Crazy Numbers
Ninja loves playing with numbers. So one day, he wants to arrange a few numbers in the ‘N’ number of rows. The first row contains 1 number, the second row has two numbers, the third row has 3 digits, and so on.

Ninja starts placing numbers in increasing order, with absolute difference 1, starting from 1 and continuing till he encounters 9, and then he again restarts from 1.

You are given an integer ‘N’ denoting the given number of rows. Can you print the pattern that Ninja wants to create?

Pattern for N = 4:

1 23 456 7891

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 the integers denoting the required pattern in the following ‘N’ lines.

Note:

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

Constraints:

1

Question

Crazy Numbers

Ninja loves playing with numbers. So one day, he wants to arrange a few numbers in the ‘N’ number of rows. The first row contains 1 number, the second row has two numbers, the third row has 3 digits, and so on.

Ninja starts placing numbers in increasing order, with absolute difference 1, starting from 1 and continuing till he encounters 9, and then he again restarts from 1.

You are given an integer ‘N’ denoting the given number of rows. Can you print the pattern that Ninja wants to create?

Pattern for N = 4:

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 the integers denoting the required pattern in the following ‘N’ lines.

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 <= 500

Time Limit: 1 sec

Accepted Answer

Brute force

The key here is to traverse all the lines sequentially, and for each line, we print the required character at the given index.

The steps are as follows:

We will define matrix ‘ans’ to store the final pattern.
We will initialize a variable ‘k’ to 1, which will be the starting value for the pattern.
We will iterate over all the rows, i.e., i = 0 to i = N - 1:
- We will run a for loop starting from j = N - i -1 with the condition that j is less than or equal to N:
  - If k is less than 10, we will insert the value in the matrix ‘ans’.
  - Otherwise, we will reset the value of k to 1 and then insert the value in the matrix ‘ans’.
We will return the matrix ‘ans’ as the final answer.

Space Complexity: O(n^2)Explanation:

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

We are using a matrix, to store the pattern. Hence, the overall space complexity is O(N^2).

Time Complexity: O(n^2)Explanation:

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

As we are printing every character of each line one at a time. The first line contains one character, the second line two, and so on. Hence the time required to print N lines is 1 + 2 + 3 + … + N = N * (N+1)/2, which is equivalent to O(N^2).

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

Java (SE 1.8)

/*

    Time Complexity: O(N^2)

    Space Complexity: O(N^2)



	Where 'N' denotes the number of lines of the pattern.

*/



import java.util.ArrayList;



public class Solution {



	public static ArrayList> numberPattern(int n) {



		// Initializing matrix to store the pattern

		ArrayList> ans = new ArrayList>(n);



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

			ans.add(new ArrayList(n));

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

				ans.get(i).add(-1);

			}

		}



		int k = 1;



		// For loop to create the pattern

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

			for (int j = n - i - 1; j < n; j++) {



				// If k becomes larger than 9 then reset it to 1

				if (k > 9) {

					k = 1;

				}

				// Storing integer values of the character

				ans.get(i).set(j, k);

				// Increment the value of K to print the next digit

				k++;

			}

		}

		// Return the matrix

		return ans;



	}

}

C++ (g++ 5.4)

/*

    Time Complexity: O(N^2)

    Space Complexity: O(N^2)



    Where 'N' denotes the number of lines of the pattern

*/



vector < vector < int > > numberPattern(int n) {



  // Initializing matrix to store the pattern

  vector < vector < int > > ans(n, vector < int > (n, -1));



  int k = 1;



  // For loop to create the pattern

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

    for (int j = n - i - 1; j < n; j++) {



      // If k becomes larger than 9 then reset it to 1

      if (k > 9) {

        k = 1;

      }

      // Storing integer values of the character

      ans[i][j] = k;

      // Increment the value of K to print the next digit 

      k++;

    }

  }

  // Return the matrix

  return ans;

}

Python (3.5)

'''

    Time Complexity: O(N^2)

    Space Complexity: O(N^2)



    Where 'N' denotes the number of lines of the pattern.



'''





def numberPattern(n):



    # Initializing matrix to store the pattern

    ans = [[-1 for j in range(n)] for i in range(n)]



    k = 1



    # For loop to create the pattern

    for i in range(n):



        for j in range(n - i - 1, n):



            # If k becomes larger than 9 then reset it to 1

            if k > 9:

                k = 1



            # Storing integer values of the character

            ans[i][j] = k



            # Increment the value of K to print the next digit

            k += 1



            

    # Return the matrix

    return ans

Accepted Answer

int N = 5;
int counter = 1;
for (int i = 1; i 9)
          counter = 1;
    }
    Console.WriteLine();
}

Here Time complexity TC=> O(N^2)
And SC => O(1)