Consecutive Numbers Sum

• Initialize count = 0.
• Iterate with an outer loop for start from 1 to n.
• Inside the outer loop, initialize currentSum = 0.
• Start an inner loop for j from start to n.
• Add j to currentSum.
• If currentSum == n, increment count and break the inner loop.
• If currentSum > n, break the inner loop.
• Return count.

We can use two nested loops. The outer loop iterates through all possible starting numbers start from 1 up to n. The inner loop adds consecutive numbers to start, forming a currentSum.

• If currentSum equals n, we've found a valid sequence, so we increment our counter and break the inner loop to try the next start.
• If currentSum exceeds n, the current sequence is too large, so we break the inner loop and move to the next start. This method is very intuitive but its inefficiency makes it impractical for large values of n.

1class Solution {2    public int consecutiveNumbersSum(int n) {3        int count = 0;4        for (int start = 1; start <= n; start++) {5            long currentSum = 0;6            for (int j = start; j <= n; j++) {7                currentSum += j;8                if (currentSum == n) {9                    count++;10                    break;11                }12                if (currentSum > n) {13                    break;14                }15            }16        }17        return count;18    }19}

• Initialize count = 0.
• Loop for k (the length of the sequence) starting from 1.
• The loop condition can be k * (k + 1) / 2 <= n, as this ensures the starting number x can be at least 1.
• Inside the loop, calculate the sum of the arithmetic series part: seriesSum = k * (k - 1) / 2.
• Calculate the remaining part that would be k*x: remainder = n - seriesSum.
• If remainder > 0 and remainder % k == 0, it means a valid positive integer x exists. Increment count.
• After the loop finishes, return count.

From the equation n = k*x + k*(k-1)/2, we can solve for x: x = (n - k*(k-1)/2) / k. For a given length k, a valid solution exists if:

1. x is an integer. This means (n - k*(k-1)/2) must be divisible by k.
2. x is positive. This means n - k*(k-1)/2 > 0, which implies n > k*(k-1)/2. This also gives us an upper bound for k. Since k*(k-1)/2 is roughly k^2/2, k can go up to approximately sqrt(2n).

The algorithm iterates k from 1 upwards as long as k*(k+1)/2 <= n. For each k, it checks if (n - k*(k-1)/2) is divisible by k. If it is, we've found a valid way.

1class Solution {2    public int consecutiveNumbersSum(int n) {3        int count = 0;4        // k is the number of terms in the series5        for (long k = 1; k * (k + 1) / 2 <= n; k++) {6            // From n = x*k + k*(k-1)/2, we get x*k = n - k*(k-1)/2.7            // We need to check if (n - k*(k-1)/2) is a positive multiple of k.8            // The loop condition k*(k+1)/2 <= n ensures x >= 1.9            if ((n - k * (k - 1) / 2) % k == 0) {10                count++;11            }12        }13        return count;14    }15}

• First, find the largest odd divisor of n. Keep dividing n by 2 until it becomes odd.
• Let the resulting odd number be m.
• Now, find the number of divisors of m. Initialize result = 1.
• Iterate through odd numbers d from 3 up to sqrt(m).
• If d divides m:
  • Count how many times d is a factor of m. Let this be exponent.
  • Update m by dividing it by d repeatedly: m = m / d.
  • Update the result: result = result * (exponent + 1).
• After the loop, if m > 1, it means the remaining m is a prime factor. Multiply result by 2.
• Return result.

Starting from the equation n = k*x + k*(k-1)/2, we can rearrange it to 2n = k(2x + k - 1). Let A = k and B = 2x + k - 1. We are looking for pairs of factors (A, B) of 2n. Notice that B - A = (2x + k - 1) - k = 2x - 1, which is always an odd number. This implies that one of A and B must be odd, and the other must be even.

Let the prime factorization of n be n = 2^p * m, where m is an odd number. Then 2n = 2^(p+1) * m. Any factor of 2n is a product of a power of 2 (from 2^(p+1)) and a factor of m. For A and B to have different parity, one must contain all the factors of 2, i.e., 2^(p+1), and the other must be purely odd.

This means that for every odd divisor d of n, we can form a pair of factors of 2n with different parity. This pair corresponds to a unique valid solution for x and k. Therefore, the number of ways is simply the number of odd divisors of n, which is the same as the number of divisors of its largest odd part, m.

The algorithm is to first find m by dividing n by 2 until it's odd, and then count the divisors of m.

1class Solution {2    public int consecutiveNumbersSum(int n) {3        // The problem is equivalent to finding the number of odd divisors of n.4        5        // Step 1: Remove all factors of 2 from n to get its largest odd part.6        while ((n & 1) == 0) { // while n is even7            n >>= 1; // n = n / 28        }9        10        int result = 1;11        // Step 2: Find the number of divisors of the remaining odd number n.12        // We do this by finding its prime factorization: n = p1^a1 * p2^a2 * ...13        // The number of divisors is (a1+1)*(a2+1)*...14        for (int i = 3; i * i <= n; i += 2) {15            int count = 0;16            while (n % i == 0) {17                n /= i;18                count++;19            }20            result *= (count + 1);21        }22        23        // If n is still greater than 1, it means the remaining n is a prime factor.24        if (n > 1) {25            result *= 2;26        }27        28        return result;29    }30}