Find politeness of a number

Given an integer n. Find politeness of number n. The politeness of a number is defined as the number of ways it can be expressed as the sum of consecutive integers. 

Examples:

Input: n = 15
Output: 3
Explanation:
There are only three ways to express
15 as sum of consecutive integers i.e.,
15 = 1 + 2 + 3 + 4 + 5
15 = 4 + 5 + 6
15 = 7 + 8
Hence answer is 3
Input: n = 9;
Output: 2
There are two ways of representation:
9 = 2 + 3 + 4
9 = 4 + 5

Naive approach:

Run a loop one inside another and find the sum of every consecutive integer up to n. The time complexity of this approach will be O(n²) which will not be sufficient for large value of n.

EfficientApproach:

Use factorization. We factorize the number n and count the number of odd factors. A total number of odd factors (except 1) is equal to the politeness of the number. Refer this for proof of this fact. In general, if a number can be represented as a^p * b^q * c^r ... where a, b, c, ... are prime factors of n. If a = 2 (even) then discard it and count total number of odd factors which can be written as [(q + 1) * (r + 1) * ...] - 1 (Here 1 is subtracted because single term in representation is not allowed). 
How does the above formula work? The fact is if a number is expressed as a^p * b^q * c^r ... where a, b, c, ... are prime factors of n, then a number of divisors is (p+1)*(q+1)*(r+1) ...... 
To simplify, let there be one factor, and the number is expressed as a^p. Divisors are 1, a, a², .... a^p. The count of divisors is p+1. Now let us take a slightly more complicated case a^pb^p. The divisors are : 
1, a, a², .... a^p
b, ba, ba², .... ba^p
b², b²a, b²a², .... b²a^p
................ 
................ 
b^q, b^qa, b^qa², .... b^qa^p
The count of the above terms is (p+1)*(q+1). Similarly, we can prove for more prime factors.
Illustration : For n = 90, decomposition of prime factors will be as follows:- 
=> 90 = 2 * 3² * 5¹. The power of odd prime factors 3, 5 are 2 and 1 respectively. Apply above formula as: (2 + 1) * (1 + 1) -1 = 5. Hence 5 will be the answer. We can crosscheck it. All odd factors are 3, 5, 9, 15 and 45.
Below is the program of the above steps:

Output:
Politeness of 90 = 5
Politeness of 15 = 3

Time complexity: O(sqrt(n)) 
Auxiliary space: O(1)
Reference:Wikipedia

Another Efficient approach:

Calculate if an AP can be generated for the given length domain [2, sqrt(2*n)]. The reason to calculate for length till sqrt(2*n) is-

max length will be for the AP 1, 2, 3...

Length for this AP is -
n= ( len * (len+1) ) / 2
len2 + len - (2*n) =0
so len?sqrt(2*n) 

so we can check for each len from 1 to sqrt(2*n) ,if AP can be generated with this len. The formula to get the first term of the AP is -

n= ( len/2) * ( (2*A1) + len-1 )

Below is the implementation of the above approach:

Politeness of 90 = 5
Politeness of 15 = 3

 Time complexity:  O(sqrt(2*n)) ? O(sqrt(n))                                                                                                                     

 Auxiliary space: O(1)                                                                                                                                                      

Example in c:

Approach:

Prompt the user to enter a number.
Read the number from the user and store it in a variable.
Initialize the politeness count to 0.
Initialize a variable to keep track of the running sum of consecutive numbers, starting from 1.
Initialize a variable to keep track of the starting number in the current consecutive sequence.
Loop through all numbers from 1 up to (num - 1).
Add the current number to the running sum.
If the running sum is greater than num, remove the starting number from the sum and increment the starting number until the sum is less than or equal to num.
If the running sum is equal to num, increment the politeness count.
Print the politeness count for the entered number.
End the program.

Enter a number: Politeness of 0: 0

Time complexity: O(N^2), where N is the given number. This is because the program needs to iterate through all possible sequences of consecutive integers that sum to N.

Space complexity: O(1), because the program uses only a fixed amount of memory regardless of the size of the input number.