Solution Explanation: Consecutive Numbers Sum

This problem asks to find the number of ways a given integer n can be expressed as the sum of consecutive positive integers. The solution uses a mathematical approach to efficiently solve this problem.

Mathematical Derivation:

Let's represent the sum of consecutive positive integers as:

n = a + (a+1) + (a+2) + ... + (a+k-1)

where:

• a is the first term in the sequence (a positive integer).
• k is the number of terms in the sequence (also a positive integer).

This is an arithmetic series, and its sum can be expressed as:

n = k * (2a + k - 1) / 2

Multiplying both sides by 2, we get:

2n = k * (2a + k - 1)

Rearranging the equation to solve for a:

2a = (2n / k) - k + 1

a = (2n / k - k + 1) / 2

Since a must be a positive integer, this equation implies the following conditions:

1. k must be a divisor of 2n: 2n must be divisible by k without any remainder.
2. 2n/k - k + 1 must be even: This ensures that a is an integer. This is equivalent to checking if 2n/k and k have the same parity (both even or both odd).

Algorithm:

The algorithm iterates through possible values of k (the number of terms). For each k, it checks if the two conditions above are met. If both conditions are true, it means we've found a valid way to express n as a sum of k consecutive positive integers. The iteration continues until k * (k + 1) > 2n, because beyond this point, k would be too large to be a valid number of terms.

Time and Space Complexity:

• Time Complexity: O(√n). The loop iterates up to the square root of 2n, which is approximately O(√n).
• Space Complexity: O(1). The algorithm uses only a few variables to store intermediate values.

Code Implementation (Python):

class Solution:
    def consecutiveNumbersSum(self, n: int) -> int:
        n <<= 1  # Equivalent to n *= 2 (bitwise left shift for efficiency)
        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
 

The other languages (Java, C++, Go, TypeScript) follow a similar structure, efficiently implementing the algorithm described above. The bitwise left shift (n <<= 1) is a common optimization for multiplying by 2.

This approach provides an efficient solution to the problem by leveraging mathematical properties to reduce the search space and avoid unnecessary computations.