#829

Hard

LeetCode Problem Workspace

Consecutive Numbers Sum

Find the number of ways to express a number as the sum of consecutive positive integers.

Problem Statement

Given an integer n, your task is to return the number of ways n can be expressed as the sum of consecutive positive integers. Each representation should involve at least two integers.

For example, the number 5 can be written as 2 + 3, while the number 9 can be written as 4 + 5 and 2 + 3 + 4. Your solution should efficiently find all valid consecutive sequences up to the maximum constraint of n = 10^9.

Examples

Example 1

Input: n = 5

Output: 2

5 = 2 + 3

Example 2

Input: n = 9

Output: 3

9 = 4 + 5 = 2 + 3 + 4

Example 3

Input: n = 15

Output: 4

15 = 8 + 7 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5

Constraints

1 <= n <= 109

Solution Approach

Mathematical Insights

Start by recognizing that a sequence of consecutive integers can be represented algebraically. The sum of k consecutive integers starting from x is given by the formula: sum = k * x + (k * (k - 1)) / 2. For each k, solve for x and check if x is a positive integer.

Efficient Enumeration

Enumerate all possible lengths for the consecutive sequences (starting from 2). For each length k, check if there is a valid starting integer x that makes the sum equal to n. This approach ensures you only check feasible cases and reduces unnecessary computations.

Stopping Condition

Stop iterating once k exceeds the square root of n. This is because the sum of any sequence longer than the square root of n would exceed n, making further checks unnecessary.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach. A direct enumeration of possible sequence lengths results in a time complexity of O(sqrt(n)). Space complexity is O(1) since only a few variables are needed to store intermediate values.

What Interviewers Usually Probe

Can the candidate quickly recognize the mathematical pattern behind consecutive sums?
Does the candidate efficiently handle large inputs (up to 10^9)?
Is the candidate able to optimize the solution by reducing unnecessary checks or calculations?

Common Pitfalls or Variants

Common pitfalls

Forgetting that the starting number must be positive, which could lead to invalid sequences.
Misunderstanding the sum formula for consecutive integers, leading to incorrect solutions.
Not optimizing the solution by stopping once k exceeds sqrt(n), resulting in inefficient solutions.

Follow-up variants

Can this problem be generalized to find all sequences for a range of numbers?
How would the approach change if negative integers were allowed?
What if the sequence must include at least three integers instead of two?

FAQ

What is the Consecutive Numbers Sum problem?

The problem asks you to determine how many ways a number can be written as the sum of consecutive positive integers, with at least two integers in the sum.

How do I approach solving this problem?

Start by using the sum formula for consecutive integers and enumerate possible lengths for the sequences. Check if a valid starting integer exists for each length.

What is the time complexity of this solution?

The time complexity is O(sqrt(n)), as we iterate through possible sequence lengths up to the square root of n.

Why do we stop when k exceeds sqrt(n)?

Once the length of the sequence exceeds the square root of n, the sum would exceed n, making further checks unnecessary.

What are some common pitfalls in this problem?

Common mistakes include overlooking the requirement for positive starting integers, misusing the sum formula, and not optimizing by stopping early.

terminal

Solution

Solution 1: Mathematical Derivation

Consecutive positive integers form an arithmetic sequence with a common difference $d = 1$. Let's assume the first term of the sequence is $a$, and the number of terms is $k$. Then, $n = (a + a + k - 1) \times k / 2$, which simplifies to $n \times 2 = (a \times 2 + k - 1) \times k$. From this, we can deduce that $k$ must divide $n \times 2$ evenly, and $(n \times 2) / k - k + 1$ must be an even number.

1

2

3

4

5

6

7

8

9

class Solution:
    def consecutiveNumbersSum(self, n: int) -> int:
        n <<= 1
        ans, k = 0, 1
        while k * (k + 1) <= n:
            if n % k == 0 and (n // k - k + 1) % 2 == 0:
                ans += 1
            k += 1
        return ans

Continue Practicing

#869 reordered power of 2 #906 super palindromes #970 powerful integers