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[] TrianglePattern(int N) {

		// Declaring an array of string for storing the pattern.
		String result[] = new String[N];

		// Result index marker
		int index = 0;

		// Keeping track of the number of rows.
		for (int row = 1; row <= N; row++) {

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

			// Keeping track of the number of spaces in each row.
			for (int space = 1; space <= N - row; space++) {
				ans += ' ';
			}

			// 'currCol' represents our current column in our current row.
			int val = row;

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

				// We append the value of val in the current row.

				ans += val;
				val++;
			}

			val = 2 * (row - 1);

			for (int currCol = val; currCol >= row; currCol--) {

				// We append the value of 'currCol' in the current row.
				ans += currCol;
			}

			result[index++] = 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 TrianglePattern(int N)

{

    

    // Declaring an array of string for storing the pattern.

    vector result;



    // Keeping track of the number of rows and initializing its value to 'N'.

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

    {

        

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

        string ans="";



        // Keeping track of the number of spaces in each row.

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

        {

            ans+=' ';

        }



         

        // 'currCol' represents our current columnn in our current row.

        int val = row;



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

        {

            

            // We append the value of â€˜val' in the current row.

            

            ans += to_string(val);

            val++;

        }



        val = 2 * (row - 1);



        for(int currCol = val; currCol >= row; currCol--)

        {

            

            // We append the value of 'currCol' in the current row.

            ans += to_string(currCol);

        }



        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.

Input Format:

Output Format:

Constraints:

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