How do you solve Sum Multiples on LeetCode?

Sum Multiples (LeetCode #2652) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: Understanding how to avoid double counting when summing multiples of multiple numbers is crucial.

What is the brute force solution for Sum Multiples?

The brute force approach for Sum Multiples is Brute Force, with O(n) time complexity and O(1) space complexity. We can simply iterate through each number from 1 to n and check if it is divisible by 3, 5, or 7. If it is, we add it to our sum. This is straightforward but not the most efficient. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Sum Multiples?

Sum Multiples uses the following concepts: Math. The recommended patterns to study are: Arithmetic Series, Mathematical Summation.

What are the interview tips for Sum Multiples?

Always start with the simplest approach to demonstrate your understanding before optimizing. Explain your thought process clearly, especially when transitioning from brute force to an optimal solution. Practice identifying patterns in problems, as this can help you recognize when to apply certain techniques.

What are common mistakes when solving Sum Multiples?

Forgetting to subtract the sums of the least common multiples, leading to incorrect results. Not considering edge cases, such as very small values of n.

#2652

Sum Multiples

Easy

MathArithmetic SeriesMathematical Summation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(1)Space O(1)

Instead of checking each number individually, we can calculate the sum of multiples of 3, 5, and 7 directly using the formula for the sum of an arithmetic series. This is much faster and avoids unnecessary checks.

⚙️

Algorithm

4 steps

1Step 1: Define a helper function to calculate the sum of multiples of a number 'x' up to 'n'.
2Step 2: Use the formula sum = x * (k * (k + 1)) / 2, where k = n // x.
3Step 3: Calculate the sum for 3, 5, and 7, and subtract the sums for their least common multiples (15, 21, 35) to avoid double counting.
4Step 4: Return the total sum.

solution.py9 lines

1# Full working Python code
2
3def sum_of_multiples(x, n):
4    k = n // x
5    return x * k * (k + 1) // 2
6
7def sum_multiples(n):
8    return (sum_of_multiples(3, n) + sum_of_multiples(5, n) + sum_of_multiples(7, n) - 
9            sum_of_multiples(15, n) - sum_of_multiples(21, n) - sum_of_multiples(35, n))

ℹ

Complexity note: The time complexity is O(1) because we are performing a constant number of calculations regardless of the size of n. The space complexity is O(1) as we are using a fixed amount of space for our variables.

1Understanding how to avoid double counting when summing multiples of multiple numbers is crucial.
2Using arithmetic series formulas can drastically reduce computation time.

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