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829. Consecutive Numbers Sum

Description

Given an integer n, return the number of ways you can write n as the sum of consecutive positive integers.

 

Example 1:

Input: n = 5
Output: 2
Explanation: 5 = 2 + 3

Example 2:

Input: n = 9
Output: 3
Explanation: 9 = 4 + 5 = 2 + 3 + 4

Example 3:

Input: n = 15
Output: 4
Explanation: 15 = 8 + 7 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5

 

Constraints:

  • 1 <= n <= 109

Solutions

Solution 1: Mathematical Derivation

Consecutive positive integers form an arithmetic sequence with a common difference $d = 1$. Let's assume the first term of the sequence is $a$, and the number of terms is $k$. Then, $n = (a + a + k - 1) \times k / 2$, which simplifies to $n \times 2 = (a \times 2 + k - 1) \times k$. From this, we can deduce that $k$ must divide $n \times 2$ evenly, and $(n \times 2) / k - k + 1$ must be an even number.

Given that $a \geq 1$, it follows that $n \times 2 = (a \times 2 + k - 1) \times k \geq k \times (k + 1)$.

In summary, we can conclude:

  1. $k$ must divide $n \times 2$ evenly;
  2. $k \times (k + 1) \leq n \times 2$;
  3. $(n \times 2) / k - k + 1$ must be an even number.

We start enumerating from $k = 1$, and we can stop when $k \times (k + 1) > n \times 2$. During the enumeration, we check if $k$ divides $n \times 2$ evenly, and if $(n \times 2) / k - k + 1$ is an even number. If both conditions are met, it satisfies the criteria, and we increment the answer by one.

After finishing the enumeration, we return the answer.

The time complexity is $O(\sqrt{n})$, where $n$ is the given positive integer. The space complexity is $O(1)$.

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class Solution:
    def consecutiveNumbersSum(self, n: int) -> int:
        n <<= 1
        ans, k = 0, 1
        while k * (k + 1) <= n:
            if n % k == 0 and (n // k - k + 1) % 2 == 0:
                ans += 1
            k += 1
        return ans

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class Solution {

    public int consecutiveNumbersSum(int n) {
        n <<= 1;
        int ans = 0;
        for (int k = 1; k * (k + 1) <= n; ++k) {
            if (n % k == 0 && (n / k + 1 - k) % 2 == 0) {
                ++ans;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int consecutiveNumbersSum(int n) {
        n <<= 1;
        int ans = 0;
        for (int k = 1; k * (k + 1) <= n; ++k) {
            if (n % k == 0 && (n / k + 1 - k) % 2 == 0) {
                ++ans;
            }
        }
        return ans;
    }
};

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func consecutiveNumbersSum(n int) int {
    n <<= 1
    ans := 0
    for k := 1; k*(k+1) <= n; k++ {
        if n%k == 0 && (n/k+1-k)%2 == 0 {
            ans++
        }
    }
    return ans
}

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function consecutiveNumbersSum(n: number): number {
    let ans = 0;
    n <<= 1;
    for (let k = 1; k * (k + 1) <= n; ++k) {
        if (n % k === 0 && (Math.floor(n / k) + 1 - k) % 2 === 0) {
            ++ans;
        }
    }
    return ans;
}

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