829. Consecutive Numbers Sum - Detailed Explanation

Problem Statement

Given a positive integer n, find the number of ways to represent n as a sum of consecutive positive integers. In other words, count all possible sequences of one or more consecutive numbers that add up to n.

Example Inputs, Outputs, and Explanations

Example Input: n = 5
Output: 2
Explanation:

• The number 5 can be represented as itself: 5.
• It can also be represented as 2 + 3.

Example Input: n = 9
Output: 3
Explanation:

• 9 by itself.
• 4 + 5 equals 9.
• 2 + 3 + 4 equals 9.

Example Input: n = 15
Output: 4
Explanation:

• 15 by itself.
• 7 + 8 equals 15.
• 4 + 5 + 6 equals 15.
• 1 + 2 + 3 + 4 + 5 equals 15.

Constraints

• 1 ≤ n ≤ 10⁹

Hints

• Hint 1: Think about the formula for the sum of an arithmetic sequence. For a sequence starting at a with k consecutive numbers, the sum is:
  [ \text{sum} = a + (a+1) + \dots + (a+k-1) = k \times a + \frac{k \times (k-1)}{2} ] Set this equal to n and solve for a.

• Hint 2: For a valid sequence, a must be a positive integer. This means that ((n - k \times (k-1)/2)) must be divisible by k and the resulting a must be at least 1.

Approach: Brute Force Iterative Method

Explanation

We iterate over all possible lengths k (starting from 1) of consecutive sequences such that the sum of the first k natural numbers (i.e. (k \times (k-1)/2)) is less than n. For each k, we compute the potential starting number:

[ a = \frac{n - \frac{k \times (k-1)}{2}}{k} ]

If a is a positive integer, then the sequence starting at a with length k is a valid representation.

Walkthrough with n = 15

• k = 1:
  (a = \frac{15 - 0}{1} = 15)
  Valid sequence: [15]

• k = 2:
  (a = \frac{15 - 1}{2} = 7)
  Valid sequence: [7, 8]

• k = 3:
  (a = \frac{15 - 3}{3} = 4)
  Valid sequence: [4, 5, 6]

• k = 4:
  (a = \frac{15 - 6}{4} = \frac{9}{4}) (not an integer)
  Invalid sequence.

• k = 5:
  (a = \frac{15 - 10}{5} = 1)
  Valid sequence: [1, 2, 3, 4, 5]

• k = 6:
  (a = \frac{15 - 15}{6} = 0) (not positive)
  Stop further iterations.

Thus, the total count is 4.

Complexity Analysis

• Time Complexity:
  The loop runs while (k \times (k-1)/2 < n). This roughly implies k is up to (O(\sqrt{n})), so the overall time complexity is O(√n).

• Space Complexity:
  Only constant extra space is used, i.e., O(1).

Approach: Mathematical Divisor Counting Method

Explanation

There is a number theory insight stating that the number of ways to write n as a sum of consecutive positive integers is equal to the number of odd divisors of n. The reasoning involves rewriting the arithmetic sequence sum and analyzing the resulting factors. Although this approach is mathematically elegant, it requires a deeper understanding of divisor properties and is less straightforward than the iterative method.

Complexity Analysis

• Time Complexity:
  Counting divisors is typically O(√n) since you only need to check divisors up to the square root of n.

• Space Complexity:
  This approach also uses constant extra space, i.e., O(1).

Python Code Implementation