Leetcode #446: Arithmetic Slices II - Subsequence
In this guide, we solve Leetcode #446 Arithmetic Slices II - Subsequence in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given an integer array nums, return the number of all the arithmetic subsequences of nums. A sequence of numbers is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: nums = [2,4,6,8,10]
Output: 7
Explanation: All arithmetic subsequence slices are:
[2,4,6]
[4,6,8]
[6,8,10]
[2,4,6,8]
[4,6,8,10]
[2,4,6,8,10]
[2,6,10]
Python Solution
class Solution:
def numberOfArithmeticSlices(self, nums: List[int]) -> int:
f = [defaultdict(int) for _ in nums]
ans = 0
for i, x in enumerate(nums):
for j, y in enumerate(nums[:i]):
d = x - y
ans += f[j][d]
f[i][d] += f[j][d] + 1
return ans
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.