leetcode - 446. Arithmetic Slices II - Subsequence

Description

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.

For example, [1, 3, 5, 7, 9], [7, 7, 7, 7], and [3, -1, -5, -9] are arithmetic sequences.
For example, [1, 1, 2, 5, 7] is not an arithmetic sequence.

A subsequence of an array is a sequence that can be formed by removing some elements (possibly none) of the array.

For example, [2,5,10] is a subsequence of [1,2,1,2,4,1,5,10].

The test cases are generated so that the answer fits in 32-bit integer.

Example 1:

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]

Example 2:

Input: nums = [7,7,7,7,7]
Output: 16
Explanation: Any subsequence of this array is arithmetic.

Constraints:

1  <= nums.length <= 1000
-2^31 <= nums[i] <= 2^31 - 1

Solution

Solved after help.

Use dp[i][diff] to denote the number of subsequence that contains 2 numbers, then the number of arithmetic subsequences is: ${\sum }_{j=0}^{j=i?1}dp[j][diff]$, where diff = nums[i] - nums[j]. Because the number of arithmetic subsequences should be the same as the number of subsequence that contains 2 numbers, by just adding 1 number to every subsequence.

For the number of subsequence that contains 2 numbers, it’s 1 from the start.

Time complexity: $o(n^2)$
Space complexity: $o(n^2)$

Code

class Solution:
    def numberOfArithmeticSlices(self, nums: List[int]) -> int:
        dp = {i: {} for i in range(1, len(nums))}
        res = 0
        for i in range(1, len(nums)):
            for j in range(i):
                diff = nums[i] - nums[j]
                dp[i][diff] = dp[i].get(diff, 0) + 1
                if diff in dp.get(j, {}):
                    dp[i][diff] += dp[j][diff]
                    res += dp[j][diff]
        return res