How do you solve Consecutive Numbers Sum on LeetCode?

Consecutive Numbers Sum (LeetCode #829) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(sqrt(n)) time complexity and O(1) space complexity. Key insight: The number of ways to express n as a sum of consecutive integers is related to the divisors of n.

What is the brute force solution for Consecutive Numbers Sum?

The brute force approach for Consecutive Numbers Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible sequences of consecutive integers that could sum up to the given number n. We can start from each integer and keep adding the next integers until we either exceed n or find a valid sum. The optimal Optimal Solution approach improves this to O(sqrt(n)).

What algorithm or data structure is used to solve Consecutive Numbers Sum?

Consecutive Numbers Sum uses the following concepts: Math, Enumeration. The recommended patterns to study are: Mathematical Enumeration, Two Pointers.

What are the interview tips for Consecutive Numbers Sum?

Always start with a brute force approach to understand the problem better. Look for patterns in the problem that can lead to a more efficient solution. Practice deriving mathematical formulas from problem statements to improve your problem-solving skills.

What are common mistakes when solving Consecutive Numbers Sum?

Not considering all possible starting points for consecutive sums. Failing to check if the sum exceeds n before counting.

#829

Consecutive Numbers Sum

Hard

MathEnumerationMathematical EnumerationTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(sqrt(n))
Space	O(1)	O(1)

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Intuition

Time O(sqrt(n))Space O(1)

The optimal approach leverages the mathematical properties of consecutive sums. We can express n as a sum of k consecutive integers starting from a number x, which leads to a formula that can be rearranged to find valid sequences efficiently.

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Algorithm

5 steps

1Step 1: Initialize a count variable to 0.
2Step 2: Loop through possible lengths k from 1 to sqrt(2*n).
3Step 3: For each k, check if (n - k*(k-1)/2) is divisible by k.
4Step 4: If it is divisible, increment the count.
5Step 5: Return the count.

solution.py8 lines

1def consecutiveNumbersSum(n):
2    count = 0
3    k = 1
4    while k * (k + 1) // 2 <= n:
5        if (n - k * (k - 1) // 2) % k == 0:
6            count += 1
7        k += 1
8    return count

ℹ

Complexity note: The time complexity is O(sqrt(n)) because we only loop through potential lengths k up to the square root of n, which is much more efficient than the brute force approach.

1The number of ways to express n as a sum of consecutive integers is related to the divisors of n.
2Each valid sequence can be represented with a starting point and a length, leading to a mathematical formula.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.