Suppose we have a list of numbers called nums, we have to find the number of arithmetic subsequences of length ≥ 3. As we know an arithmetic sequence is a list of numbers where the difference between one number and the next is the same.
So, if the input is like nums = [6, 12, 13, 8, 10, 14], then the output will be 3, as we have subsequences like: [6, 8, 10], [6, 10, 14], [12, 13, 14].
To solve this, we will follow these steps:
dp := a new map
n := size of nums
res := 0
for i in range 0 to n, do
for j in range 0 to i, do
diff := nums[i] - nums[j]
prev := dp[(i, diff)] if that is not present, then 0
prevprev := dp[(j, diff)] if that is not present, then 0
dp[i, diff] := prev + prevprev + 1
res := res + prevprev
return res
Example
class Solution:
def solve(self, nums):
dp = {}
n = len(nums)
res = 0
for i in range(n):
for j in range(i):
diff = nums[i] - nums[j]
prev = dp.get((i, diff), 0)
prevprev = dp.get((j, diff), 0)
dp[(i, diff)] = prev + prevprev + 1
res += prevprev
return res
ob = Solution()
nums = [6, 12, 13, 8, 10, 14]
print(ob.solve(nums))Input
[6, 12, 13, 8, 10, 14]
Output
3