Consecutive Numbers Sum

Consecutive Numbers Sum - Problem

Math Enumeration Hard

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

A consecutive sequence means the integers are in order: 1, 2, 3 or 5, 6, 7, 8, etc.

Example: n = 9 can be written as:

9 (single number)
4 + 5 (two consecutive numbers)
2 + 3 + 4 (three consecutive numbers)

So the answer is 3 ways.

Input & Output

Example 1 — Basic Case

$ Input: n = 9

› Output: 3

💡 Note: Three ways: 9 (single), 4+5 (two consecutive), 2+3+4 (three consecutive)

Example 2 — Small Number

$ Input: n = 15

› Output: 4

💡 Note: Four ways: 15, 7+8, 4+5+6, 1+2+3+4+5

Example 3 — Power of 2

$ Input: n = 8

› Output: 1

💡 Note: Only one way: 8 (powers of 2 can only be expressed as single number)

Constraints

1 ≤ n ≤ 10⁹

Visualization

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Asked in

G Google 25 M Microsoft 18

The key insight is using the arithmetic sequence formula: sum of k consecutive numbers starting at a is k×a + k×(k-1)/2 = n. We solve for a and check if it's a positive integer. Best approach uses mathematical formula with Time: O(√n), Space: O(1).

Common Approaches

✓ Brute Force - Try All Sequences

⏱️ Time: O(n²) Space: O(1)

For each possible sequence length k (from 1 to n), try all starting numbers and check if their sum equals n. Use the arithmetic sequence sum formula.

Mathematical Formula

⏱️ Time: O(√n) Space: O(1)

For k consecutive numbers starting at a, the sum is k*a + k*(k-1)/2 = n. Solve for a: a = (n - k*(k-1)/2)/k. Check if a is a positive integer.

Brute Force - Try All Sequences — Algorithm Steps

For each sequence length k from 1 to n
For each starting number a from 1 to n
Calculate sum = k*a + k*(k-1)/2
If sum equals n, increment count

Visualization

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Step-by-Step Walkthrough

Try k=1

Single number: 9 = 9 ✓

Try k=2

Two numbers: 4+5 = 9 ✓

Try k=3

Three numbers: 2+3+4 = 9 ✓

Code -

solution.c — C

#include <stdio.h>

int solution(int n) {
    int count = 0;
    for (int k = 1; k <= n; k++) {  // sequence length
        for (int a = 1; a <= n; a++) {  // starting number
            int total = k * a + k * (k - 1) / 2;
            if (total == n) {
                count++;
            } else if (total > n) {
                break;
            }
        }
    }
    return count;
}

int main() {
    int n;
    scanf("%d", &n);
    int result = solution(n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops: outer loop runs n times, inner loop runs up to n times

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables for counting

✓ Linear Space

28.0K Views

Medium Frequency

~15 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try All Sequences — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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