829. Consecutive Numbers Sum

// LeetCode URL: https://leetcode.com/problems/consecutive-numbers-sum/submissions/

/**
 * Given a number N, we can possibly write it as a sum of 1 number, 2 numbers, 3
 * numbers, 4 numbers and so on. Let's assume the fist number in this series be
 * x. Hence, we should have
 *
 * x + (x+1) + (x+2)+...+ k terms = N
 *
 * kx + k*(k-1)/2 = N implies 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 starting 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 which implies k*(k-1) < 2N which can be approximated to
 *
 * k*k < 2N ==> k < sqrt(2N)
 *
 * Time Complexity: O(sqrt(N))
 *
 * Space Complexity: O(1)
 */
class Solution {
    public int consecutiveNumbersSum(int N) {
        if (N <= 0) {
            return 0;
        }
        if (N <= 2) {
            return 1;
        }

        int count = 1;
        double endCond = Math.sqrt(2 * N);

        for (int k = 2; k < endCond; k++) {
            if ((N - (k * (k - 1) / 2)) % k == 0) {
                count++;
            }
        }

        return count;
    }
}