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

# Given a positive integer N,
# how many ways can we write it as a sum of
# consecutive positive integers?
#
# Example 1:
#
# Input: 5
# Output: 2
# Explanation: 5 = 5 = 2 + 3
# Example 2:
#
# Input: 9
# Output: 3
# Explanation: 9 = 9 = 4 + 5 = 2 + 3 + 4
# Example 3:
#
# Input: 15
# Output: 4
# Explanation: 15 = 15 = 8 + 7 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5
# Note: 1 <= N <= 10 ^ 9.


class Solution(object):
    def consecutiveNumbersSum(self, N):
        """
        :type N: int
        :rtype: int
        """
        # x + x+1 + x+2 + ... + x+l-1 = N = 2^k * M
        # => l*x + (l-1)*l/2 = N
        # => x = (N -(l-1)*l/2)/l= 2^k * M/l - (l-1)/2 is integer
        # => the answer is the number of all odd factors of M
        # if prime factorization of N is 2^k * p1^a * p2^b * ..
        # => answer is the number of all odd factors = (a+1) * (b+1) * ...
        result = 1
        while N % 2 == 0:
            N /= 2
        i = 3
        while i*i <= N:
            count = 0
            while N % i == 0:
                N /= i
                count += 1
            result *= count+1
            i += 2
        if N > 1:
            result *= 2
        return result

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

# Given a positive integer N,
# how many ways can we write it as a sum of
# consecutive positive integers?
#
# Example 1:
#
# Input: 5
# Output: 2
# Explanation: 5 = 5 = 2 + 3
# Example 2:
#
# Input: 9
# Output: 3
# Explanation: 9 = 9 = 4 + 5 = 2 + 3 + 4
# Example 3:
#
# Input: 15
# Output: 4
# Explanation: 15 = 15 = 8 + 7 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5
# Note: 1 <= N <= 10 ^ 9.


class Solution(object):
    def consecutiveNumbersSum(self, N):
        """
        :type N: int
        :rtype: int
        """
        # x + x+1 + x+2 + ... + x+l-1 = N = 2^k * M
        # => l*x + (l-1)*l/2 = N
        # => x = (N -(l-1)*l/2)/l= 2^k * M/l - (l-1)/2 is integer
        # => l could be 2 or any odd factor of M (excluding M)
        # => the answer is the number of all odd factors of M
        # if prime factorization of N is 2^k * p1^a * p2^b * ..
        # => answer is the number of all odd factors = (a+1) * (b+1) * ...
        result = 1
        while N % 2 == 0:
            N /= 2
        i = 3
        while i*i <= N:
            count = 0
            while N % i == 0:
                N /= i
                count += 1
            result *= count+1
            i += 2
        if N > 1:
            result *= 2
        return result

Update consecutive-numbers-sum.py