# Python 40ms O(n) with Math Intuition - Local Max, Local Min and Saddle points

• This is just a simple math question disguised in cs language. To paraphrase, it is equiv. to:

"Find all local min / max points on xy plane (where x is simply 0 - n-1".

To do that, in discrete math, you just need to find all x[i] where:

1. x[i] > x[i-1] and x[i] < x[i+1]
2. x[i] > x[i-1] and x[i] > x[i+1]

The only caveat is SADDLE points.

The code is written in a rush that can def be simplified, yet it passed all testcases within 40ms, so it is acceptable

``````class Solution(object):
def wiggleMaxLength(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
peaks, troughs, n = [], [], len(nums)
if n == 1:
return 1
for i in range(n):
if i == 0:
if nums[i] > nums[i+1]:
peaks.append(i)
if nums[i] < nums[i+1]:
troughs.append(i)
elif i == n-1:
if nums[i] > nums[i-1]:
peaks.append(i)
if nums[i] < nums[i-1]:
troughs.append(i)
else:
if nums[i] >= nums[i-1] and nums[i] > nums[i+1]:
if nums[i] == nums[i-1]:
k = i
while k > 0 and nums[k] == nums[k-1]:
k -= 1
if k == 0 or nums[k] > nums[k-1]:
peaks.append(i)
else:
peaks.append(i)
if nums[i] <= nums[i-1] and nums[i] < nums[i+1]:
if nums[i] == nums[i-1]:
k = i
while k > 0 and nums[k] == nums[k-1]:
k -= 1
if k == 0 or nums[k] < nums[k-1]:
troughs.append(i)
else:
peaks.append(i)

return len(peaks)+len(troughs)
``````

• it is equiv. to

Can you prove that?

• @StefanPochmann suppose we found all local min and max, first they satisfy the requirement of the problem - i.e. wiggling sequence, and we cannot get another "wiggle", because any wiggle requires a "kink" - i.e. our data cannot be monotonically increasing or decreasing - which means we must have a local min or max - this is however contradictory of our assumption that we have found ALL local min and max.

• @StefanPochmann should change my wording - not "equivalent", but "Finding all local mins and maxs" is "sufficient but not required" to get "wiggle subsequence"...

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