動機

如果沒有大於3的限制就很簡單 同時，又是一題只能用bottom-up的dp

Problem

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: 7Explanation: 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: 16Explanation: Any subsequence of this array is arithmetic.

 

Constraints:

  • 1  <= nums.length <= 1000
  • -231 <= nums[i] <= 231 - 1

Sol

這裡的重點是長度大於3 所以利用從buttom-up時，前一個會被建構好這件事，去取答案

class Solution:
    def numberOfArithmeticSlices(self, nums: List[int]) -> int:
        dp = defaultdict(lambda : defaultdict(int))
        ret = 0
        for i in range(len(nums)):
            for j in range(i):
                diff = nums[i]-nums[j]
                ret += dp[j][diff] # Use all comb whose size is bigger than 1 to construct this case!!
                dp[i][diff] += dp[j][diff]+1
        return ret