How do you solve Divisible and Non-divisible Sums Difference on LeetCode?

Divisible and Non-divisible Sums Difference (LeetCode #2894) 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: Using arithmetic properties can significantly reduce computation time.

What is the brute force solution for Divisible and Non-divisible Sums Difference?

The brute force approach for Divisible and Non-divisible Sums Difference is Brute Force, with O(n) time complexity and O(1) space complexity. The brute-force approach involves iterating through each number from 1 to n, checking if it is divisible by m, and summing the appropriate values. This method is straightforward but inefficient for larger values of n. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Divisible and Non-divisible Sums Difference?

Divisible and Non-divisible Sums Difference uses the following concepts: Math. The recommended patterns to study are: Arithmetic Progression, Mathematical Formulas.

What are the interview tips for Divisible and Non-divisible Sums Difference?

Always consider if there's a mathematical formula that can simplify your calculations. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice writing code for both brute force and optimal solutions to solidify your understanding.

What are common mistakes when solving Divisible and Non-divisible Sums Difference?

Not recognizing that the sum of the first n integers can be computed directly. Failing to handle edge cases where m is greater than n.

#2894

Divisible and Non-divisible Sums Difference

Easy

MathArithmetic ProgressionMathematical Formulas

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(1)Space O(1)

Instead of iterating through each number, we can use the formula for the sum of the first n integers and the properties of arithmetic sequences to calculate the sums directly. This approach is much faster and more efficient.

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Algorithm

5 steps

1Step 1: Calculate the total sum of integers from 1 to n using the formula n * (n + 1) / 2.
2Step 2: Calculate the number of multiples of m in the range [1, n] as k = n // m.
3Step 3: Calculate the sum of multiples of m using the formula m * (k * (k + 1)) / 2.
4Step 4: Calculate num1 as total_sum - num2, where total_sum is the sum from Step 1 and num2 is the sum from Step 3.
5Step 5: Return the difference num1 - num2.

solution.py6 lines

1def divisible_and_non_divisible_sum_difference(n, m):
2    total_sum = n * (n + 1) // 2
3    k = n // m
4    num2 = m * (k * (k + 1)) // 2
5    num1 = total_sum - num2
6    return num1 - num2

ℹ

Complexity note: The time complexity is O(1) because we perform a constant number of arithmetic operations regardless of the size of n. The space complexity is O(1) since we only use a fixed amount of extra space.

1Using arithmetic properties can significantly reduce computation time.
2Understanding the relationship between divisibility and sums helps in optimizing solutions.

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