Print Pattern

Problem Description

Print a sequence of numbers starting with nn, without using a loop. Replace nn with n−5n - 5n−5 until n≤0n \leq 0n≤0. Then, replace n with n+5n + 5n+5 until nn regains its initial value. Complete the function pattern(n) which takes n as input and returns a list containing the pattern.

Examples

Example 1:

Input: n = 16
Output: 16 11 6 1 -4 1 6 11 16
Explanation: The value decreases until it is greater than 0. After that it increases and stops when it becomes 16 again.

Example 2:

Input: n = 10
Output: 10 5 0 5 10
Explanation: It follows the same logic as per the above example.

Expected Time Complexity: $O(n)$

Expected Auxiliary Space: $O(n)$ for dynamic programming

Constraints

• -10^5 ≤ n ≤ 10^5

Problem Explanation

Print a sequence of numbers starting with nn, without using a loop. Replace nn with n−5n - 5n−5 until n≤0n \leq 0n≤0. Then, replace n with n+5n + 5n+5 until nn regains its initial value. Complete the function pattern(n) which takes n as input and returns a list containing the pattern.

Code Implementation

Python

Written by @Ishitamukherjee2004

def pattern(n, initial=None, result=None):
  if result is None:
      result = []
  if initial is None:
      initial = n
  if n > 0:
      result.append(n)
      return pattern(n - 5, initial, result)
  elif n < initial:
      result.append(n)
      return pattern(n + 5, initial, result)
  return result

C++

Written by @Ishitamukherjee2004

vector<int> pattern(int n) {
  vector<int> result;
  while (n > 0) {
      result.push_back(n);
      n -= 5;
  }
  while (n < result[0]) {
      result.push_back(n);
      n += 5;
  }
  return result;
}

Java

Written by @Ishitamukherjee2004

public List<Integer> pattern(int n) {
  List<Integer> result = new ArrayList<>();
  while (n > 0) {
      result.add(n);
      n -= 5;
  }
  while (n < result.get(0)) {
      result.add(n);
      n += 5;
  }
  return result;
}

Time Complexity

• The iterative approach has a time complexity of $O(n)$ because we are iterating through the sequence of numbers twice: once from n to 0, and once from 0 to n.

Space Complexity

• The space complexity is O(n) because we are storing the sequence of numbers in the result list.