How do you solve Arithmetic Slices on LeetCode?

Arithmetic Slices (LeetCode #413) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Arithmetic slices must have at least three elements.

What is the brute force solution for Arithmetic Slices?

The brute force approach for Arithmetic Slices is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subarray of the input array to see if it forms an arithmetic sequence. This is straightforward but inefficient, as it checks all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Arithmetic Slices?

Arithmetic Slices uses the following concepts: Array, Dynamic Programming, Sliding Window. The recommended patterns to study are: Dynamic Programming, Sliding Window.

What are the interview tips for Arithmetic Slices?

Always clarify the problem requirements before diving into coding. Think about edge cases, such as arrays with fewer than three elements. Explain your thought process clearly while coding.

What are common mistakes when solving Arithmetic Slices?

Not checking for the minimum length of the subarray. Confusing arithmetic sequences with other types of sequences.

#413

Arithmetic Slices

Medium

ArrayDynamic ProgrammingSliding WindowDynamic ProgrammingSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Instead of checking every subarray, we can keep track of the number of valid arithmetic slices ending at each index. This allows us to count slices in linear time.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to keep track of the current count of arithmetic slices.
2Step 2: Iterate through the array starting from the third element.
3Step 3: If the current element and the previous two elements form an arithmetic sequence, increment the current count and add it to the total count.

solution.py11 lines

1def count_arithmetic_slices(nums):
2    count = 0
3    current = 0
4    n = len(nums)
5    for i in range(2, n):
6        if nums[i] - nums[i - 1] == nums[i - 1] - nums[i - 2]:
7            current += 1
8            count += current
9        else:
10            current = 0
11    return count

ℹ

Complexity note: This complexity is linear because we only traverse the array once, keeping track of the current count of arithmetic slices.

1Arithmetic slices must have at least three elements.
2The difference between consecutive elements must be constant.

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