#413

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Arithmetic Slices

Count the number of arithmetic subarrays in a given integer array using dynamic programming.

array dynamic programming sliding window

Problem Statement

Given an integer array nums, an arithmetic subarray is a contiguous subsequence where the difference between any two consecutive elements is the same. A valid arithmetic subarray must consist of at least three elements.

Your task is to return the number of arithmetic subarrays in the given array nums. For example, if nums = [1,2,3,4], the answer would be 3, as the subarrays [1, 2, 3], [2, 3, 4], and [1, 2, 3, 4] all form arithmetic sequences.

Examples

Example 1

Input: nums = [1,2,3,4]

Output: 3

We have 3 arithmetic slices in nums: [1, 2, 3], [2, 3, 4] and [1,2,3,4] itself.

Example 2

Input: nums = [1]

Output: 0

Example details omitted.

Constraints

1 <= nums.length <= 5000
-1000 <= nums[i] <= 1000

Solution Approach

Dynamic Programming Approach

The problem can be efficiently solved using dynamic programming. Start by defining a variable to keep track of the number of arithmetic subarrays ending at the current element. For each element, check if it continues the arithmetic sequence formed by the previous elements. If it does, increment the count of subarrays ending at that element. The total number of arithmetic subarrays is the sum of all these counts.

State Transition Optimization

In this approach, we use a state transition technique where we track the length of the longest arithmetic subsequence ending at each index. For each new element, if it extends the arithmetic sequence, we update the current count and add it to the result. This allows us to count subarrays in linear time.

Sliding Window for Efficiency

By employing a sliding window approach, we can keep track of potential arithmetic subarrays efficiently. As we move through the array, we expand the window when the sequence remains arithmetic and shrink it when it breaks. This method minimizes unnecessary recalculations, ensuring that the solution is optimal.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this solution is O(n) where n is the length of the input array. Each element is processed once in constant time, making the approach scalable for larger arrays. The space complexity is O(1) as we are using only a few additional variables to track the counts and subarray lengths, independent of the input size.

What Interviewers Usually Probe

Check if the candidate can optimize the solution using dynamic programming and state transition techniques.
Assess the candidate's ability to optimize space complexity, particularly when handling large input sizes.
Look for a clear understanding of sliding window techniques and how they can apply to sequence-based problems.

Common Pitfalls or Variants

Common pitfalls

Failing to account for subarrays with fewer than three elements, which are not considered arithmetic subarrays.
Incorrectly updating the count of arithmetic subarrays, leading to over-counting or under-counting.
Using brute-force methods to check each possible subarray, resulting in an inefficient solution.

Follow-up variants

What if the array contains only two elements? How would the solution change?
Can we solve this problem with a more memory-efficient approach while maintaining optimal time complexity?
What happens if the array contains large negative or positive numbers? How would the approach handle extreme values?

FAQ

What is the primary pattern for solving the Arithmetic Slices problem?

The primary pattern for solving this problem is state transition dynamic programming, where we keep track of valid arithmetic subarrays and update the count efficiently.

How can I optimize the solution for large input sizes?

By using dynamic programming and a sliding window approach, we can optimize both the time and space complexity to handle larger arrays.

Are there any edge cases to consider in the Arithmetic Slices problem?

Yes, edge cases include arrays with fewer than three elements, as they cannot form arithmetic subarrays.

How do I prevent over-counting in this problem?

By carefully updating the count of subarrays only when the sequence continues the arithmetic pattern, you can avoid over-counting subarrays.

What is the time complexity of the Arithmetic Slices solution?

The time complexity is O(n), where n is the length of the input array, since each element is processed once in constant time.

terminal

Solution

Solution 1: Iteration and Counting

We use $d$ to represent the current difference between two adjacent elements, and $cnt$ to represent the length of the current arithmetic sequence. Initially, $d = 3000$, $cnt = 2$.

1

2

3

4

5

6

7

8

9

10

11

12

class Solution:
    def numberOfArithmeticSlices(self, nums: List[int]) -> int:
        ans = cnt = 0
        d = 3000
        for a, b in pairwise(nums):
            if b - a == d:
                cnt += 1
            else:
                d = b - a
                cnt = 0
            ans += cnt
        return ans

Continue Practicing

#689 maximum sum of 3 non overlapping subarrays #718 maximum length of repeated subarray #410 split array largest sum