829. Consecutive Numbers Sum

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.

 

Approach #1: Math. [Java]

class Solution {
    public int consecutiveNumbersSum(int N) {
        int count = 1;
        for (int k = 2; k < Math.sqrt(2*N); ++k) {
            if ((N - (k * (k - 1) / 2)) % k == 0)
                count++;
        }
        return count;
    }
}

　　

Analysis:

The though process goes like this - Given a number N, we can possibly write it as a sum of 2 numbers, 3 numbers, 4 numbers and so on. Let's assum the first number in this serise be x. Hence, we should have:

x + (x + 1) + (x + 2) + ... + (x + k) = N

kx + k * (k - 1) / 2 = N

kx = N - k * (k - 1) / 2

So, we can calculate the RHS for every value of k and if it is a multiple of k then we can construct a sum of N using k terms string from x.

Now, the question arises, till what value of k should we loop for?

That's easy. In the worst case, RHS should be greater than 0, That is 

N - k * (k - 1) / 2 > 0

k * (k - 1) < 2 * N

~ k * k < 2N ==> k < sqrt(2N)

Hence the overall complexity of the algorithm is O(sqrt(N)).

 

Reference:

https://leetcode.com/problems/consecutive-numbers-sum/discuss/129015/5-lines-C%2B%2B-solution-with-detailed-mathematical-explanation.