We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsDisclaimersPrivacyTermsContactAbout Us

Consecutive Numbers Sum

Difficulty: Hard

Problem Description

Given an integer n, return the number of ways you can write n as the sum of consecutive positive integers.

Key Insights

  • A number n can be expressed as the sum of k consecutive integers starting from a number x if there exists an integer x such that n = x + (x + 1) + ... + (x + k - 1).
  • This can be simplified to the equation: n = kx + (k(k-1))/2, where (k*(k-1))/2 is the sum of the first k-1 integers.
  • Rearranging the equation gives: x = (n - (k*(k-1))/2) / k.
  • For x to be a positive integer, (n - (k*(k-1))/2) must be divisible by k and greater than 0.
  • The maximum value of k can be derived from the condition that (k*(k-1))/2 < n.

Space and Time Complexity

Time Complexity: O(sqrt(n))
Space Complexity: O(1)

Solution

The solution involves iterating over possible values of k to find valid starting integers x that satisfy the equation for sums of consecutive integers. By checking each k, we can count the valid configurations. The key is to ensure that (n - (k*(k-1))/2) is divisible by k and greater than zero.

Code Solutions

def consecutiveNumbersSum(n: int) -> int:
    count = 0
    k = 1
    while (k * (k - 1)) // 2 < n:
        if (n - (k * (k - 1)) // 2) % k == 0:
            count += 1
        k += 1
    return count
← Back to All Questions