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
Solved after help.
Use dp[i][diff]
to denote the number of subsequence that contains 2 numbers, then the number of arithmetic subsequences is:
∑
j
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0
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\sum_{j=0}^{j=i-1}dp[j][diff]
∑j=0j=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
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n
2
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o(n^2)
o(n2)
Space complexity:
o
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n
2
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o(n^2)
o(n2)
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